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Dual Nature of Radiation and Matter

Questions
509
MHT CET
169
JEE Main
218
NEET
122
Years
1988–2026
0 of 509 attemptedPick an answer to see the solution

Q1NEET 2010 · AIPMT · PRELIMSEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Cathode rays & Thomson's e/m

A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (Where V is the potential difference between cathode and anode)

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Answer: (d)

  1. No deflection: eE=evB⇒v=EBeE = evB \Rightarrow v = \frac{E}{B}.
  2. Accelerated through VV: eV=12mv2⇒em=v22VeV = \frac{1}{2}mv^2 \Rightarrow \frac{e}{m} = \frac{v^2}{2V}.
  3. Substitute vv: em=E22VB2\frac{e}{m} = \frac{E^2}{2VB^2}.
  4. So the answer is (d) E22VB2\frac{E^2}{2VB^2}.
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Q2NEET 2008 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Electric discharge through gases

In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of

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Answer: (a)

  1. At low pressure, cathode-ray electrons are accelerated to high speed by the large field.
  2. They collide with gas atoms, excite them, and the atoms emit light on de-excitation.
  3. So the glow is due to collisions of charged particles from the cathode with gas atoms: answer (a).
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Q3NEET 2007 · AIPMTMedium

Cathode Rays, Thomson's Experiment & Gas Discharge · Cathode rays & Thomson's e/m

A beam of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move

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Answer: (a)

  1. With E⃗\vec{E} off, only the magnetic force qv⃗×B⃗q\vec{v} \times \vec{B} acts.
  2. This force is always perpendicular to v⃗\vec{v} (and v⃗⊥B⃗\vec{v} \perp \vec{B}), so it gives a constant centripetal force.
  3. So the electrons move in a circular orbit: answer (a).
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Q4NEET 2006 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Electric discharge through gases

In a discharge tube ionization of enclosed gas is produced due to collisions between

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Answer: (c)

  1. Fast electrons (from the cathode) gain energy from the field between the electrodes.
  2. When they strike neutral gas atoms/molecules they knock out electrons, ionising the gas.
  3. So the answer is (c) negative electrons and neutral atoms/molecules.
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Q5NEET 2003 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Cathode rays & Thomson's e/m

J.J. Thomson's cathode-ray tube experiment demonstrated that

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Answer: (c)

  1. Thomson measured e/me/m of cathode-ray particles and found it the same for every gas, so (d) is wrong.
  2. Cathode rays are electrons, not ions, so (a) is wrong; the nucleus was found later by Rutherford, so (b) is wrong.
  3. The measured e/me/m was about 1800 times that of the hydrogen ion (proton).
  4. So the answer is (c).
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Q6NEET 2002 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Cathode rays & Thomson's e/m

Which of the following is not the property of cathode rays?

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Answer: (b)

  1. Cathode rays are fast electrons: they carry charge, so they are deflected by an electric field.
  2. They heat the target, travel in straight lines (cast shadows) and cause fluorescence.
  3. So the property that is NOT true is (b).
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Q7NEET 2001 · AIPMTMedium

Cathode Rays, Thomson's Experiment & Gas Discharge · Cathode rays & Thomson's e/m

In Thomson mass spectrograph E⃗⊥B⃗\vec{E} \perp \vec{B}then the velocity of electron beam will be

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Answer: (a)

  1. Undeflected beam: electric and magnetic forces balance, eE=evBeE = evB.
  2. v=∣E⃗∣∣B⃗∣v = \frac{|\vec{E}|}{|\vec{B}|}
  3. So the answer is (a) ∣E⃗∣∣B⃗∣\frac{|\vec{E}|}{|\vec{B}|}.
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Q8NEET 1997 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Electron emission & accelerated electrons

The kinetic energy of an electron, which is accelerated in the potential difference of 100 volts, is

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Answer: (c)

  1. K=eV=(1.602×10−19)(100)K = eV = (1.602 \times 10^{-19})(100)
  2. K=1.602×10−17 JK = 1.602 \times 10^{-17}\text{ J}
  3. So the answer is (c).
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Q9NEET 1996 · AIPMTEasyOld syllabus

Cathode Rays, Thomson's Experiment & Gas Discharge · Electric discharge through gases

In a discharge tube at 0.02 mm, there is formation of

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Answer: (a)

  1. As pressure falls to about 0.010.01 to 0.05 mm0.05\text{ mm} of Hg, the Crookes dark space grows and fills most of the tube.
  2. At 0.02 mm0.02\text{ mm} the Crookes dark space is formed.
  3. So the answer is (a) Crooke's dark space.
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Q10NEET 1996 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Electron emission & accelerated electrons

An electron of mass mm and charge ee is accelerated from rest through a potential difference VV in vacuum. Its final velocity will be

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Answer: (a)

  1. Work done by the field becomes kinetic energy: eV=12mv2eV = \frac{1}{2}mv^2.
  2. v=2eVmv = \sqrt{\frac{2eV}{m}}
  3. So the answer is (a).
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Q11NEET 1990 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Electron emission & accelerated electrons

In which of the following, emission of electrons does not take place

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Answer: (b)

  1. Thermionic, photoelectric and secondary emission all release electrons from a surface.
  2. X-ray emission gives out photons (electromagnetic radiation), not electrons.
  3. So the answer is (b) X-rays emission.
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Q12NEET 1988 · AIPMTEasy

Cathode Rays, Thomson's Experiment & Gas Discharge · Electron emission & accelerated electrons

Thermions are

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Answer: (b)

  1. Thermions are the particles emitted from a heated metal surface (thermionic emission).
  2. These particles are electrons.
  3. So the answer is (b) electrons.
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Q13MHT CET 2026 · 15 MAY · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

The number of photons per second emitted by the source of monochromatic light of wavelength 600600 nm, when it delivers the power of 3.3×10−33.3 \times 10^{-3} W will be

(Planck's constant =6.6×10−34= 6.6 \times 10^{-34} Js, c=3×108c = 3 \times 10^{8} m/s)

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Answer: (b)

  1. Energy per photon E=hcλ=(6.6×10−34)(3×108)600×10−9=3.3×10−19 JE = \frac{hc}{\lambda} = \frac{(6.6 \times 10^{-34})(3 \times 10^{8})}{600 \times 10^{-9}} = 3.3 \times 10^{-19}\text{ J}
  2. n=PE=3.3×10−33.3×10−19=1016 s−1n = \frac{P}{E} = \frac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}\text{ s}^{-1}
  3. So the answer is (b) 101610^{16}.
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Q14JEE Main 2026 · 28 JAN · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is ________.

(Given: h=6.63×10−34h = 6.63 \times 10^{-34} J.s and c=3×108c = 3 \times 10^{8} m/s)

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Answer: (d)

  1. E=hcλ=(6.63×10−34)(3×108)663×10−9=3×10−19 JE = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{663 \times 10^{-9}} = 3 \times 10^{-19}\text{ J}
  2. n=PE=6×10−33×10−19=2×1016 s−1n = \frac{P}{E} = \frac{6 \times 10^{-3}}{3 \times 10^{-19}} = 2 \times 10^{16}\text{ s}^{-1}
  3. So the answer is (d) 2×10162 \times 10^{16}.
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Q15NEET 2026Easy

Photons: Energy, Momentum & Radiation Pressure · Particle vs wave nature of light

Match List I with List II.

List-IList-II
A. E=hνE = h\nuI. de Broglie wavelength
B. Diffraction and InterferenceII. Particle nature of light
C. λ=h/p\lambda = h/pIII. Wave nature of light
D. Compton effectIV. Energy of photon
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Answer: (d)

  1. A. E=hνE = h\nu gives the energy of a photon → IV.
  2. B. Diffraction and interference show wave nature → III.
  3. C. λ=h/p\lambda = h/p is the de Broglie wavelength → I.
  4. D. Compton effect (photon–electron collision) shows particle nature → II.
  5. So the answer is (d) A-IV, B-III, C-I, D-II.
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Q16JEE Main 2026 · 8 APR · SHIFT 2Medium

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A monochromatic source of light operating at 15 kW emits 2.5×10222.5 \times 10^{22} photons /s/\text{s}. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ______. (Take h=6.6×10−34 J.sh = 6.6 \times 10^{-34}\text{ J.s} and c=3×108 m/sc = 3 \times 10^{8}\text{ m/s} ).

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Answer: (d)

  1. Energy per photon E=Pn=15×1032.5×1022=6×10−19 JE = \dfrac{P}{n} = \dfrac{15\times10^{3}}{2.5\times10^{22}} = 6\times10^{-19}\text{ J}
  2. λ=hcE=6.6×10−34×3×1086×10−19=3.3×10−7 m=330 nm\lambda = \dfrac{hc}{E} = \dfrac{6.6\times10^{-34}\times3\times10^{8}}{6\times10^{-19}} = 3.3\times10^{-7}\text{ m} = 330\text{ nm}
  3. 330 nm330\text{ nm} is shorter than violet (≈400 nm\approx 400\text{ nm}), so it is ultraviolet.
  4. So the answer is (d) Ultraviolet.
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Q17MHT CET 2026 · 13 MAY · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

A parallel beam of light is incident normally on a plane surface absorbing 40%40\% of light and reflecting the rest. If the incident beam carries 6060 W of power the force exerted by it on the surface is (C = velocity of light = 3×1083 \times 10^{8} m/s)

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Answer: (b)

  1. Absorbed part gives P/cP/c per unit fraction, reflected part gives 2P/c2P/c: F=Pc(a+2r)F = \dfrac{P}{c}(a + 2r)
  2. a=0.4, r=0.6a = 0.4,\ r = 0.6: F=603×108(0.4+1.2)=2×10−7×1.6F = \dfrac{60}{3\times10^{8}}(0.4 + 1.2) = 2\times10^{-7}\times1.6
  3. F=3.2×10−7 NF = 3.2\times10^{-7}\text{ N}
  4. So the answer is (b) 3.2×10−7 N3.2\times10^{-7}\text{ N}.
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Q18JEE Main 2025 · 28 JAN · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Particle vs wave nature of light

Which of the following phenomena cannot be explained by wave theory of light?

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Answer: (d)

  1. Reflection, refraction and diffraction are all explained by wave theory.
  2. Compton effect needs photons carrying momentum h/λh/\lambda colliding with electrons.
  3. So the answer is (d) Compton effect.
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Q19JEE Main 2025 · 24 JAN · SHIFT 2Medium

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

The ratio of the power of a light source S1S_1 to that of the light source S2S_2 is 2. S12.\ S_1 is emitting 2×10152 \times 10^{15} photons per second at 600 nm. If the wavelength of the source S2S_2 is 300 nm, then the number of photons per second emitted by S2S_2 is __________ ×1014\times 10^{14}.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. P=nhcλ⇒n=PλhcP = n\dfrac{hc}{\lambda} \Rightarrow n = \dfrac{P\lambda}{hc}, so n1n2=P1P2⋅λ1λ2\dfrac{n_1}{n_2} = \dfrac{P_1}{P_2}\cdot\dfrac{\lambda_1}{\lambda_2}
  2. n1n2=2×600300=4\dfrac{n_1}{n_2} = 2\times\dfrac{600}{300} = 4
  3. n2=2×10154=5×1014n_2 = \dfrac{2\times10^{15}}{4} = 5\times10^{14} per second
  4. So the answer is 55.
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Q20MHT CET 2025 · 20 APR · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

A parallel beam of light is incident normally on a plane surface absorbing 50%50\% of the light and reflecting the rest. If the incident beam carries 90 W of power, the force exerted by it on the surface is ( C=C = velocity of light in air =3×108 m/s= 3 \times 10^{8}\text{ m/s})

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Answer: (a)

  1. F=Pc(a+2r)F = \dfrac{P}{c}(a + 2r) with a=0.5, r=0.5a = 0.5,\ r = 0.5
  2. F=903×108(0.5+1.0)=3×10−7×1.5F = \dfrac{90}{3\times10^{8}}(0.5 + 1.0) = 3\times10^{-7}\times1.5
  3. F=4.5×10−7 NF = 4.5\times10^{-7}\text{ N}
  4. So the answer is (a) 4.5×10−7 N4.5\times10^{-7}\text{ N}.
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Q21JEE Main 2025 · 3 APR · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is

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Answer: (d)

  1. Intensity at r=2 mr = 2\text{ m}: I=P4πr2=4504π×4≈8.95 W/m2I = \dfrac{P}{4\pi r^2} = \dfrac{450}{4\pi\times4} \approx 8.95\text{ W/m}^2
  2. Perfect reflector: prad=2Ic=2×8.953×108p_{rad} = \dfrac{2I}{c} = \dfrac{2\times8.95}{3\times10^{8}}
  3. prad≈6×10−8 Pap_{rad} \approx 6\times10^{-8}\text{ Pa}
  4. So the answer is (d) 6×10−86\times10^{-8} Pascals.
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Q22JEE Main 2025 · 4 APR · SHIFT 1Difficult

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

A small mirror of mass mm is suspended by a massless thread of length ll. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror

(c=c = speed of light in vacuum and g=g = acceleration due to gravity)

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Answer: (d)

  1. Mirror reflects the pulse, so momentum given =2Ec= \dfrac{2E}{c} and v=2Emcv = \dfrac{2E}{mc}
  2. Energy conservation: 12mv2=mgl(1−cos⁡θ)≈mglθ22\tfrac12 mv^2 = mgl(1-\cos\theta) \approx mgl\dfrac{\theta^2}{2}, so v=θglv = \theta\sqrt{gl}
  3. θ=vgl=2Emcgl\theta = \dfrac{v}{\sqrt{gl}} = \dfrac{2E}{mc\sqrt{gl}}
  4. So the answer is (d).
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Q23NEET 2024Easy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

If cc is the velocity of light in free space, the correct statements about photon among the following are:

A. The energy of a photon is E=hνE = h\nu.

B. The velocity of a photon is cc.

C. The momentum of a photon, p=hνcp = \frac{h\nu}{c}.

D. In a photon-electron collision, both total energy and total momentum are conserved. E. Photon possesses positive charge.

Choose the correct answer from the options given below:

Show answer and solution

Answer: (b)

  1. A: E=hνE = h\nu, true. B: a photon travels at cc in free space, true.
  2. C: p=Ec=hνcp = \frac{E}{c} = \frac{h\nu}{c}, true. D: energy and momentum are conserved in photon-electron collisions (Compton effect), true.
  3. E: a photon is uncharged, false.
  4. So the answer is (b) A, B, C and D only.
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Q24JEE Main 2024 · 4 APR · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Given below are two statements: one is labelled as Assertion A\mathbf{A} and the other is labelled as Reason R.

Assertion A: Number of photons increases with increase in frequency of light.

Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.

In the light of the above statements, choose the most appropriate answer from the options given below:

Show answer and solution

Answer: (a)

  1. Number of photons per second n=Phνn = \frac{P}{h\nu}: it depends on intensity (power), and for fixed power it actually decreases as ν\nu rises. So A is false.
  2. Kmax=hν−ϕK_{max} = h\nu - \phi increases with ν\nu, so R is true.
  3. So the answer is (a) A is not correct but R is correct.
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Q25JEE Main 2024 · 1 FEB · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Monochromatic light of frequency 6×1014 Hz6 \times 10^{14}\text{ Hz} is produced by a laser. The power emitted is 2×10−3 W2 \times 10^{-3}\text{ W}.

How many photons per second on an average, are emitted by the source?

(Given h=6.63×10−34 Jsh = 6.63 \times 10^{-34}\text{ Js})

Show answer and solution

Answer: (a)

  1. E=hν=(6.63×10−34)(6×1014)=3.98×10−19 JE = h\nu = (6.63 \times 10^{-34})(6 \times 10^{14}) = 3.98 \times 10^{-19}\text{ J}
  2. n=PE=2×10−33.98×10−19≈5×1015 s−1n = \frac{P}{E} = \frac{2 \times 10^{-3}}{3.98 \times 10^{-19}} \approx 5 \times 10^{15}\text{ s}^{-1}
  3. So the answer is (a) 5×10155 \times 10^{15}.
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Q26JEE Main 2024 · 29 JAN · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Two sources of light emit with a power of 200 W200\text{ W}. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm300\text{ nm} and 500 nm500\text{ nm} respectively, will be:

Show answer and solution

Answer: (b)

  1. n=Pλhcn = \frac{P\lambda}{hc}, so for equal power n∝λn \propto \lambda.
  2. n1n2=300500=35\frac{n_1}{n_2} = \frac{300}{500} = \frac{3}{5}
  3. So the answer is (b) 3:53 : 5.
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Q27JEE Main 2024 · 30 JAN · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

If the total energy transferred to a surface in time tt is 6.48×105 J6.48 \times 10^{5}\text{ J}, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

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Answer: (a)

  1. For complete absorption, momentum delivered p=Ecp = \dfrac{E}{c}
  2. p=6.48×1053×108=2.16×10−3 kg m/sp = \dfrac{6.48\times10^{5}}{3\times10^{8}} = 2.16\times10^{-3}\text{ kg m/s}
  3. So the answer is (a) 2.16×10−3 kg m/s2.16\times10^{-3}\text{ kg m/s}.
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Q28JEE Main 2024 · 6 APR · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Particle vs wave nature of light

Which of the following phenomena is not explained by wave nature of light?

A. reflection

B. diffraction

C. photoelectric effect

D. interference

E. polarization

Choose the most appropriate answer from the options given below:

Show answer and solution

Answer: (a)

  1. Reflection, diffraction, interference and polarization are all explained by wave theory.
  2. Photoelectric effect (instant emission, threshold frequency) needs the photon picture.
  3. So the answer is (a) C only.
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Q29JEE Main 2023 · 30 JAN · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

A small object at rest, absorbs a light pulse of power 20 mW20\text{ mW} and duration 300 ns300\text{ ns}. Assuming speed of light as 3×108 m/s3 \times 10^{8}\text{ m/s}, the momentum of the object becomes equal to:

Show answer and solution

Answer: (d)

  1. Energy absorbed E=Pt=20×10−3×300×10−9=6×10−9 JE = Pt = 20\times10^{-3}\times 300\times10^{-9} = 6\times10^{-9}\text{ J}
  2. On absorption p=Ec=6×10−93×108=2×10−17 kg m/sp = \dfrac{E}{c} = \dfrac{6\times10^{-9}}{3\times10^{8}} = 2\times10^{-17}\text{ kg m/s}
  3. So the answer is (d) 2×10−17 kg m/s2\times10^{-17}\text{ kg m/s}.
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Q30MHT CET 2023 · 13 MAY · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Particle vs wave nature of light

Dual nature of light is exhibited by

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Answer: (a)

  1. Diffraction is explained only by the wave nature of light.
  2. Photoelectric effect is explained only by the particle (photon) nature.
  3. Together they show the dual nature; reflection, refraction and interference are all wave-explainable.
  4. So the answer is (a).
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Q31JEE Main 2023 · 31 JAN · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

If a source of electromagnetic radiation having power 15 kW15\text{ kW} produces 101610^{16} photons per second, the radiation belongs to a part of spectrum is.

(Take Planck constant h=6×10−34 Jsh = 6 \times 10^{-34}\text{ Js})

Show answer and solution

Answer: (a)

  1. Energy per photon E=Pn=15×1031016=1.5×10−12 JE = \dfrac{P}{n} = \dfrac{15\times10^{3}}{10^{16}} = 1.5\times10^{-12}\text{ J}
  2. λ=hcE=6×10−34×3×1081.5×10−12=1.2×10−13 m\lambda = \dfrac{hc}{E} = \dfrac{6\times10^{-34}\times3\times10^{8}}{1.5\times10^{-12}} = 1.2\times10^{-13}\text{ m}
  3. E≈9 MeVE \approx 9\text{ MeV} and λ∼10−13 m\lambda \sim 10^{-13}\text{ m}: gamma-ray region.
  4. So the answer is (a) Gamma rays.
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Q32JEE Main 2022 · 30 JUN · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A source of monochromatic light liberates 9 ×\times 1020 photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be:

Show answer and solution

Answer: (a)

  1. At the same power, n=Pλhc⇒n∝λn = \frac{P\lambda}{hc} \Rightarrow n \propto \lambda.
  2. n2=n1λ2λ1=9×1020×800600n_2 = n_1 \frac{\lambda_2}{\lambda_1} = 9 \times 10^{20} \times \frac{800}{600}
  3. n2=12×1020n_2 = 12 \times 10^{20} photons per second
  4. So the answer is (a).
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Q33JEE Main 2022 · 26 JUL · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A parallel beam of light of wavelength 900 nm900\text{ nm} and intensity 100 Wm−2100\text{ Wm}^{-2} is incident on a surface perpendicular to the beam. The number of photons crossing 1 cm21\text{ cm}^{2} area perpendicular to the beam in one second is:

Show answer and solution

Answer: (b)

  1. Power through 1 cm21\text{ cm}^2: P=IA=100×10−4=10−2 WP = IA = 100\times10^{-4} = 10^{-2}\text{ W}
  2. Photon energy hcλ=6.63×10−34×3×108900×10−9=2.21×10−19 J\dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^{8}}{900\times10^{-9}} = 2.21\times10^{-19}\text{ J}
  3. n=10−22.21×10−19≈4.5×1016n = \dfrac{10^{-2}}{2.21\times10^{-19}} \approx 4.5\times10^{16} per second
  4. So the answer is (b) 4.5×10164.5\times10^{16}.
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Q34MHT CET 2021 · 24 SEP · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

A photon has wavelength 3 nm3\text{ nm}, then its momentum and energy respectively will be [h=6.63×10−34 Js,c=[h = 6.63 \times 10^{-34}\text{ Js}, c = velocity of light =3×108 m/s]= 3 \times 10^{8}\text{ m/s}]

Show answer and solution

Answer: (c)

  1. p=hλ=6.63×10−343×10−9=2.21×10−25 kg m s−1p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{3 \times 10^{-9}} = 2.21 \times 10^{-25}\text{ kg m s}^{-1}
  2. E=pc=(2.21×10−25)(3×108)=6.63×10−17 JE = pc = (2.21 \times 10^{-25})(3 \times 10^{8}) = 6.63 \times 10^{-17}\text{ J}
  3. So the answer is (c).
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Q35MHT CET 2021 · 22 SEP · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

When a photon enters glass from air, which one of the following quantity does not change?

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Answer: (a)

  1. Photon energy E=hνE = h\nu depends only on frequency.
  2. Frequency is set by the source and does not change on refraction; speed, wavelength and momentum h/λh/\lambda change.
  3. So the answer is (a) Energy.
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Q36JEE Main 2021 · 24 FEB · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

Given below are two statements:

Statement I: Two photons having equal linear momenta have equal wavelengths. Statement II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.

In the light of the above statements, choose the correct answer from the options given below.

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Answer: (d)

  1. p=hλp = \frac{h}{\lambda}: equal momenta mean equal wavelengths, so Statement I is true.
  2. p=hλp = \frac{h}{\lambda} and E=hcλE = \frac{hc}{\lambda}: decreasing λ\lambda increases both, so Statement II is false.
  3. So the answer is (d).
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Q37MHT CET 2021 · 20 SEP · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

When light of wavelength 'λ\lambda' is incident on a photosensitive surface, photons of power 'P' are emitted. The number of photon 'n' emitted in time 't' is [h = Planck's constant, c = velocity of light in vacuum]

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Answer: (c)

  1. Energy in time tt: PtPt. Energy per photon: hcλ\frac{hc}{\lambda}.
  2. n=Pthc/λ=Pλthcn = \frac{Pt}{hc/\lambda} = \frac{P\lambda t}{hc}
  3. So the answer is (c).
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Q38NEET 2021Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600nm, when it delivers the power of 3.3 ×\times 10−-3 watt will be: (h = 6.6 ×\times 10−-34 Js)

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Answer: (d)

  1. E=hcλ=(6.6×10−34)(3×108)6×10−7=3.3×10−19 JE = \frac{hc}{\lambda} = \frac{(6.6 \times 10^{-34})(3 \times 10^{8})}{6 \times 10^{-7}} = 3.3 \times 10^{-19}\text{ J}
  2. n=PE=3.3×10−33.3×10−19=1016 s−1n = \frac{P}{E} = \frac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}\text{ s}^{-1}
  3. So the answer is (d) 101610^{16}.
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Q39JEE Main 2021 · 25 FEB · SHIFT 2Medium

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The wavelength of an X-ray beam is 10A˚\mathring{A}. The mass of a fictitious particle having the same energy as that of the X-ray photons is x3h\frac{x}{3}h kg. The value of x is __________. (h = Planck's constant)

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. Equate energies: mc2=hcλ⇒m=hλcmc^2 = \dfrac{hc}{\lambda} \Rightarrow m = \dfrac{h}{\lambda c}
  2. m=h10−9×3×108=h0.3=103hm = \dfrac{h}{10^{-9}\times 3\times10^{8}} = \dfrac{h}{0.3} = \dfrac{10}{3}h
  3. So x=10x = 10.
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Q40JEE Main 2020 · 3 SEP · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is:

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Answer: (a)

  1. For equal power, photons per second n∝λn \propto \lambda; both travel at cc, so number density is also ∝λ\propto \lambda.
  2. nXnV=1 nm500 nm=1500\frac{n_X}{n_V} = \frac{1\text{ nm}}{500\text{ nm}} = \frac{1}{500}
  3. So the answer is (a) 1500\frac{1}{500}.
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Q41NEET 2020 · PHASE 1Medium

Photons: Energy, Momentum & Radiation Pressure · Radiation pressure & force of light

Light with an average flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence having surface area 20 cm2. The energy received by the surface during time span of 1 minute is:

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Answer: (b)

  1. Energy E=(flux)×A×tE = (\text{flux})\times A\times t
  2. E=20 W/cm2×20 cm2×60 s=24000 JE = 20\text{ W/cm}^2\times 20\text{ cm}^2\times 60\text{ s} = 24000\text{ J}
  3. So the answer is (b) 24×103 J24\times10^{3}\text{ J}.
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Q42JEE Main 2019 · 10 APR · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is: [Given Planck's constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108 m/s]

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Answer: (a)

  1. Photons per second n=Phc/λ=Pλhcn = \dfrac{P}{hc/\lambda} = \dfrac{P\lambda}{hc}
  2. n=2×10−3×500×10−96.6×10−34×3×108=10−91.98×10−25n = \dfrac{2\times10^{-3}\times 500\times10^{-9}}{6.6\times10^{-34}\times 3\times10^{8}} = \dfrac{10^{-9}}{1.98\times10^{-25}}
  3. n≈5×1015n \approx 5\times10^{15} per second
  4. So the answer is (a) 5×10155\times10^{15}.
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Q43JEE Main 2017 · 9 APR · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are: [Take Planck's constant h == 6.62 ×\times 10−-34 Js]

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Answer: (a)

  1. Pulse energy E=Pt=500×60×10−3=30 JE = Pt = 500\times 60\times10^{-3} = 30\text{ J}
  2. Photon energy hcλ=6.62×10−34×3×108660×10−9≈3×10−19 J\dfrac{hc}{\lambda} = \dfrac{6.62\times10^{-34}\times3\times10^{8}}{660\times10^{-9}} \approx 3\times10^{-19}\text{ J}
  3. N=303×10−19=1020N = \dfrac{30}{3\times10^{-19}} = 10^{20}
  4. So the answer is (a) 102010^{20}.
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Q44NEET 2012 · AIPMT · PRELIMSMedium

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A 200 W sodium street lamp emits yellow light of wavelength 0.6 μ\mum. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

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Answer: (a)

  1. Light output =25%= 25\% of 200 W=50 W200\text{ W} = 50\text{ W}
  2. Photon energy hcλ=6.6×10−34×3×1080.6×10−6=3.3×10−19 J\dfrac{hc}{\lambda} = \dfrac{6.6\times10^{-34}\times3\times10^{8}}{0.6\times10^{-6}} = 3.3\times10^{-19}\text{ J}
  3. n=503.3×10−19≈1.5×1020n = \dfrac{50}{3.3\times10^{-19}} \approx 1.5\times10^{20} per second
  4. So the answer is (a) 1.5×10201.5\times10^{20}.
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Q45NEET 2010 · AIPMT · PRELIMSEasy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A source S1 is producing, 1015 photons per second of wavelength 5000 A˚\mathring{A}. Another source S2 is producing 1.02 ×\times 1015 photons per second of wavelength 5100 A˚\mathring{A}. Then, (power of S2)/(power of S1) is equal to

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Answer: (a)

  1. Power P=nhcλ⇒P∝nλP = n\frac{hc}{\lambda} \Rightarrow P \propto \frac{n}{\lambda}.
  2. P2P1=n2n1⋅λ1λ2=1.02×50005100\frac{P_2}{P_1} = \frac{n_2}{n_1} \cdot \frac{\lambda_1}{\lambda_2} = 1.02 \times \frac{5000}{5100}
  3. P2P1=1.021.02=1.00\frac{P_2}{P_1} = \frac{1.02}{1.02} = 1.00
  4. So the answer is (a) 1.00.
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Q46JEE Main 2010 · AIEEEMedium

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

If a source of power 4 kW4\text{ kW} produces 102010^{20} photons/second, the radiation belongs to a part of the spectrum called

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Answer: (a)

  1. Energy per photon E=4×1031020=4×10−17 J≈250 eVE = \dfrac{4\times10^{3}}{10^{20}} = 4\times10^{-17}\text{ J} \approx 250\text{ eV}
  2. λ=hcE=6.6×10−34×3×1084×10−17≈5×10−9 m=5 nm\lambda = \dfrac{hc}{E} = \dfrac{6.6\times10^{-34}\times3\times10^{8}}{4\times10^{-17}} \approx 5\times10^{-9}\text{ m} = 5\text{ nm}
  3. λ≈50 A˚\lambda \approx 50\text{ Å} lies in the X-ray region.
  4. So the answer is (a) X-rays.
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Q47NEET 2009 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per sec. on the average at a target irradiated by this beam is

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Answer: (a)

  1. E=hcλ=(6.63×10−34)(3×108)667×10−9≈2.98×10−19 JE = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{667 \times 10^{-9}} \approx 2.98 \times 10^{-19}\text{ J}
  2. n=PE=9×10−32.98×10−19≈3×1016 s−1n = \frac{P}{E} = \frac{9 \times 10^{-3}}{2.98 \times 10^{-19}} \approx 3 \times 10^{16}\text{ s}^{-1}
  3. So the answer is (a) 3×10163 \times 10^{16}.
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Q48JEE Main 2007 · AIEEEEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

Photon of frequency ν\nu has a momentum associated with it. If cc is the velocity of light, the momentum is

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Answer: (a)

  1. E=hνE = h\nu and for a photon p=Ecp = \frac{E}{c}.
  2. p=hνcp = \frac{h\nu}{c}
  3. So the answer is (a) hν/ch\nu/c.
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Q49NEET 2007 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

Monochromatic light of frequency 6.0 ×\times 1014 Hz is produced by a laser. The power emitted is 2 ×\times 10−-3 W. The number of photons emitted, on the average, by the source per second is

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Answer: (d)

  1. E=hν=(6.63×10−34)(6×1014)=3.98×10−19 JE = h\nu = (6.63 \times 10^{-34})(6 \times 10^{14}) = 3.98 \times 10^{-19}\text{ J}
  2. n=PE=2×10−33.98×10−19≈5×1015 s−1n = \frac{P}{E} = \frac{2 \times 10^{-3}}{3.98 \times 10^{-19}} \approx 5 \times 10^{15}\text{ s}^{-1}
  3. So the answer is (d) 5×10155 \times 10^{15}.
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Q50NEET 2006 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The momentum of a photon of energy 1 MeV in kg m/s will be

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Answer: (a)

  1. E=1 MeV=1.6×10−13 JE = 1\text{ MeV} = 1.6 \times 10^{-13}\text{ J}
  2. p=Ec=1.6×10−133×108=5.3×10−22 kg m/sp = \frac{E}{c} = \frac{1.6 \times 10^{-13}}{3 \times 10^{8}} = 5.3 \times 10^{-22}\text{ kg m/s}
  3. So the answer is (a) 5×10−225 \times 10^{-22}.
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Q51NEET 2002 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The value of Planck's constant is

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Answer: (d)

  1. From E=hνE = h\nu: h=Eνh = \frac{E}{\nu}, unit Js−1=J s\frac{\text{J}}{\text{s}^{-1}} = \text{J s}.
  2. h=6.63×10−34 J sh = 6.63 \times 10^{-34}\text{ J s}; J/s (watt) and kg m2^2 are wrong units.
  3. So the answer is (d) 6.63×10−346.63 \times 10^{-34} J-sec.
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Q52NEET 2001 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Particle vs wave nature of light

Which one among the following shows particle nature of light?

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Answer: (a)

  1. Interference, refraction and polarization are explained by wave theory.
  2. Photoelectric effect needs light as photons of energy hνh\nu.
  3. So the answer is (a) photoelectric effect.
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Q53NEET 2000 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

By photoelectric effect, Einstein proved

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Answer: (a)

  1. Einstein explained the photoelectric effect by treating light as photons of energy hνh\nu.
  2. This confirmed the quantum relation E=hνE = h\nu.
  3. So the answer is (a).
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Q54NEET 2000 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The energy of a photon of light is 3 eV3\text{ eV}. Then the wavelength of photon must be

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Answer: (b)

  1. λ=hcE\lambda = \frac{hc}{E} with hc≈12375 eV A˚hc \approx 12375\text{ eV Å}.
  2. λ=123753 A˚=4125 A˚=412.5 nm\lambda = \frac{12375}{3}\text{ Å} = 4125\text{ Å} = 412.5\text{ nm}
  3. So the answer is (b) 412.5 nm412.5\text{ nm}.
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Q55NEET 1996 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

If a photon has velocity cc and frequency ν\nu, then which of the following represents its wavelength?

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Answer: (c)

  1. λ=cν\lambda = \frac{c}{\nu} and E=hν⇒ν=EhE = h\nu \Rightarrow \nu = \frac{E}{h}.
  2. λ=hcE\lambda = \frac{hc}{E}
  3. So the answer is (c).
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Q56NEET 1996 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The velocity of photons is proportional to (where ν\nu = frequency)

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Answer: (b)

  1. All photons travel at the same speed c=3×108 m/sc = 3 \times 10^{8}\text{ m/s} in vacuum, whatever their frequency.
  2. So the speed is independent of ν\nu (proportional to ν0\nu^0); none of the given options is physically correct.
  3. The source answer is (b) ν2\nu^2.
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Q57NEET 1993 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

Momentum of photon wavelength λ\lambda is

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Answer: (a)

  1. Photon momentum p=hλp = \frac{h}{\lambda}.
  2. With λ=cν\lambda = \frac{c}{\nu}: p=hνcp = \frac{h\nu}{c}.
  3. So the answer is (a) hνc\frac{h\nu}{c}.
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Q58NEET 1991 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The wavelength of a 1 keV photon is 1.24×10−91.24 \times 10^{-9} m. What is the frequency of 1 MeV photon?

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Answer: (b)

  1. For the 1 keV photon: ν1=cλ=3×1081.24×10−9=2.42×1017 Hz\nu_1 = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{1.24 \times 10^{-9}} = 2.42 \times 10^{17}\text{ Hz}.
  2. E=hν⇒ν∝EE = h\nu \Rightarrow \nu \propto E; 1 MeV is 1000 times 1 keV.
  3. ν2=1000×2.42×1017≈2.4×1020 Hz\nu_2 = 1000 \times 2.42 \times 10^{17} \approx 2.4 \times 10^{20}\text{ Hz}
  4. So the answer is (b).
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Q59NEET 1990 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The momentum of a photon of an electromagnetic radiation is 3.3×10−29 kg ms−13.3 \times 10^{-29}\text{ kg ms}^{-1}. What is the frequency of the associated waves?

[h=6.6×10−34 Jsh = 6.6 \times 10^{-34}\text{ Js}; c=3×108 ms−1c = 3 \times 10^{8}\text{ ms}^{-1}]

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Answer: (a)

  1. p=hνc⇒ν=pchp = \frac{h\nu}{c} \Rightarrow \nu = \frac{pc}{h}
  2. ν=(3.3×10−29)(3×108)6.6×10−34=1.5×1013 Hz\nu = \frac{(3.3 \times 10^{-29})(3 \times 10^{8})}{6.6 \times 10^{-34}} = 1.5 \times 10^{13}\text{ Hz}
  3. So the answer is (a).
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Q60NEET 1990 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Number of photons from a source

A radio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is

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Answer: (a)

  1. E=hν=(6.63×10−34)(880×103)=5.83×10−28 JE = h\nu = (6.63 \times 10^{-34})(880 \times 10^{3}) = 5.83 \times 10^{-28}\text{ J}
  2. n=PE=1045.83×10−28≈1.72×1031 s−1n = \frac{P}{E} = \frac{10^{4}}{5.83 \times 10^{-28}} \approx 1.72 \times 10^{31}\text{ s}^{-1}
  3. So the answer is (a) 1.72×10311.72 \times 10^{31}.
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Q61NEET 1988 · AIPMTEasy

Photons: Energy, Momentum & Radiation Pressure · Photon energy, wavelength & momentum

The energy of a photon of wavelength λ\lambda is

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Answer: (b)

  1. E=hνE = h\nu and ν=cλ\nu = \frac{c}{\lambda}.
  2. E=hcλE = \frac{hc}{\lambda}
  3. So the answer is (b).
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Q62NEET 2026 · RE-NEETEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

A ray of light with wavelength λ\lambda is incident on three different photoelectric cells namely 1,2 and 3. The threshold wavelength of these photoelectric cells are λ1,λ2\lambda_1, \lambda_2, and λ3\lambda_3, respectively and the magnitude of stopping potentials of these cells are V1,V2V_1, V_2 and V3V_3, respectively. The relation between λ\lambda and threshold wavelengths are λ1<λ,λ2>λ\lambda_1 < \lambda, \lambda_2 > \lambda and λ3≫λ\lambda_3 \gg \lambda. The correct option is:

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Answer: (b)

  1. Cell 1: λ>λ1\lambda > \lambda_1, photon energy below work function, no emission, so V1=0V_1 = 0.
  2. eV=hcλ−hcλ0eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}: larger threshold λ0\lambda_0 means smaller work function and larger VV.
  3. λ3≫λ2\lambda_3 \gg \lambda_2, so V3>V2V_3 > V_2.
  4. So the answer is (b) V1=0, V2<V3V_1 = 0,\ V_2 < V_3.
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Q63NEET 2026Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

For a metal of work function 6.6 eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect?

(Take Planck's constant as 6.6×10−34 J s6.6 \times 10^{-34}\text{ J s})

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Answer: (c)

  1. Threshold wavelength λ0=hcϕ=1240 eV nm6.6 eV≈188 nm\lambda_0 = \dfrac{hc}{\phi} = \dfrac{1240\text{ eV nm}}{6.6\text{ eV}} \approx 188\text{ nm}
  2. Only λ>188 nm\lambda > 188\text{ nm} fails to cause emission.
  3. Of the options only 200 nm200\text{ nm} exceeds 188 nm188\text{ nm}.
  4. So the answer is (c) 200 nm.
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Q64JEE Main 2026 · 22 JAN · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

Light is incident on a metallic plate having work function 110×10−20 J110 \times 10^{-20}\text{ J}. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ______ rad/s. (h=6.63×10−34 J.s)(h = 6.63 \times 10^{-34}\text{ J.s}).

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Answer: (d)

  1. Zero KE means hν=ϕh\nu = \phi, so ν=ϕh=110×10−206.63×10−34≈1.66×1015 Hz\nu = \dfrac{\phi}{h} = \dfrac{110\times10^{-20}}{6.63\times10^{-34}} \approx 1.66\times10^{15}\text{ Hz}
  2. Angular frequency ω=2πν=2π×1.66×1015\omega = 2\pi\nu = 2\pi\times1.66\times10^{15}
  3. ω≈1.04×1016 rad/s\omega \approx 1.04\times10^{16}\text{ rad/s}
  4. So the answer is (d) 1.04×10161.04\times10^{16}.
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Q65MHT CET 2026 · 16 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

For a photosensitive material, work function is 'W0W_0' and stopping potential is 'V'. The wavelength of the incident radiation is (h=Planck's constant, c = velocity of light, e = electronic charge)

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Answer: (b)

  1. Einstein's equation: hcλ=W0+Kmax=W0+eV\frac{hc}{\lambda} = W_0 + K_{max} = W_0 + eV
  2. λ=hcW0+eV\lambda = \frac{hc}{W_0 + eV}
  3. So the answer is (b) hc(W0+eV)\frac{hc}{(W_0 + eV)}.
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Q66JEE Main 2026 · 5 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19 J\alpha \times 10^{-19}\text{ J}. The value of α\alpha is ______.

(h=6.62×10−34 J s, e=1.6×10−19 C\left(h = 6.62 \times 10^{-34}\text{ J s},\ e = 1.6 \times 10^{-19}\text{ C}\right. and c=3×108 m/s)\left.c = 3 \times 10^{8}\text{ m/s}\right)

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Answer: (c)

  1. ϕ=hcλ−eV0\phi = \frac{hc}{\lambda} - eV_0
  2. hcλ=6.62×10−34×3×108331×10−9=6.0×10−19 J\frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{331 \times 10^{-9}} = 6.0 \times 10^{-19}\text{ J}
  3. eV0=1.6×10−19×0.2=0.32×10−19 JeV_0 = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19}\text{ J}
  4. ϕ=(6.0−0.32)×10−19=5.68×10−19 J\phi = (6.0 - 0.32) \times 10^{-19} = 5.68 \times 10^{-19}\text{ J}, so the answer is (c) 5.68.
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Q67MHT CET 2026 · 18 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

In case of photoelectric emission from certain metal, the cutoff frequency is ν\nu. If the radiation of frequency 3ν3\nu is incident on the metal plate, the maximum possible velocity of the emitted electrons will be (mm = mass of electron, hh = Planck's constant)

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Answer: (c)

  1. 12mvmax2=h(3ν)−hν=2hν\frac{1}{2}mv_{max}^2 = h(3\nu) - h\nu = 2h\nu
  2. vmax=4hνm=2hνmv_{max} = \sqrt{\frac{4h\nu}{m}} = 2\sqrt{\frac{h\nu}{m}}
  3. So the answer is (c) 2(hνm)2\left(\sqrt{\frac{h\nu}{m}}\right).
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Q68MHT CET 2026 · 13 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Energy of the incident photons on the photosensitive metal surface is 3W and then 5W where 'W' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

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Answer: (b)

  1. 12mv2=E−W\frac{1}{2}mv^2 = E - W, so v∝E−Wv \propto \sqrt{E - W}.
  2. K1=3W−W=2WK_1 = 3W - W = 2W, K2=5W−W=4WK_2 = 5W - W = 4W
  3. v1v2=2W4W=12\frac{v_1}{v_2} = \sqrt{\frac{2W}{4W}} = \frac{1}{\sqrt{2}}
  4. So the answer is (b) 1:21 : \sqrt{2}.
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Q69MHT CET 2026 · 14 MAY · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When photons of energy hνh\nu fall on a metal plate of work function W0W_0, photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of ejected photoelectrons will be

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Answer: (b)

  1. K=hν−W0K = h\nu - W_0, so hν=K+W0h\nu = K + W_0.
  2. New: K′=2hν−W0=hν+(hν−W0)=K+hνK' = 2h\nu - W_0 = h\nu + (h\nu - W_0) = K + h\nu
  3. So the answer is (b) K+hνK + h\nu.
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Q70MHT CET 2026 · 13 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

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Answer: (c)

  1. K=hν−ϕK = h\nu - \phi.
  2. New: K′=2hν−ϕ=2(hν−ϕ)+ϕ=2K+ϕK' = 2h\nu - \phi = 2(h\nu - \phi) + \phi = 2K + \phi
  3. Since ϕ>0\phi > 0, K′>2KK' > 2K.
  4. So the answer is (c) more than double.
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Q71JEE Main 2026 · 21 JAN · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

A light wave described by E=60[sin⁡(3×1015)t+sin⁡(12×1015)t]E = 60\left[\sin\left(3 \times 10^{15}\right)t + \sin\left(12 \times 10^{15}\right)t\right] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately)

______ eV. (h=6.6×10−34\left(h = 6.6 \times 10^{-34}\right. J.s. and e=1.6×10−19 C)\left.e = 1.6 \times 10^{-19}\text{ C}\right)

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Answer: (d)

  1. The wave has two angular frequencies; the larger one, ω=12×1015\omega = 12 \times 10^{15} rad/s, gives the maximum KE.
  2. E=hω2π=6.6×10−34×12×10152π=1.26×10−18E = \frac{h\omega}{2\pi} = \frac{6.6 \times 10^{-34} \times 12 \times 10^{15}}{2\pi} = 1.26 \times 10^{-18} J =7.88= 7.88 eV
  3. Kmax=7.88−2.8≈5.1K_{max} = 7.88 - 2.8 \approx 5.1 eV
  4. So the answer is (d) 5.1 eV.
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Q72MHT CET 2026 · 19 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metallic surface is illuminated with radiation of wavelength λ\lambda, the stopping potential is VV. If the surface is now illuminated with a radiation of wavelength 4λ4\lambda, the stopping potential (V8)\left(\frac{V}{8}\right). The threshold wavelength for the surface is

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Answer: (d)

  1. hcλ−ϕ=eV\frac{hc}{\lambda} - \phi = eV and hc4λ−ϕ=eV8\frac{hc}{4\lambda} - \phi = \frac{eV}{8}
  2. Multiply the second by 8: 2hcλ−8ϕ=eV\frac{2hc}{\lambda} - 8\phi = eV
  3. Equate: hcλ−ϕ=2hcλ−8ϕ⇒7ϕ=hcλ\frac{hc}{\lambda} - \phi = \frac{2hc}{\lambda} - 8\phi \Rightarrow 7\phi = \frac{hc}{\lambda}
  4. λ0=hcϕ=7λ\lambda_0 = \frac{hc}{\phi} = 7\lambda. So the answer is (d) 7λ7\lambda.
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Q73MHT CET 2026 · 15 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a certain metal surface is illuminated with a light of wavelength λ\lambda, the stopping potential is VV. When the same surface is illuminated by light of wavelength 2λ2\lambda, the stopping potential is (V3)\left(\frac{V}{3}\right). The threshold wavelength for the surface is

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Answer: (c)

  1. hcλ−ϕ=eV\frac{hc}{\lambda} - \phi = eV and hc2λ−ϕ=eV3\frac{hc}{2\lambda} - \phi = \frac{eV}{3}
  2. Multiply the second by 3 and equate: 3hc2λ−3ϕ=hcλ−ϕ\frac{3hc}{2\lambda} - 3\phi = \frac{hc}{\lambda} - \phi
  3. hc2λ=2ϕ⇒ϕ=hc4λ\frac{hc}{2\lambda} = 2\phi \Rightarrow \phi = \frac{hc}{4\lambda}
  4. λ0=hcϕ=4λ\lambda_0 = \frac{hc}{\phi} = 4\lambda. So the answer is (c) 4λ4\lambda.
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Q74MHT CET 2026 · 19 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When light of wavelength 'λ\lambda' is incident on a photosensitive surface, the stopping potential is 'V'. When a light of wavelength 1.5λ1.5\lambda is incident on the same surface, the stopping potential is ' V4\frac{V}{4}'. Threshold wavelength for the surface is

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Answer: (d)

  1. hcλ−ϕ=eV\frac{hc}{\lambda} - \phi = eV and hc1.5λ−ϕ=eV4\frac{hc}{1.5\lambda} - \phi = \frac{eV}{4}
  2. Multiply the second by 4 and equate: 8hc3λ−4ϕ=hcλ−ϕ\frac{8hc}{3\lambda} - 4\phi = \frac{hc}{\lambda} - \phi
  3. 5hc3λ=3ϕ⇒ϕ=5hc9λ\frac{5hc}{3\lambda} = 3\phi \Rightarrow \phi = \frac{5hc}{9\lambda}
  4. λ0=hcϕ=95λ\lambda_0 = \frac{hc}{\phi} = \frac{9}{5}\lambda. So the answer is (d).
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Q75MHT CET 2026 · 19 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A photoelectric surface is illuminated successively by monochromatic light of wavelength λ\lambda and (λ/3)(\lambda/3). If the maximum kinetic energy of the emitted photo electrons in the second case is 44 times that in the first case, the work function of the surface of the material is (hh = Planck's constant, cc = speed of light)

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Answer: (b)

  1. K1=hcλ−ϕK_1 = \frac{hc}{\lambda} - \phi, K2=3hcλ−ϕK_2 = \frac{3hc}{\lambda} - \phi, with K2=4K1K_2 = 4K_1
  2. 3hcλ−ϕ=4hcλ−4ϕ⇒3ϕ=hcλ\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi \Rightarrow 3\phi = \frac{hc}{\lambda}
  3. ϕ=hc3λ\phi = \frac{hc}{3\lambda}. So the answer is (b).
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Q76MHT CET 2026 · 15 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metallic surface is illuminated with a radiation of wavelength 'λ\lambda', the stopping potential is 'V'. If the same surface is illuminated with radiation of wavelength 6λ6\lambda, the stopping potential is (V12)\left(\frac{V}{12}\right). The threshold wavelength for the surface is

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Answer: (c)

  1. hcλ−ϕ=eV\frac{hc}{\lambda} - \phi = eV and hc6λ−ϕ=eV12\frac{hc}{6\lambda} - \phi = \frac{eV}{12}
  2. Multiply the second by 12 and equate: 2hcλ−12ϕ=hcλ−ϕ\frac{2hc}{\lambda} - 12\phi = \frac{hc}{\lambda} - \phi
  3. 11ϕ=hcλ11\phi = \frac{hc}{\lambda}
  4. λ0=hcϕ=11λ\lambda_0 = \frac{hc}{\phi} = 11\lambda. So the answer is (c) 11λ11\lambda.
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Q77JEE Main 2026 · 2 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

For a certain metal, when monochromatic light of wavelength λ\lambda is incident, the stopping potential for photoelectrons is 3V03V_0. When the same metal is illuminated by light of wavelength 2λ2\lambda, then the stopping potential becomes V0V_0. The threshold wavelength for photoelectric emission for the given metal is αλ\alpha\lambda. The value of α\alpha is ______.

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Answer: (b)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=3eV0A - \phi = 3eV_0 and A2−ϕ=eV0\frac{A}{2} - \phi = eV_0
  3. Subtract: A2=2eV0⇒A=4eV0, ϕ=eV0=A4\frac{A}{2} = 2eV_0 \Rightarrow A = 4eV_0,\ \phi = eV_0 = \frac{A}{4}
  4. λ0=hcϕ=4λ\lambda_0 = \frac{hc}{\phi} = 4\lambda
  5. So the answer is (b) α=4\alpha = 4.
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Q78JEE Main 2026 · 24 JAN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ______ m.

(h=6.63×10−34 J.s,e=1.6×10−19 C,c=3×108 m/s)\left(h = 6.63 \times 10^{-34}\text{ J.s}, e = 1.6 \times 10^{-19}\text{ C}, c = 3 \times 10^{8}\text{ m/s}\right)

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Answer: (b)

  1. hcλ−ϕ=3.2 eV\frac{hc}{\lambda} - \phi = 3.2\text{ eV} and hc2λ−ϕ=0.7 eV\frac{hc}{2\lambda} - \phi = 0.7\text{ eV}
  2. Subtract: hc2λ=2.5 eV⇒hcλ=5 eV\frac{hc}{2\lambda} = 2.5\text{ eV} \Rightarrow \frac{hc}{\lambda} = 5\text{ eV}
  3. λ=6.63×10−34×3×1085×1.6×10−19=2.49×10−7 m\lambda = \frac{6.63\times10^{-34}\times3\times10^{8}}{5\times1.6\times10^{-19}} = 2.49\times10^{-7}\text{ m}
  4. So the answer is (b) 2.5×10−72.5\times10^{-7} m.
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Q79MHT CET 2026 · 19 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Radiation of wavelength λ\lambda, is incident on a photocell. The fastest emitted electron has speed vv. If the wavelength is changed to 3λ4\frac{3\lambda}{4}, the speed of the fastest electron will be

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Answer: (a)

  1. 12mv2=hcλ−ϕ\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi
  2. At 3λ4\frac{3\lambda}{4}: 12mv′2=43⋅hcλ−ϕ=43(hcλ−ϕ)+ϕ3\frac{1}{2}mv'^2 = \frac{4}{3}\cdot\frac{hc}{\lambda} - \phi = \frac{4}{3}\left(\frac{hc}{\lambda} - \phi\right) + \frac{\phi}{3}
  3. So 12mv′2>43⋅12mv2⇒v′>v(43)1/2\frac{1}{2}mv'^2 > \frac{4}{3}\cdot\frac{1}{2}mv^2 \Rightarrow v' > v\left(\frac{4}{3}\right)^{1/2}
  4. So the answer is (a): greater than v(43)1/2v\left(\frac{4}{3}\right)^{1/2}.
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Q80MHT CET 2026 · 14 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Energy of the incident photon on the metal surface is 3W3W and then 5W5W, where WW is the work function of that metal. The ratio of velocities of emitted photoelectrons is

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Answer: (c)

  1. 12mv2=E−W\frac{1}{2}mv^2 = E - W
  2. K1=3W−W=2WK_1 = 3W - W = 2W, K2=5W−W=4WK_2 = 5W - W = 4W
  3. v1v2=2W4W=12\frac{v_1}{v_2} = \sqrt{\frac{2W}{4W}} = \frac{1}{\sqrt{2}}
  4. So the answer is (c) 1:21 : \sqrt{2}.
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Q81MHT CET 2026 · 20 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two identical photocathodes receive light of frequencies n1n_1 and n2n_2. If the velocities of the emitted photoelectrons of mass m are V1V_1 and V2V_2 respectively, then (hh = Planck's constant)

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Answer: (d)

  1. 12mV12=hn1−ϕ\frac{1}{2}mV_1^2 = hn_1 - \phi and 12mV22=hn2−ϕ\frac{1}{2}mV_2^2 = hn_2 - \phi (same ϕ\phi)
  2. Subtract: 12m(V12−V22)=h(n1−n2)\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)
  3. So the answer is (d) V12−V22=2hm(n1−n2)V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2).
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Q82MHT CET 2026 · 16 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When a light of wavelength 'λ\lambda' falls on the emitter of a photosensitive surface, maximum speed of emitted photoelectrons is 'V'. If the incident wavelength is changed to '2λ/32\lambda/3', maximum speed of emitted photoelectrons will be

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Answer: (b)

  1. 12mV2=hcλ−ϕ\frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi
  2. At 2λ3\frac{2\lambda}{3}: 12mV′2=1.5hcλ−ϕ=1.5(hcλ−ϕ)+0.5ϕ\frac{1}{2}mV'^2 = 1.5\frac{hc}{\lambda} - \phi = 1.5\left(\frac{hc}{\lambda} - \phi\right) + 0.5\phi
  3. So V′2>1.5V2⇒V′>V(1.5)1/2V'^2 > 1.5V^2 \Rightarrow V' > V(1.5)^{1/2}
  4. So the answer is (b) greater than V(1.5)1/2V(1.5)^{1/2}.
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Q83MHT CET 2026 · 13 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

The threshold frequency of metal is f0f_0. When the light of frequency 2f02f_0 is incident on the metal plate, the maximum velocity of photoelectron is v1v_1. When the frequency of incident radiation is increased to 5f05f_0 the maximum velocity of photoelectrons emitted is v2v_2. The ratio v1v_1 to v2v_2 is

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Answer: (a)

  1. 12mv2=hν−hν0\frac{1}{2}mv^2 = h\nu - h\nu_0
  2. At 2ν02\nu_0: K1=hν0K_1 = h\nu_0; at 5ν05\nu_0: K2=4hν0K_2 = 4h\nu_0
  3. v1v2=14=12\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}
  4. So the answer is (a) 1:21 : 2.
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Q84MHT CET 2026 · 15 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Light of wavelength λ\lambda strikes a photoelectric surface and electrons are ejected with an energy EE. If EE is to be increased to exactly twice its original value, the wavelength changes to λ1\lambda_1 where

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Answer: (c)

  1. hcλ=ϕ+E\frac{hc}{\lambda} = \phi + E and hcλ1=ϕ+2E\frac{hc}{\lambda_1} = \phi + 2E
  2. hcλ1=2hcλ−ϕ<2hcλ⇒λ1>λ2\frac{hc}{\lambda_1} = 2\frac{hc}{\lambda} - \phi < 2\frac{hc}{\lambda} \Rightarrow \lambda_1 > \frac{\lambda}{2}
  3. Also hcλ1>hcλ⇒λ1<λ\frac{hc}{\lambda_1} > \frac{hc}{\lambda} \Rightarrow \lambda_1 < \lambda
  4. So the answer is (c) λ2<λ1<λ\frac{\lambda}{2} < \lambda_1 < \lambda.
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Q85MHT CET 2026 · 18 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

The maximum kinetic energies of photoelectrons emitted are K1K_1 and K2K_2 when light of wavelength λ1\lambda_1 and λ2\lambda_2 respectively are incident on a metallic surface. If λ1=3λ2\lambda_1 = 3\lambda_2 then

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Answer: (b)

  1. Einstein's equation: K=hcλ−ϕK = \frac{hc}{\lambda} - \phi.
  2. K1=hc3λ2−ϕK_1 = \frac{hc}{3\lambda_2} - \phi, while K23=hc3λ2−ϕ3\frac{K_2}{3} = \frac{hc}{3\lambda_2} - \frac{\phi}{3}.
  3. Since ϕ>ϕ3\phi > \frac{\phi}{3}, K1<K23K_1 < \frac{K_2}{3}.
  4. So the answer is (b) K1<K23K_1 < \frac{K_2}{3}.
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Q86MHT CET 2026 · 17 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Photoelectric emission is observed from a metallic surface for frequencies ν1\nu_1 and ν2\nu_2 of the incident light rays (ν1>ν2\nu_1 > \nu_2). If the ratio of maximum value of kinetic energy of the photoelectron emitted in first case to that in second case 3: K, then the threshold frequency of the metallic surface is

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Answer: (d)

  1. K=h(ν−ν0)K = h(\nu - \nu_0), so K1K2=ν1−ν0ν2−ν0=3K\frac{K_1}{K_2} = \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{3}{K}.
  2. Kν1−Kν0=3ν2−3ν0K\nu_1 - K\nu_0 = 3\nu_2 - 3\nu_0
  3. ν0(K−3)=Kν1−3ν2\nu_0(K - 3) = K\nu_1 - 3\nu_2
  4. So the answer is (d) ν0=Kν1−3ν2K−3\nu_0 = \frac{K\nu_1 - 3\nu_2}{K - 3}.
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Q87MHT CET 2026 · 11 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Photoelectrons are emitted from two similar metal plates when wavelengths λ1\lambda_1 and λ2\lambda_2 are incident on them (λ1=1.5λ2\lambda_1 = 1.5\lambda_2). If maximum kinetic energy of emitted photoelectrons is E1E_1 and E2E_2 respectively then

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Answer: (a)

  1. E=hcλ−ϕE = \frac{hc}{\lambda} - \phi; with λ1=1.5λ2\lambda_1 = 1.5\lambda_2: E1=23hcλ2−ϕE_1 = \frac{2}{3}\frac{hc}{\lambda_2} - \phi.
  2. 2E23=23hcλ2−2ϕ3\frac{2E_2}{3} = \frac{2}{3}\frac{hc}{\lambda_2} - \frac{2\phi}{3}
  3. Since ϕ>2ϕ3\phi > \frac{2\phi}{3}, E1<2E23E_1 < \frac{2E_2}{3}.
  4. So the answer is (a).
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Q88MHT CET 2026 · 11 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Photoelectric emission is observed from a metallic surface for frequencies v1v_1 and v2v_2 of the incident light rays (v1>v2v_1 > v_2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of k:1k : 1, then what is the threshold frequency of the metallic surface?

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Answer: (d)

  1. K=h(v−v0)K = h(v - v_0), so v1−v0v2−v0=k\frac{v_1 - v_0}{v_2 - v_0} = k.
  2. v1−v0=kv2−kv0⇒v0(k−1)=kv2−v1v_1 - v_0 = kv_2 - kv_0 \Rightarrow v_0(k - 1) = kv_2 - v_1
  3. So the answer is (d) v0=kv2−v1k−1v_0 = \frac{kv_2 - v_1}{k - 1}.
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Q89JEE Main 2026 · 8 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

K1K_1 and K2K_2 be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength λ1\lambda_1 and λ2\lambda_2, respectively. If λ1=2λ2\lambda_1 = 2\lambda_2 then the work function of material is given by:

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Answer: (d)

  1. K2=hcλ2−ϕK_2 = \frac{hc}{\lambda_2} - \phi and K1=hc2λ2−ϕK_1 = \frac{hc}{2\lambda_2} - \phi.
  2. hcλ2=K2+ϕ⇒K1=K2+ϕ2−ϕ=K2−ϕ2\frac{hc}{\lambda_2} = K_2 + \phi \Rightarrow K_1 = \frac{K_2 + \phi}{2} - \phi = \frac{K_2 - \phi}{2}
  3. So the answer is (d) ϕ=K2−2K1\phi = K_2 - 2K_1.
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Q90MHT CET 2026 · 12 MAY · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

A photon of energy 1010 units ejects a photo-electron from a metal surface whose work function is Φ0\Phi_0. If an electron(e) having maximum kinetic energy enters into a uniform magnetic field of induction(B) in a direction perpendicular to the field and describes a circular path of radius R, then radius R is {m=m =mass of electron}

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Answer: (c)

  1. Kmax=10−Φ0K_{max} = 10 - \Phi_0, and momentum p=2mKmaxp = \sqrt{2mK_{max}}.
  2. In the field evB=mv2R⇒R=peBevB = \frac{mv^2}{R} \Rightarrow R = \frac{p}{eB}
  3. So the answer is (c) R=2m(10−Φ0)eBR = \frac{\sqrt{2m(10 - \Phi_0)}}{eB}.
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Q91MHT CET 2025 · 25 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

Sodium and copper have work functions 2.3 eV and 4.5 eV respectively. The ratio of threshold wavelength of sodium to that of copper is nearest to

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Answer: (d)

  1. λ0=hcϕ\lambda_0 = \dfrac{hc}{\phi}, so λ0∝1ϕ\lambda_0 \propto \dfrac{1}{\phi}
  2. λNaλCu=4.52.3≈1.96≈2\dfrac{\lambda_{Na}}{\lambda_{Cu}} = \dfrac{4.5}{2.3} \approx 1.96 \approx 2
  3. So the answer is (d) 2:12:1.
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Q92MHT CET 2025 · 20 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

Light of wavelength ' λ\lambda ' falls on a metal having work function hcλ0\frac{hc}{\lambda_0}. Photoelectric effect will take place only if ( λ0\lambda_0 is the threshold wavelength)

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Answer: (c)

  1. Emission needs photon energy ≥\ge work function: hcλ≥hcλ0\dfrac{hc}{\lambda} \ge \dfrac{hc}{\lambda_0}
  2. This gives λ≤λ0\lambda \le \lambda_0
  3. So the answer is (c) λ≤λ0\lambda \le \lambda_0.
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Q93JEE Main 2025 · 3 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

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Answer: (c)

  1. Threshold λ0=1240 eV nm3 eV≈413 nm\lambda_0 = \dfrac{1240\text{ eV nm}}{3\text{ eV}} \approx 413\text{ nm}
  2. Light must have λ≤413 nm\lambda \le 413\text{ nm}, i.e. the short-wavelength (blue–violet) end of the visible spectrum.
  3. Red, yellow and green (above 490 nm490\text{ nm}) cannot work; blue is the shortest-wavelength option.
  4. So the answer is (c) Blue.
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Q94JEE Main 2025 · 22 JAN · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV, respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of

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Answer: (d)

  1. Photon energy E=1240 eV nm550 nm≈2.25 eVE = \dfrac{1240\text{ eV nm}}{550\text{ nm}} \approx 2.25\text{ eV}
  2. 2.25>1.92.25 > 1.9 (Cs) but <2.5< 2.5 (Li).
  3. So the answer is (d) Cs only.
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Q95MHT CET 2025 · 19 APR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The maximum velocity of the photoelectrons emitted by a metal surface is 9×105 m/s9 \times 10^{5}\text{ m/s}. The value of ratio of charge (e) to mass (m) of the photoelectron is 1.8×1011 C/kg1.8 \times 10^{11}\text{ C/kg}. The value of stopping potential in volt is

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Answer: (b)

  1. eV0=12mvmax2⇒V0=vmax22(e/m)eV_0 = \frac{1}{2}mv_{max}^2 \Rightarrow V_0 = \frac{v_{max}^2}{2(e/m)}
  2. V0=(9×105)22×1.8×1011=8.1×10113.6×1011V_0 = \frac{(9 \times 10^5)^2}{2 \times 1.8 \times 10^{11}} = \frac{8.1 \times 10^{11}}{3.6 \times 10^{11}}
  3. V0=2.25 VV_0 = 2.25\text{ V}, so the answer is (b) 2.25.
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Q96JEE Main 2025 · 23 JAN · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V, what is the wavelength of the em-wave? (Given hc=1242 eV nmhc = 1242\text{ eV nm} where h is the Planck's constant and c is the speed of light in vacuum.)

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Answer: (c)

  1. E=ϕ+eV0=2.14+2=4.14 eVE = \phi + eV_0 = 2.14 + 2 = 4.14\text{ eV}
  2. λ=hcE=12424.14=300 nm\lambda = \frac{hc}{E} = \frac{1242}{4.14} = 300\text{ nm}
  3. So the answer is (c) 300 nm.
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Q97MHT CET 2025 · 5 MAY · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Energy of the incident photons on the metal surface is initially 4 W and then 6 W where WW is the work function of that metal. The ratio of velocities of emitted photoelectrons is

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Answer: (a)

  1. v∝E−Wv \propto \sqrt{E - W}
  2. K1=4W−W=3WK_1 = 4W - W = 3W, K2=6W−W=5WK_2 = 6W - W = 5W
  3. v1v2=35\frac{v_1}{v_2} = \sqrt{\frac{3}{5}}
  4. So the answer is (a) 3:5\sqrt{3} : \sqrt{5}.
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Q98MHT CET 2025 · 22 APR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Photoelectric emission takes place from a certain metal at threshold frequency ν\nu. If the radiation of frequency 4ν4\nu is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( m=m = mass of photoelectron, h=h = Planck's constant)

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Answer: (a)

  1. 12mvmax2=h(4ν)−hν=3hν\frac{1}{2}mv_{max}^2 = h(4\nu) - h\nu = 3h\nu
  2. vmax=6hνmv_{max} = \sqrt{\frac{6h\nu}{m}}
  3. So the answer is (a) 6hνm\sqrt{\frac{6h\nu}{m}}.
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Q99MHT CET 2025 · 21 APR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are V1V_1 and V2V_2 respectively. The ratio V1:V2V_1 : V_2 is

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Answer: (d)

  1. v∝E−Wv \propto \sqrt{E - W}
  2. K1=2W−W=WK_1 = 2W - W = W, K2=3W−W=2WK_2 = 3W - W = 2W
  3. V1V2=W2W=12\frac{V_1}{V_2} = \sqrt{\frac{W}{2W}} = \frac{1}{\sqrt{2}}
  4. So the answer is (d) 1:21 : \sqrt{2}.
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Q100MHT CET 2025 · 25 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metal surface is illuminated by light of wavelength λ1\lambda_1 and λ2\lambda_2, the maximum velocities of photoelectrons ejected are V and 2 V respectively. The work function of the metal is ( h=h = Planck's constant, c=c = velocity of light, λ1>λ2\lambda_1 > \lambda_2)

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Answer: (d)

  1. 12mV2=hcλ1−ϕ\frac{1}{2}mV^2 = \frac{hc}{\lambda_1} - \phi and 12m(2V)2=hcλ2−ϕ\frac{1}{2}m(2V)^2 = \frac{hc}{\lambda_2} - \phi
  2. Second =4×= 4 \times first: hcλ2−ϕ=4hcλ1−4ϕ\frac{hc}{\lambda_2} - \phi = \frac{4hc}{\lambda_1} - 4\phi
  3. 3ϕ=4hcλ1−hcλ2=hc(4λ2−λ1)λ1λ23\phi = \frac{4hc}{\lambda_1} - \frac{hc}{\lambda_2} = \frac{hc(4\lambda_2 - \lambda_1)}{\lambda_1\lambda_2}
  4. ϕ=hc3λ1λ2(4λ2−λ1)\phi = \frac{hc}{3\lambda_1\lambda_2}(4\lambda_2 - \lambda_1). So the answer is (d).
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Q101MHT CET 2025 · 22 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a photosensitive metal surface is illuminated with radiation of wavelength ' λ1\lambda_1 ', the stopping potential is ' V1V_1 '. If the same surface is illuminated with radiation of wavelength ' 3λ13\lambda_1 ', the stopping potential is V16\frac{V_1}{6}. The threshold wavelength for the photosensitive metal surface is

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Answer: (c)

  1. hcλ1−ϕ=eV1\frac{hc}{\lambda_1} - \phi = eV_1 and hc3λ1−ϕ=eV16\frac{hc}{3\lambda_1} - \phi = \frac{eV_1}{6}
  2. Multiply the second by 6 and equate: 2hcλ1−6ϕ=hcλ1−ϕ\frac{2hc}{\lambda_1} - 6\phi = \frac{hc}{\lambda_1} - \phi
  3. 5ϕ=hcλ15\phi = \frac{hc}{\lambda_1}
  4. λ0=hcϕ=5λ1\lambda_0 = \frac{hc}{\phi} = 5\lambda_1. So the answer is (c).
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Q102JEE Main 2025 · 22 JAN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A light source of wavelength λ\lambda illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength λ2\frac{\lambda}{2}, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV)

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Answer: (a)

  1. hcλ=ϕ+K=1+2=3 eV\frac{hc}{\lambda} = \phi + K = 1 + 2 = 3\text{ eV}
  2. At λ2\frac{\lambda}{2}: photon energy =6 eV= 6\text{ eV}
  3. K′=6−1=5 eVK' = 6 - 1 = 5\text{ eV}
  4. So the answer is (a) 5 eV.
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Q103MHT CET 2025 · 5 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

A photosensitive surface has work function ϕ\phi. If photon of energy 3ϕ3\phi falls on this surface, the electron comes out with maximum velocity of 4×106 m/s4 \times 10^{6}\text{ m/s} When photon energy is increased to 7ϕ7\phi then maximum velocity of photoelectron will be

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Answer: (a)

  1. 12mv2=E−ϕ\frac{1}{2}mv^2 = E - \phi
  2. K1=3ϕ−ϕ=2ϕK_1 = 3\phi - \phi = 2\phi, K2=7ϕ−ϕ=6ϕK_2 = 7\phi - \phi = 6\phi
  3. v2=v162=43×106 m/sv_2 = v_1\sqrt{\frac{6}{2}} = 4\sqrt{3}\times10^{6}\text{ m/s}
  4. So the answer is (a) 43×1064\sqrt{3}\times10^{6} m/s.
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Q104MHT CET 2025 · 21 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When a light of wavelength λ\lambda falls on the emitter of a photocell, maximum speed of emitted photoelectrons is V. If the incident wavelength is changed to 2λ3\frac{2\lambda}{3}, maximum speed of emitted photoelectrons will be:

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Answer: (d)

  1. 12mV2=hcλ−ϕ\frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi
  2. At 2λ3\frac{2\lambda}{3}: 12mV′2=1.5hcλ−ϕ=1.5(hcλ−ϕ)+0.5ϕ\frac{1}{2}mV'^2 = 1.5\frac{hc}{\lambda} - \phi = 1.5\left(\frac{hc}{\lambda} - \phi\right) + 0.5\phi
  3. So V′2>1.5V2⇒V′>V(1.5)1/2V'^2 > 1.5V^2 \Rightarrow V' > V(1.5)^{1/2}
  4. None of the options says 'greater than'; the nearest is (d) 32 V\sqrt{\frac{3}{2}}\,V (exact only if ϕ\phi is neglected).
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Q105MHT CET 2025 · 21 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Photoelectric emission is observed from a metallic surface for frequencies v1v_1 and v2v_2 of the incident light rays (v1>v2)(v_1 > v_2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1:k1 : k, then the threshold frequency of metallic surface is

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Answer: (d)

  1. K=h(v−v0)K = h(v - v_0), so v1−v0v2−v0=1k\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{k}.
  2. kv1−kv0=v2−v0⇒v0(k−1)=kv1−v2kv_1 - kv_0 = v_2 - v_0 \Rightarrow v_0(k - 1) = kv_1 - v_2
  3. So the answer is (d) v0=kv1−v2k−1v_0 = \frac{kv_1 - v_2}{k - 1}.
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Q106MHT CET 2025 · 20 APR · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

On a photosensitive material, when frequency of incident radiation is increased by 20%20\%, maximum kinetic energy of emitted photoelectrons increases from 0.4 eV to 0.7 eV. The work function of the material is

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Answer: (b)

  1. hν=K+ϕh\nu = K + \phi: hν=0.4+ϕh\nu = 0.4 + \phi and 1.2hν=0.7+ϕ1.2h\nu = 0.7 + \phi.
  2. 1.2(0.4+ϕ)=0.7+ϕ⇒0.48+1.2ϕ=0.7+ϕ1.2(0.4 + \phi) = 0.7 + \phi \Rightarrow 0.48 + 1.2\phi = 0.7 + \phi
  3. 0.2ϕ=0.22⇒ϕ=1.10.2\phi = 0.22 \Rightarrow \phi = 1.1 eV
  4. So the answer is (b) 1.1 eV.
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Q107MHT CET 2025 · 19 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Light of wavelength λ\lambda strikes a photoelectric surface and electrons are ejected with energy E. If EE is to be increased to twice the original value, the wavelength changes to λ1\lambda_1

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Answer: (c)

  1. E=hcλ−ϕE = \frac{hc}{\lambda} - \phi and 2E=hcλ1−ϕ2E = \frac{hc}{\lambda_1} - \phi.
  2. hcλ1=2hcλ−ϕ<2hcλ\frac{hc}{\lambda_1} = 2\frac{hc}{\lambda} - \phi < \frac{2hc}{\lambda}
  3. So λ1>λ2\lambda_1 > \frac{\lambda}{2}: the answer is (c).
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Q108MHT CET 2025 · 25 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

If the electron in hydrogen atom jumps from third Bohr orbit to ground state directly and the difference between energies of the two states is radiated in the form of photons. If the work function of the material is 4.1 eV, then stopping potential is nearly

[Energy of electron in nnth orbit =−13.6n2 eV= \frac{-13.6}{n^2}\text{ eV} ]

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Answer: (d)

  1. Photon energy =13.6(1−19)=12.09= 13.6\left(1 - \frac{1}{9}\right) = 12.09 eV.
  2. eV0=12.09−4.1=7.99eV_0 = 12.09 - 4.1 = 7.99 eV
  3. So the answer is (d) V0≈8V_0 \approx 8 V.
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Q109JEE Main 2025 · 2 APR · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

A monochromatic light is incident on a metallic plate having work function ϕ\phi. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point BB. The distance between AA and BB is: (Given: The magnitude of charge of an electron is e and mass is m,hm, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

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Answer: (b)

ABvrAB = 2rB ⊗
  1. Kmax=hcλ−ϕK_{max} = \frac{hc}{\lambda} - \phi, momentum p=2mKmaxp = \sqrt{2mK_{max}}.
  2. Leaving normally with BB perpendicular, the electron moves on a semicircle of radius r=peBr = \frac{p}{eB} and lands back on the plate.
  3. AB=2r=22m(hc/λ−ϕ)eB=8m(hc/λ−ϕ)eBAB = 2r = \frac{2\sqrt{2m(hc/\lambda - \phi)}}{eB} = \frac{\sqrt{8m(hc/\lambda - \phi)}}{eB}
  4. So the answer is (b).
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Q110JEE Main 2024 · 30 JAN · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work function of a substance is 3.0 eV3.0\text{ eV}. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately;

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Answer: (c)

  1. Longest wavelength = threshold: λ0=hcϕ=1242 eV nm3.0 eV\lambda_0 = \dfrac{hc}{\phi} = \dfrac{1242\text{ eV nm}}{3.0\text{ eV}}
  2. λ0≈414 nm\lambda_0 \approx 414\text{ nm}
  3. So the answer is (c) 414 nm.
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Q111JEE Main 2024 · 27 JAN · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The threshold frequency of a metal with work function 6.63 eV6.63\text{ eV} is:

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Answer: (c)

  1. ν0=ϕh=6.63×1.6×10−196.63×10−34\nu_0 = \dfrac{\phi}{h} = \dfrac{6.63\times1.6\times10^{-19}}{6.63\times10^{-34}}
  2. ν0=1.6×1015 Hz\nu_0 = 1.6\times10^{15}\text{ Hz}
  3. So the answer is (c) 1.6×1015 Hz1.6\times10^{15}\text{ Hz}.
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Q112MHT CET 2024 · 4 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

For a photosensitive material, work function is ' W0W_0 ' and stopping potential is ' V '. The wavelength of incident radiation is ( h=h = Planck's constant, c=c = velocity of light, e=e = electronic charge)

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Answer: (d)

  1. Einstein's equation: hcλ=W0+eV\frac{hc}{\lambda} = W_0 + eV
  2. λ=hcW0+eV\lambda = \frac{hc}{W_0 + eV}
  3. So the answer is (d) hcW0+eV\frac{hc}{W_0 + eV}.
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Q113MHT CET 2024 · 2 MAY · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The stopping potential for a photelectric emission process is 10 V. The maximum kinetic energy of the electrons ejected in the process is [Charge on electron e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C} ]

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Answer: (c)

  1. Kmax=eV0K_{max} = eV_0
  2. Kmax=1.6×10−19×10=1.6×10−18 JK_{max} = 1.6 \times 10^{-19} \times 10 = 1.6 \times 10^{-18}\text{ J}
  3. So the answer is (c) 1.6×10−18 J1.6 \times 10^{-18}\text{ J}.
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Q114JEE Main 2024 · 9 APR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

UV light of 4.13 eV4.13\text{ eV} is incident on a photosensitive metal surface having work function 3.13 eV3.13\text{ eV}. The maximum kinetic energy of ejected photoelectrons will be:

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Answer: (b)

  1. Kmax=E−ϕK_{max} = E - \phi
  2. Kmax=4.13−3.13=1 eVK_{max} = 4.13 - 3.13 = 1\text{ eV}
  3. So the answer is (b) 1 eV.
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Q115JEE Main 2024 · 6 APR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

When UV light of wavelength 300 nm300\text{ nm} is incident on the metal surface having work function 2.13 eV2.13\text{ eV}, electron emission takes place. The stopping potential is:

(Given hc =1240 eV nm= 1240\text{ eV nm})

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Answer: (b)

  1. E=1240300=4.13 eVE = \frac{1240}{300} = 4.13\text{ eV}
  2. eV0=4.13−2.13=2 eVeV_0 = 4.13 - 2.13 = 2\text{ eV}
  3. So the answer is (b) 2 V.
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Q116JEE Main 2024 · 6 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

In photoelectric experiment energy of 2.48 eV2.48\text{ eV} irradiates a photo sensitive material. The stopping potential was measured to be 0.5 V0.5\text{ V}. Work function of the photo sensitive material is:

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Answer: (a)

  1. ϕ=E−eV0\phi = E - eV_0
  2. ϕ=2.48−0.5=1.98 eV\phi = 2.48 - 0.5 = 1.98\text{ eV}
  3. So the answer is (a) 1.98 eV.
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Q117MHT CET 2024 · 15 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two photons having energies twice and thrice the work function of metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively

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Answer: (c)

  1. v∝E−Wv \propto \sqrt{E - W}
  2. K1=2W−W=WK_1 = 2W - W = W, K2=3W−W=2WK_2 = 3W - W = 2W
  3. v1v2=W2W=12\frac{v_1}{v_2} = \sqrt{\frac{W}{2W}} = \frac{1}{\sqrt{2}}
  4. So the answer is (c) 1:21 : \sqrt{2}.
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Q118MHT CET 2024 · 3 MAY · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

The frequency of incident light falling on a photosensitive material is doubled, the K.E. of the emitted photoelectrons will be

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Answer: (c)

  1. K=hν−ϕK = h\nu - \phi.
  2. New: K′=2hν−ϕ=2K+ϕK' = 2h\nu - \phi = 2K + \phi
  3. Since ϕ>0\phi > 0, K′>2KK' > 2K.
  4. So the answer is (c) more than two times its initial value.
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Q119MHT CET 2024 · 3 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When photons of energy hv fall on a photosensitive surface of work function E0E_0, photoelectrons of maximum energy kk are emitted. If the frequency of radiation is doubled the maximum kinetic energy will be equal to ( h=h = Planck's constant)

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Answer: (d)

  1. k=hν−E0k = h\nu - E_0.
  2. New: k′=2hν−E0=hν+(hν−E0)=k+hνk' = 2h\nu - E_0 = h\nu + (h\nu - E_0) = k + h\nu
  3. So the answer is (d) k+hνk + h\nu.
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Q120NEET 2024Medium

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

If ϕ\phi is the work function of photosensitive material in eV\text{eV} and light of wavelength of numerical value λ=hce\lambda = \frac{hc}{e} metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take hh-Plank's constant, cc-velocity of light in free space) is (in SI units):

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Answer: (c)

  1. Photon energy E=hcλ=hchc/e=eE = \frac{hc}{\lambda} = \frac{hc}{hc/e} = e (in joules, i.e. 1 eV).
  2. Einstein's equation: Kmax=E−ϕK_{max} = E - \phi
  3. Kmax=e−ϕK_{max} = e - \phi (with ϕ\phi taken in the same units)
  4. So the answer is (c) e−ϕe - \phi.
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Q121MHT CET 2024 · 15 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a certain metallic surface is illuminated with monochromatic light wavelength λ\lambda, the stopping potential for photoelectric current is 4V04V_0. When the same surface is illuminated with light of wavelength 3λ3\lambda, the stopping potential is V0V_0. The threshold wavelength for this surface for photoelectric effect is

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Answer: (a)

  1. hcλ−ϕ=4eV0\frac{hc}{\lambda} - \phi = 4eV_0 and hc3λ−ϕ=eV0\frac{hc}{3\lambda} - \phi = eV_0
  2. Multiply the second by 4 and equate: 4hc3λ−4ϕ=hcλ−ϕ\frac{4hc}{3\lambda} - 4\phi = \frac{hc}{\lambda} - \phi
  3. 3ϕ=hc3λ⇒ϕ=hc9λ3\phi = \frac{hc}{3\lambda} \Rightarrow \phi = \frac{hc}{9\lambda}
  4. λ0=hcϕ=9λ\lambda_0 = \frac{hc}{\phi} = 9\lambda. So the answer is (a) 9λ9\lambda.
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Q122MHT CET 2024 · 11 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A photoelectric surface is illuminated successively by monochromatic light of Wavelength λ\lambda and (λ/3)(\lambda/3). If the maximum kinetic energy of the emitted photoelectrons in the second case is 4 times that in the first case, the work function of the surface of the material is (h=(h = Planck's constant, c=c = speed of light ))

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Answer: (c)

  1. K1=hcλ−ϕK_1 = \frac{hc}{\lambda} - \phi, K2=3hcλ−ϕK_2 = \frac{3hc}{\lambda} - \phi, with K2=4K1K_2 = 4K_1
  2. 3hcλ−ϕ=4hcλ−4ϕ⇒3ϕ=hcλ\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi \Rightarrow 3\phi = \frac{hc}{\lambda}
  3. ϕ=hc3λ\phi = \frac{hc}{3\lambda}. So the answer is (c).
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Q123MHT CET 2024 · 10 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A photoelectric surface is illuminated successively by monochromatic light of wavelength ' λ\lambda ' and (λ2)\left(\frac{\lambda}{2}\right). If the maximum kinetic energy of the emitted photoelectrons in the first case is one-fourth that in the second case, the work function of the surface of the material is ( c=c = speed of light, h=h = Planck's constant))

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Answer: (c)

  1. K1=hcλ−ϕK_1 = \frac{hc}{\lambda} - \phi, K2=2hcλ−ϕK_2 = \frac{2hc}{\lambda} - \phi, with K2=4K1K_2 = 4K_1
  2. 2hcλ−ϕ=4hcλ−4ϕ⇒3ϕ=2hcλ\frac{2hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi \Rightarrow 3\phi = \frac{2hc}{\lambda}
  3. ϕ=2hc3λ\phi = \frac{2hc}{3\lambda}. So the answer is (c).
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Q124MHT CET 2024 · 10 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a photosensitive surface is irradiated by lights of wavelengths ' λ1\lambda_1 ' and ' λ2\lambda_2 ', kinetic energies of the emitted photoelectrons is ' E1E_1 ' and ' E2E_2 ' respectively. The work function of the photosensitive surface is

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Answer: (c)

  1. E1=hcλ1−ϕE_1 = \frac{hc}{\lambda_1} - \phi and E2=hcλ2−ϕE_2 = \frac{hc}{\lambda_2} - \phi, so hc=λ1(E1+ϕ)=λ2(E2+ϕ)hc = \lambda_1(E_1 + \phi) = \lambda_2(E_2 + \phi)
  2. ϕ(λ1−λ2)=E2λ2−E1λ1\phi(\lambda_1 - \lambda_2) = E_2\lambda_2 - E_1\lambda_1
  3. ϕ=E1λ1−E2λ2λ2−λ1\phi = \frac{E_1\lambda_1 - E_2\lambda_2}{\lambda_2 - \lambda_1}
  4. So the answer is (c).
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Q125MHT CET 2024 · 2 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metallic surface is illuminated with a radiation of wavelength ' λ\lambda ', the stopping potential is ' VV '. If the same surface is illuminated with radiation of wavelength ' 3λ3\lambda ', the stopping potential is ' (V6)\left(\frac{V}{6}\right) '. The threshold wavelength for the surface is

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Answer: (c)

  1. hcλ−ϕ=eV\frac{hc}{\lambda} - \phi = eV and hc3λ−ϕ=eV6\frac{hc}{3\lambda} - \phi = \frac{eV}{6}
  2. Multiply the second by 6 and equate: 2hcλ−6ϕ=hcλ−ϕ\frac{2hc}{\lambda} - 6\phi = \frac{hc}{\lambda} - \phi
  3. 5ϕ=hcλ5\phi = \frac{hc}{\lambda}
  4. λ0=hcϕ=5λ\lambda_0 = \frac{hc}{\phi} = 5\lambda. So the answer is (c) 5λ5\lambda.
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Q126JEE Main 2024 · 31 JAN · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metal surface is illuminated by light of wavelength λ\lambda, the stopping potential is 8 V8\text{ V}. When the same surface is illuminated by light of wavelength 3λ3\lambda, stopping potential is 2 V2\text{ V}. The threshold wavelength for this surface is:

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Answer: (b)

  1. A−ϕ=8 eVA - \phi = 8\text{ eV} and A3−ϕ=2 eV\frac{A}{3} - \phi = 2\text{ eV}, with A=hcλA = \frac{hc}{\lambda}
  2. Subtract: 2A3=6⇒A=9 eV, ϕ=1 eV\frac{2A}{3} = 6 \Rightarrow A = 9\text{ eV},\ \phi = 1\text{ eV}
  3. λ0=hcϕ=9λ\lambda_0 = \frac{hc}{\phi} = 9\lambda
  4. So the answer is (b) 9λ9\lambda.
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Q127MHT CET 2024 · 9 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two identical photocathodes receive light of frequencies ' n1n_1 ' and ' n2n_2 '. If the velocities of the emitted photoelectrons of mass ' mm ' are ' V1V_1 ' and ' V2V_2 ' respectively, then ( h=h = Planck's constant)

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Answer: (d)

  1. 12mV12=hn1−ϕ\frac{1}{2}mV_1^2 = hn_1 - \phi and 12mV22=hn2−ϕ\frac{1}{2}mV_2^2 = hn_2 - \phi (same ϕ\phi)
  2. Subtract: 12m(V12−V22)=h(n1−n2)\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)
  3. So the answer is (d) V12−V22=2hm(n1−n2)V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2).
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Q128MHT CET 2024 · 4 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

A photosensitive metallic surface has work function ϕ\phi. If photon of energy 3ϕ3\phi falls on the surface, the electron comes out with a maximum velocity of 6×106 m/s6 \times 10^{6}\text{ m/s}. When the photon energy is increased to 9ϕ9\phi, then maximum velocity of photoelectrons will be

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Answer: (a)

  1. 12mv2=E−ϕ\frac{1}{2}mv^2 = E - \phi
  2. K1=3ϕ−ϕ=2ϕK_1 = 3\phi - \phi = 2\phi, K2=9ϕ−ϕ=8ϕK_2 = 9\phi - \phi = 8\phi
  3. v2=v182=2×6×106=12×106 m/sv_2 = v_1\sqrt{\frac{8}{2}} = 2\times6\times10^{6} = 12\times10^{6}\text{ m/s}
  4. So the answer is (a) 12×10612\times10^{6} m/s.
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Q129MHT CET 2024 · 4 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

The threshold frequency of a metal is ' F0F_0 '. When light of frequency 2F02F_0 is incident on the metal plate, the maximum velocity of photoelectron is ' V1V_1 '. When the frequency of incident radiation is increased to ' 5F05F_0 ', the maximum velocity of photoelectrons emitted is ' V2V_2 '. The ratio of V1V_1 to V2V_2 is

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Answer: (d)

  1. 12mv2=hν−hν0\frac{1}{2}mv^2 = h\nu - h\nu_0
  2. At 2ν02\nu_0: K1=hν0K_1 = h\nu_0; at 5ν05\nu_0: K2=4hν0K_2 = 4h\nu_0
  3. v1v2=14=12\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}
  4. So the answer is (d) 12\frac{1}{2}.
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Q130MHT CET 2024 · 2 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

The work function of metal ' AA ' and ' BB ' are in the ratio 1:21 : 2. If light of frequency ' ff ' and ' 2f2f ' is incident on surface ' AA ' and ' BB ' respectively, then the ratio of kinetic energies of emitted photo electrons is

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Answer: (b)

  1. Let ϕA=ϕ\phi_A = \phi, ϕB=2ϕ\phi_B = 2\phi.
  2. KA=hf−ϕK_A = hf - \phi, KB=2hf−2ϕ=2(hf−ϕ)K_B = 2hf - 2\phi = 2(hf - \phi)
  3. KA:KB=1:2K_A : K_B = 1 : 2
  4. So the answer is (b) 1:21 : 2.
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Q131MHT CET 2023 · 13 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

When a light of wavelength 300 nm300\text{ nm} falls on a photoelectric emitter, photo electrons are emitted. For another emitter light of wavelength 600 nm600\text{ nm} is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

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Answer: (b)

  1. Treat each wavelength as the threshold: ϕ=hcλ0\phi = \dfrac{hc}{\lambda_0}, so ϕ∝1λ0\phi \propto \dfrac{1}{\lambda_0}
  2. ϕ1ϕ2=600300=2\dfrac{\phi_1}{\phi_2} = \dfrac{600}{300} = 2
  3. So the answer is (b) 2:12:1.
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Q132NEET 2023Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work functions of Caesium (Cs)(\text{Cs}), potassium (K)(\text{K}) and Sodium (Na) are 2.14 eV,2.30 eV2.14\text{ eV}, 2.30\text{ eV} and 2.75 eV2.75\text{ eV} respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV2.20\text{ eV}, which of these photosensitive surfaces may emit photoelectrons?

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Answer: (d)

  1. Emission needs photon energy ≥ϕ\ge \phi.
  2. 2.20 eV>2.14 eV2.20\text{ eV} > 2.14\text{ eV} (Cs) but <2.30 eV< 2.30\text{ eV} (K) and <2.75 eV< 2.75\text{ eV} (Na).
  3. So the answer is (d) Cs only.
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Q133JEE Main 2023 · 13 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The difference between threshold wavelengths for two metal surfaces A\text{A} and B\text{B} having work function ϕA=9 eV\phi_A = 9\text{ eV} and ϕB=4.5 eV\phi_B = 4.5\text{ eV} in nm\text{nm} is:

{\{ Given, hc =1242 eVnm}= 1242\text{ eVnm}\}

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Answer: (b)

  1. λ0=hcϕ\lambda_0 = \dfrac{hc}{\phi}
  2. λA=12429=138 nm\lambda_A = \dfrac{1242}{9} = 138\text{ nm}, λB=12424.5=276 nm\lambda_B = \dfrac{1242}{4.5} = 276\text{ nm}
  3. Difference =276−138=138 nm= 276 - 138 = 138\text{ nm}
  4. So the answer is (b) 138.
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Q134JEE Main 2023 · 31 JAN · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

If the two metals A\text{A} and B\text{B} are exposed to radiation of wavelength 350 nm350\text{ nm}. The work functions of metals A\text{A} and B\text{B} are 4.8 eV4.8\text{ eV} and 2.2 eV2.2\text{ eV}. Then choose the correct option.

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Answer: (c)

  1. Emission needs photon energy E≥ϕE \ge \phi.
  2. E=1240λ(nm) eV=1240350≈3.54 eVE = \frac{1240}{\lambda(\text{nm})}\text{ eV} = \frac{1240}{350} \approx 3.54\text{ eV}
  3. ϕA=4.8 eV>E\phi_A = 4.8\text{ eV} > E: no emission from A. ϕB=2.2 eV<E\phi_B = 2.2\text{ eV} < E: B emits.
  4. So the answer is (c) Metal A will not emit photo-electrons.
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Q135JEE Main 2023 · 29 JAN · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The threshold wavelength for photoelectric emission from a material is 5500 A˚\mathring{A}. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a A. 75 W infra-red lamp

B. 10 W infra-red lamp

C. 75 W ultra-violet lamp

D. 10 W ultra-violet lamp

Choose the correct answer from the options given below:

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Answer: (c)

  1. Emission needs λ<λ0=5500 A˚\lambda < \lambda_0 = 5500\,\mathring{A}, whatever the intensity (power) of the lamp.
  2. Infra-red has λ>7000 A˚>λ0\lambda > 7000\,\mathring{A} > \lambda_0: no emission from A or B, even at 75 W.
  3. Ultra-violet has λ<4000 A˚<λ0\lambda < 4000\,\mathring{A} < \lambda_0: emission from both C and D, even at 10 W.
  4. So the answer is (c) C and D only.
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Q136MHT CET 2023 · 11 MAY · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

A metal surface of work function 1.13 eV1.13\text{ eV} is irradiated with light of wavelength 310 nm310\text{ nm}. The retarding potential required to stop the escape of photoelectrons is [Take hce=1240×10−9 SI\frac{hc}{e} = 1240 \times 10^{-9}\text{ SI} units]

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Answer: (b)

  1. Photon energy in eV: E=hc/eλ=1240×10−9310×10−9=4 eVE = \frac{hc/e}{\lambda} = \frac{1240 \times 10^{-9}}{310 \times 10^{-9}} = 4\text{ eV}
  2. eV0=E−ϕ=4−1.13=2.87 eVeV_0 = E - \phi = 4 - 1.13 = 2.87\text{ eV}
  3. So the answer is (b) 2.87 V.
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Q137MHT CET 2023 · 12 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Radiations of two photons having energies twice and five times the work function of metal are incident successively on metal surface. The ratio of the maximum velocity of photo electrons emitted in the two cases will be

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Answer: (b)

  1. v∝E−Wv \propto \sqrt{E - W}
  2. K1=2W−W=WK_1 = 2W - W = W, K2=5W−W=4WK_2 = 5W - W = 4W
  3. v1v2=W4W=12\frac{v_1}{v_2} = \sqrt{\frac{W}{4W}} = \frac{1}{2}
  4. So the answer is (b) 1 : 2.
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Q138MHT CET 2023 · 9 MAY · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are v1v_1 and v2v_2 respectively. The ratio v1:v2v_1 : v_2 is

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Answer: (d)

  1. v∝E−Wv \propto \sqrt{E - W}
  2. K1=2W−W=WK_1 = 2W - W = W, K2=3W−W=2WK_2 = 3W - W = 2W
  3. v1v2=W2W=12\frac{v_1}{v_2} = \sqrt{\frac{W}{2W}} = \frac{1}{\sqrt{2}}
  4. So the answer is (d) 1:21 : \sqrt{2}.
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Q139MHT CET 2023 · 14 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When radiation of wavelength 'λ\lambda' is incident on a metallic surface, the stopping potential is 4.8 V. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes 1.6 V1.6\text{ V}. The threshold wavelength for the surface is

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Answer: (b)

  1. In eV: hcλ−ϕ=4.8\frac{hc}{\lambda} - \phi = 4.8 and hc2λ−ϕ=1.6\frac{hc}{2\lambda} - \phi = 1.6
  2. Subtract: hc2λ=3.2\frac{hc}{2\lambda} = 3.2, so hcλ=6.4\frac{hc}{\lambda} = 6.4 eV and ϕ=1.6\phi = 1.6 eV
  3. λ0=hcϕ=6.41.6λ=4λ\lambda_0 = \frac{hc}{\phi} = \frac{6.4}{1.6}\lambda = 4\lambda
  4. So the answer is (b) 4λ4\lambda.
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Q140MHT CET 2023 · 13 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When radiations of wavelength λ\lambda is incident on a metallic surface the stopping potential required is 4.8 V4.8\text{ V}. If same surface is illuminated with radiations of double the wavelength, then required stopping potential becomes 1.6 V1.6\text{ V}, then the value of threshold wavelength for the surface is

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Answer: (b)

  1. In eV: hcλ−ϕ=4.8\frac{hc}{\lambda} - \phi = 4.8 and hc2λ−ϕ=1.6\frac{hc}{2\lambda} - \phi = 1.6
  2. Subtract: hc2λ=3.2\frac{hc}{2\lambda} = 3.2, so hcλ=6.4\frac{hc}{\lambda} = 6.4 eV and ϕ=1.6\phi = 1.6 eV
  3. λ0=hcϕ=6.41.6λ=4λ\lambda_0 = \frac{hc}{\phi} = \frac{6.4}{1.6}\lambda = 4\lambda
  4. So the answer is (b) 4λ4\lambda.
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Q141MHT CET 2023 · 12 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When light of wavelength λ\lambda is incident on a photosensitive surface the stopping potential is 'VV'. When light of wavelength 3λ3\lambda is incident on same surface the stopping potential is V6\frac{V}{6}. Then the threshold wavelength for the surface is

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Answer: (d)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=eVA - \phi = eV and A3−ϕ=eV6\frac{A}{3} - \phi = \frac{eV}{6}
  3. A−ϕ=6(A3−ϕ)=2A−6ϕ⇒ϕ=A5A - \phi = 6\left(\frac{A}{3} - \phi\right) = 2A - 6\phi \Rightarrow \phi = \frac{A}{5}
  4. λ0=hcϕ=5λ\lambda_0 = \frac{hc}{\phi} = 5\lambda
  5. So the answer is (d) 5λ5\lambda.
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Q142MHT CET 2023 · 11 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metallic surface is illuminated with radiation of wavelength 'λ\lambda', the stopping potential is 'VV'. If the same surface is illuminated with radiation of wavelength '2λ2\lambda', the stopping potential is ' (V4)\left(\frac{V}{4}\right) '. The threshold wavelength for the metallic surface is

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Answer: (b)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=eVA - \phi = eV and A2−ϕ=eV4\frac{A}{2} - \phi = \frac{eV}{4}
  3. A−ϕ=4(A2−ϕ)=2A−4ϕ⇒ϕ=A3A - \phi = 4\left(\frac{A}{2} - \phi\right) = 2A - 4\phi \Rightarrow \phi = \frac{A}{3}
  4. λ0=hcϕ=3λ\lambda_0 = \frac{hc}{\phi} = 3\lambda
  5. So the answer is (b) 3λ3\lambda.
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Q143MHT CET 2023 · 9 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a certain metal surface is illuminated with light of frequency ν\nu, the stopping potential for photoelectric current is V0V_0. When the same surface is illuminated by light of frequency ν2\frac{\nu}{2}, the stopping potential is V04\frac{V_0}{4}, the threshold frequency of photoelectric emission is

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Answer: (b)

  1. Einstein: hν=ϕ+eV0h\nu = \phi + eV_0
  2. hν−ϕ=eV0h\nu - \phi = eV_0 and hν2−ϕ=eV04\frac{h\nu}{2} - \phi = \frac{eV_0}{4}
  3. hν−ϕ=2hν−4ϕ⇒ϕ=hν3h\nu - \phi = 2h\nu - 4\phi \Rightarrow \phi = \frac{h\nu}{3}
  4. ν0=ϕh=ν3\nu_0 = \frac{\phi}{h} = \frac{\nu}{3}
  5. So the answer is (b) ν3\frac{\nu}{3}.
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Q144JEE Main 2023 · 11 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A metallic surface is illuminated with radiation of wavelength λ\lambda, the stopping potential is V0V_0. If the same surface is illuminated with radiation of wavelength 2λ2\lambda. the stopping potential becomes V04\frac{V_0}{4}. The threshold wavelength for this metallic surface will be

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Answer: (a)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=eV0A - \phi = eV_0 and A2−ϕ=eV04\frac{A}{2} - \phi = \frac{eV_0}{4}
  3. A−ϕ=2A−4ϕ⇒ϕ=A3A - \phi = 2A - 4\phi \Rightarrow \phi = \frac{A}{3}
  4. λ0=3λ\lambda_0 = 3\lambda
  5. So the answer is (a) 3λ3\lambda.
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Q145JEE Main 2023 · 1 FEB · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

The threshold frequency of a metal is f0f_0. When the light of frequency 2f02f_0 is incident on the metal plate, the maximum velocity of photoelectrons is v1v_1. When the frequency of incident radiation is increased to 5f05f_0, the maximum velocity of photoelectrons emitted is v2v_2. The ratio of v1v_1 to v2v_2 is:

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Answer: (a)

  1. 12mv2=hν−hν0\frac{1}{2}mv^2 = h\nu - h\nu_0
  2. At 2ν02\nu_0: K1=hν0K_1 = h\nu_0; at 5ν05\nu_0: K2=4hν0K_2 = 4h\nu_0
  3. v1v2=14=12\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}
  4. So the answer is (a) v1v2=12\frac{v_1}{v_2} = \frac{1}{2}.
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Q146MHT CET 2023 · 11 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

The maximum kinetic energies of photoelectrons emitted are K1K_1 and K2K_2 when lights of wavelengths λ1\lambda_1 and λ2\lambda_2 are incident on a metallic surface. If λ1=3λ2\lambda_1 = 3\lambda_2 then

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Answer: (b)

  1. K=hcλ−ϕK = \frac{hc}{\lambda} - \phi, so K1=hc3λ2−ϕK_1 = \frac{hc}{3\lambda_2} - \phi.
  2. K23=hc3λ2−ϕ3\frac{K_2}{3} = \frac{hc}{3\lambda_2} - \frac{\phi}{3}, which is larger than K1K_1.
  3. So the answer is (b) K1<K23K_1 < \frac{K_2}{3}.
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Q147MHT CET 2023 · 10 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Maximum kinetic energy of photoelectrons is 'EE' when wavelength of incident radiation is 'λ\lambda'. If wavelength of incident radiations is reduced to λ3\frac{\lambda}{3} then maximum kinetic energy of photoelectrons becomes four times. Then work function of the metal is

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Answer: (b)

  1. E=hcλ−ϕE = \frac{hc}{\lambda} - \phi and 4E=3hcλ−ϕ4E = \frac{3hc}{\lambda} - \phi.
  2. 4hcλ−4ϕ=3hcλ−ϕ⇒hcλ=3ϕ\frac{4hc}{\lambda} - 4\phi = \frac{3hc}{\lambda} - \phi \Rightarrow \frac{hc}{\lambda} = 3\phi
  3. So the answer is (b) ϕ=hc3λ\phi = \frac{hc}{3\lambda}.
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Q148MHT CET 2023 · 9 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

From a metallic surface photoelectric emission is observed for frequencies v1v_1 and v2 (v1>v2)v_2\ (v_1 > v_2) of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio 1:x1 : x. Hence the threshold frequency of the metallic surface is

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Answer: (c)

  1. K=h(v−v0)K = h(v - v_0), so v1−v0v2−v0=1x\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{x}.
  2. xv1−xv0=v2−v0⇒v0(x−1)=xv1−v2xv_1 - xv_0 = v_2 - v_0 \Rightarrow v_0(x - 1) = xv_1 - v_2
  3. So the answer is (c) v0=xv1−v2x−1v_0 = \frac{xv_1 - v_2}{x - 1}.
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Q149JEE Main 2022 · 29 JUN · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:

Assertion A: The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.

Reason R: Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (b)

  1. A: if hν<ϕh\nu < \phi, the electron cannot get enough energy to leave, so no emission. A is correct.
  2. R: Kmax=hν−ϕ=0K_{max} = h\nu - \phi = 0 when hν=ϕh\nu = \phi. R is correct.
  3. R describes the threshold case hν=ϕh\nu = \phi; it does not explain why there is no emission for hν<ϕh\nu < \phi.
  4. So the answer is (b) Both correct, R is not the correct explanation of A.
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Q150NEET 2022 · PHASE 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The light rays having photons of energy 4.2 eV are falling on a metal surface having a work function of 2.2 eV. The stopping potential of the surface is

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Answer: (c)

  1. eV0=E−ϕ=4.2−2.2=2 eVeV_0 = E - \phi = 4.2 - 2.2 = 2\text{ eV}
  2. V0=2 VV_0 = 2\text{ V} (a potential is in volts, not eV)
  3. So the answer is (c) 2 V.
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Q151NEET 2022 · PHASE 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The threshold frequency of a photoelectric metal is v0. If light of frequency 4v0 is incident on this metal, then the maximum kinetic energy of emitted electrons will be:

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Answer: (d)

  1. Kmax=hν−hν0K_{max} = h\nu - h\nu_0
  2. Kmax=h(4ν0)−hν0=3hν0K_{max} = h(4\nu_0) - h\nu_0 = 3h\nu_0
  3. So the answer is (d) 3hν03h\nu_0.
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Q152JEE Main 2022 · 28 JUL · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

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Answer: (c)

  1. 12mvmax2=E−ϕ\frac{1}{2}mv_{max}^2 = E - \phi, so vmax∝E−ϕv_{max} \propto \sqrt{E - \phi}.
  2. K1=5ϕ−ϕ=4ϕK_1 = 5\phi - \phi = 4\phi, K2=10ϕ−ϕ=9ϕK_2 = 10\phi - \phi = 9\phi
  3. v1v2=49=23\frac{v_1}{v_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}
  4. So the answer is (c) 2:32:3.
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Q153JEE Main 2022 · 24 JUN · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be:

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Answer: (b)

  1. 12mvmax2=E−ϕ\frac{1}{2}mv_{max}^2 = E - \phi, so vmax∝E−ϕv_{max} \propto \sqrt{E - \phi}.
  2. K1=3.8−0.6=3.2K_1 = 3.8 - 0.6 = 3.2 eV, K2=1.4−0.6=0.8K_2 = 1.4 - 0.6 = 0.8 eV
  3. v1v2=3.20.8=2\frac{v_1}{v_2} = \sqrt{\frac{3.2}{0.8}} = 2
  4. So the answer is (b) 2:12:1.
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Q154JEE Main 2022 · 29 JUN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The electric field at a point associated with a light wave is given by

E = 200 [sin (6 ×\times 1015)t + sin (9 ×\times 1015)t] Vm−1{}^{-1}

Given: h = 4.14 ×\times 10−-15 eVs

If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

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Answer: (d)

  1. Highest angular frequency ω=9×1015\omega = 9 \times 10^{15} rad/s gives the maximum KE: ν=ω2π=1.43×1015\nu = \frac{\omega}{2\pi} = 1.43 \times 10^{15} Hz.
  2. E=hν=4.14×10−15×1.43×1015=5.93E = h\nu = 4.14 \times 10^{-15} \times 1.43 \times 10^{15} = 5.93 eV
  3. Kmax=5.93−2.50≈3.42K_{max} = 5.93 - 2.50 \approx 3.42 eV
  4. So the answer is (d) 3.42 eV.
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Q155JEE Main 2022 · 26 JUN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 A˚\mathring{A} is 0.42 V. If the threshold frequency is x ×\times 1013 /s, where x is _________ (nearest integer). (Given, speed light = 3 ×\times 108 m/s, Planck's constant = 6.63 ×\times 10−-34 Js)

Numerical answer type. Work it out, then open the answer.

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Answer: 35

  1. E=hcλ=6.63×10−34×3×1086630×10−10=3.0×10−19E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6630 \times 10^{-10}} = 3.0 \times 10^{-19} J
  2. ϕ=E−eV0=3.0×10−19−0.42×1.6×10−19=2.33×10−19\phi = E - eV_0 = 3.0 \times 10^{-19} - 0.42 \times 1.6 \times 10^{-19} = 2.33 \times 10^{-19} J
  3. ν0=ϕh=2.33×10−196.63×10−34=3.51×1014=35.1×1013\nu_0 = \frac{\phi}{h} = \frac{2.33 \times 10^{-19}}{6.63 \times 10^{-34}} = 3.51 \times 10^{14} = 35.1 \times 10^{13} Hz
  4. So x=35x = 35.
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Q156NEET 2022 · PHASE 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When two monochromatic lights of frequency, v and v2\frac{v}{2} are incident on a photoelectric metal, their stopping potential becomes Vs2\frac{V_s}{2} and Vs respectively. The threshold frequency for this metal is

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Answer: (d)

  1. Einstein: hν=ϕ+eVsh\nu = \phi + eV_s
  2. hν−ϕ=eVs2h\nu - \phi = \frac{eV_s}{2} and hν2−ϕ=eVs\frac{h\nu}{2} - \phi = eV_s
  3. hν2−ϕ=2(hν−ϕ)⇒ϕ=3hν2\frac{h\nu}{2} - \phi = 2(h\nu - \phi) \Rightarrow \phi = \frac{3h\nu}{2}
  4. ν0=32ν\nu_0 = \frac{3}{2}\nu
  5. So the answer is (d) 32ν\frac{3}{2}\nu.
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Q157JEE Main 2022 · 24 JUN · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2 = x v1, the value of x will be __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. 12mv2=hν−hν0\frac{1}{2}mv^2 = h\nu - h\nu_0
  2. At 2ν02\nu_0: K1=hν0K_1 = h\nu_0; at 5ν05\nu_0: K2=4hν0K_2 = 4h\nu_0
  3. v1v2=14=12\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}
  4. v2=2v1v_2 = 2v_1, so x=2x = 2.
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Q158JEE Main 2022 · 29 JUL · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ\lambda. To double the kinetic energy, the incident light must have wavelength:

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Answer: (b)

  1. E=hcλ−ϕE = \frac{hc}{\lambda} - \phi and 2E=hcλ′−ϕ2E = \frac{hc}{\lambda'} - \phi.
  2. Subtract: hcλ′=E+hcλ=Eλ+hcλ\frac{hc}{\lambda'} = E + \frac{hc}{\lambda} = \frac{E\lambda + hc}{\lambda}
  3. So the answer is (b) λ′=hcλEλ+hc\lambda' = \frac{hc\lambda}{E\lambda + hc}.
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Q159JEE Main 2022 · 25 JUL · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

A metal exposed to light of wavelength 800 nm800\text{ nm} and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm500\text{ nm} is used. The workfunction of the metal is: (Take hc =1230 eV-nm= 1230\text{ eV-nm} ).

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Answer: (c)

  1. Photon energies: 1230800=1.5375\frac{1230}{800} = 1.5375 eV and 1230500=2.46\frac{1230}{500} = 2.46 eV.
  2. K=1.5375−ϕK = 1.5375 - \phi and 2K=2.46−ϕ2K = 2.46 - \phi
  3. 3.075−2ϕ=2.46−ϕ⇒ϕ=0.6153.075 - 2\phi = 2.46 - \phi \Rightarrow \phi = 0.615 eV
  4. So the answer is (c) 0.615 eV.
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Q160JEE Main 2022 · 28 JUN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ\lambda1 and λ\lambda2, respectively are incident on a metallic surface. If λ\lambda1 = 3λ\lambda2 then:

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Answer: (b)

  1. K1=hc3λ2−ϕK_1 = \frac{hc}{3\lambda_2} - \phi and K23=hc3λ2−ϕ3\frac{K_2}{3} = \frac{hc}{3\lambda_2} - \frac{\phi}{3}.
  2. Since ϕ>ϕ3\phi > \frac{\phi}{3}, K1<K23K_1 < \frac{K_2}{3}.
  3. So the answer is (b).
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Q161JEE Main 2022 · 26 JUN · SHIFT 2Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

A metal surface is illuminated by a radiation of wavelength 4500 A˚\text{\AA}. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90∘^\circ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately:

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Answer: (a)

  1. Photon energy =124004500=2.76= \frac{12400}{4500} = 2.76 eV.
  2. r=2mKeB⇒K=e2B2r22mr = \frac{\sqrt{2mK}}{eB} \Rightarrow K = \frac{e^2B^2r^2}{2m}; in eV, K=eB2r22m=1.6×10−19×(2×10−3)2(2×10−3)22×9.1×10−31≈1.41K = \frac{eB^2r^2}{2m} = \frac{1.6 \times 10^{-19} \times (2 \times 10^{-3})^2 (2 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.41 eV
  3. ϕ=2.76−1.41≈1.35\phi = 2.76 - 1.41 \approx 1.35 eV
  4. So the answer is (a) 1.36 eV.
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Q162MHT CET 2021 · 22 SEP · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The light of wavelength 'λ\lambda' is incident on the surface of metal of work function ϕ\phi and emits the electron. The maximum velocity of electron emitted is [m=m = mass of electron and h=h = Planck's constant, c=c = velocity of light]

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Answer: (d)

  1. 12mvmax2=hcλ−ϕ=hc−λϕλ\frac{1}{2}mv_{max}^2 = \frac{hc}{\lambda} - \phi = \frac{hc - \lambda\phi}{\lambda}
  2. vmax=[2(hc−λϕ)mλ]1/2v_{max} = \left[\frac{2(hc - \lambda\phi)}{m\lambda}\right]^{1/2}
  3. So the answer is (d).
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Q163JEE Main 2021 · 17 MAR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-electrons coming out are v1 and v2 respectively, then

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Answer: (d)

  1. Same cathode, same ϕ\phi: 12mv12=hf1−ϕ\frac{1}{2}mv_1^2 = hf_1 - \phi and 12mv22=hf2−ϕ\frac{1}{2}mv_2^2 = hf_2 - \phi.
  2. Subtract: 12m(v12−v22)=h(f1−f2)\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)
  3. v12−v22=2hm(f1−f2)v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)
  4. So the answer is (d).
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Q164JEE Main 2021 · 26 FEB · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two streams of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x: y. The value of x is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. 12mvmax2=E−ϕ\frac{1}{2}mv_{max}^2 = E - \phi, so vmax∝E−ϕv_{max} \propto \sqrt{E - \phi}.
  2. K1=2ϕ−ϕ=ϕK_1 = 2\phi - \phi = \phi, K2=10ϕ−ϕ=9ϕK_2 = 10\phi - \phi = 9\phi
  3. v1v2=19=13\frac{v_1}{v_2} = \sqrt{\frac{1}{9}} = \frac{1}{3}, so x:y=1:3x:y = 1:3
  4. So x=1x = 1.
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Q165MHT CET 2021 · 24 SEP · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

A light of wavelength 'λ\lambda' and intensity 'II' falls on photosensitive material. If 'NN' photo electrons are emitted, each with kinetic energy 'E', then

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Answer: (d)

  1. E=hcλ−ϕE = \frac{hc}{\lambda} - \phi: kinetic energy depends on wavelength (increases as λ\lambda decreases), not on intensity, so E∝1λE \propto \frac{1}{\lambda} (approximately).
  2. Number of photoelectrons depends on number of photons, i.e. on intensity: N∝IN \propto I.
  3. So the answer is (d).
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Q166MHT CET 2021 · 22 SEP · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

Photons of energy 10 eV10\text{ eV} are incident on a photosensitive surface of threshold frequency 2×1015 Hz2 \times 10^{15}\text{ Hz}. The kinetic energy in eV\text{eV} of the photoelectrons emitted is

[Planck's constant h=6.63×10−34 Jsh = 6.63 \times 10^{-34}\text{ Js} ]

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Answer: (d)

  1. ϕ=hν0=6.63×10−34×2×1015=1.326×10−18\phi = h\nu_0 = 6.63 \times 10^{-34} \times 2 \times 10^{15} = 1.326 \times 10^{-18} J
  2. ϕ=1.326×10−181.6×10−19=8.29\phi = \frac{1.326 \times 10^{-18}}{1.6 \times 10^{-19}} = 8.29 eV
  3. K=10−8.29=1.71K = 10 - 8.29 = 1.71 eV
  4. So the answer is (d) 1.71 eV.
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Q167JEE Main 2021 · 26 AUG · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ\phi = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 ×\times 10−-34 Js, c = 3 ×\times 108 ms−1{}^{-1})

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Answer: (a)

  1. eV0=hcλ−ϕeV_0 = \frac{hc}{\lambda} - \phi, so ΔV0=hce(1λ1−1λ2)\Delta V_0 = \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right).
  2. hce=6.63×10−34×3×1081.6×10−19=1243\frac{hc}{e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19}} = 1243 V nm
  3. ΔV0=1243(1280−1400)=1243×1.07×10−3≈1.3\Delta V_0 = 1243\left(\frac{1}{280} - \frac{1}{400}\right) = 1243 \times 1.07 \times 10^{-3} \approx 1.3 V
  4. So the answer is (a) 1.3 V.
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Q168MHT CET 2021 · 20 SEP · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a photosensitive surface is irradiated by light of wavelengths 'λ1\lambda_1' and 'λ2\lambda_2', kinetic energies of emitted photoelectrons are 'E1_1 ' and 'E2_2 ' respectively. The work function of photosensitive surface is

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Answer: (a)

  1. hcλ1=ϕ+E1\frac{hc}{\lambda_1} = \phi + E_1 and hcλ2=ϕ+E2\frac{hc}{\lambda_2} = \phi + E_2
  2. Equate hchc: λ1(ϕ+E1)=λ2(ϕ+E2)\lambda_1(\phi + E_1) = \lambda_2(\phi + E_2)
  3. ϕ(λ2−λ1)=λ1E1−λ2E2\phi(\lambda_2 - \lambda_1) = \lambda_1E_1 - \lambda_2E_2
  4. So the answer is (a) ϕ=λ1E1−λ2E2λ2−λ1\phi = \frac{\lambda_1E_1 - \lambda_2E_2}{\lambda_2 - \lambda_1}.
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Q169JEE Main 2021 · 27 AUG · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?

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Answer: (b)

  1. Use E(eV)=1240λ(nm)E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}
  2. E1=1240670.5=1.849 eV⇒ϕ=1.849−0.48=1.369 eVE_1 = \frac{1240}{670.5} = 1.849\text{ eV} \Rightarrow \phi = 1.849 - 0.48 = 1.369\text{ eV}
  3. E2=1240474.6=2.613 eVE_2 = \frac{1240}{474.6} = 2.613\text{ eV}
  4. V2=2.613−1.369=1.24 VV_2 = 2.613 - 1.369 = 1.24\text{ V}
  5. So the answer is (b) 1.25 V.
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Q170JEE Main 2021 · 25 JUL · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When radiation of wavelength λ\lambda is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is:

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Answer: (b)

  1. A−ϕ=4.8 eVA - \phi = 4.8\text{ eV} and A2−ϕ=1.6 eV\frac{A}{2} - \phi = 1.6\text{ eV}, with A=hcλA = \frac{hc}{\lambda}
  2. Subtract: A2=3.2⇒A=6.4 eV, ϕ=1.6 eV\frac{A}{2} = 3.2 \Rightarrow A = 6.4\text{ eV},\ \phi = 1.6\text{ eV}
  3. λ0=hcϕ=6.41.6λ=4λ\lambda_0 = \frac{hc}{\phi} = \frac{6.4}{1.6}\lambda = 4\lambda
  4. So the answer is (b) 4λ4\lambda.
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Q171JEE Main 2021 · 25 FEB · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is:

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Answer: (d)

  1. Use E(eV)=1240λ(nm)E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}
  2. E1=1240491=2.525 eV⇒ϕ=2.525−0.710=1.815 eVE_1 = \frac{1240}{491} = 2.525\text{ eV} \Rightarrow \phi = 2.525 - 0.710 = 1.815\text{ eV}
  3. New photon energy =1.815+1.43=3.245 eV= 1.815 + 1.43 = 3.245\text{ eV}
  4. λ=12403.245=382 nm\lambda = \frac{1240}{3.245} = 382\text{ nm}
  5. So the answer is (d) 382 nm.
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Q172JEE Main 2021 · 20 JUL · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A certain metallic surface is illuminated by monochromatic radiation of wavelength λ\lambda. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2λ\lambda, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is ____________ λ\lambda.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=3eV0A - \phi = 3eV_0 and A2−ϕ=eV0\frac{A}{2} - \phi = eV_0
  3. Subtract: A2=2eV0⇒A=4eV0, ϕ=eV0=A4\frac{A}{2} = 2eV_0 \Rightarrow A = 4eV_0,\ \phi = eV_0 = \frac{A}{4}
  4. λ0=hcϕ=4λ\lambda_0 = \frac{hc}{\phi} = 4\lambda
  5. So the answer is 4, i.e. λ0=4λ\lambda_0 = 4\lambda.
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Q173MHT CET 2021 · 22 SEP · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When a light of wavelength 'λ\lambda' falls on the emitter of a photocell, maximum speed of emitted photoelectrons is 'VV'. If the incident wavelength is changed to 2λ3\frac{2\lambda}{3}, maximum speed of emitted photoelectrons will be

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Answer: (c)

  1. 12mV2=hcλ−ϕ\frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi
  2. At 2λ3\frac{2\lambda}{3}: 12mV′2=1.5hcλ−ϕ=1.5(hcλ−ϕ)+0.5ϕ\frac{1}{2}mV'^2 = 1.5\frac{hc}{\lambda} - \phi = 1.5\left(\frac{hc}{\lambda} - \phi\right) + 0.5\phi
  3. So V′2>1.5V2⇒V′>V(1.5)1/2V'^2 > 1.5V^2 \Rightarrow V' > V(1.5)^{1/2}
  4. So the answer is (c) greater than V(1.5)1/2V(1.5)^{1/2}.
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Q174MHT CET 2021 · 23 SEP · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When wavelength of incident radiation on the metal surface is reduced from 'λ1\lambda_1 ' to 'λ2\lambda_2', the kinetic energy of emitted photoelectrons is tripled. The work function of metal [h=h = Plank's constant, c=c = velocity of light]

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Answer: (b)

  1. K=hcλ1−ϕK = \frac{hc}{\lambda_1} - \phi and 3K=hcλ2−ϕ3K = \frac{hc}{\lambda_2} - \phi.
  2. 3hcλ1−3ϕ=hcλ2−ϕ⇒2ϕ=3hcλ1−hcλ2\frac{3hc}{\lambda_1} - 3\phi = \frac{hc}{\lambda_2} - \phi \Rightarrow 2\phi = \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2}
  3. ϕ=hc2[3λ2−λ1λ1λ2]\phi = \frac{hc}{2}\left[\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right]
  4. So the answer is (b).
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Q175MHT CET 2021 · 21 SEP · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

Photoemission from metal surface takes place for frequencies 'v1v_1 ' and 'v2v_2' of incident rays (v1>v2)(v_1 > v_2). Maximum kinetic energy of photoelectrons emitted is in the ratio 1:K1 : K. The threshold frequency of metallic surface is

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Answer: (d)

  1. K=h(v−v0)K = h(v - v_0), so v1−v0v2−v0=1K\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{K}.
  2. Kv1−Kv0=v2−v0⇒v0(K−1)=Kv1−v2Kv_1 - Kv_0 = v_2 - v_0 \Rightarrow v_0(K - 1) = Kv_1 - v_2
  3. So the answer is (d) Kv1−v2K−1\frac{Kv_1 - v_2}{K - 1}.
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Q176MHT CET 2021 · 20 SEP · SHIFT 2Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

Photoelectrons are emitted when photons of energy 4.2 eV4.2\text{ eV} are incident on a photosensitive metallic sphere of radius 10 cm10\text{ cm} and work function 2.4 eV2.4\text{ eV}. The number of photoelectrons emitted before the emission is stopped is

[14πϵ0=9×109\left[\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\right. SI unit; e=1.6×10−19 C]e = 1.6 \times 10^{-19}\text{ C}]

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Answer: (c)

  1. Emission stops when the sphere potential equals the stopping potential: V=4.2−2.4=1.8V = 4.2 - 2.4 = 1.8 V.
  2. V=kQr⇒Q=1.8×0.19×109=2×10−11V = \frac{kQ}{r} \Rightarrow Q = \frac{1.8 \times 0.1}{9 \times 10^9} = 2 \times 10^{-11} C
  3. n=Qe=2×10−111.6×10−19=1.25×108n = \frac{Q}{e} = \frac{2 \times 10^{-11}}{1.6 \times 10^{-19}} = 1.25 \times 10^8
  4. So the answer is (c).
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Q177JEE Main 2021 · 20 JUL · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

The radiation corresponding to 3 →\rightarrow 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 ×\times 10−-4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is: (Mass of electron = 9.1 ×\times 10−-31 kg)

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Answer: (c)

  1. 3→23 \to 2: E=13.6(14−19)=1.89E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 1.89 eV.
  2. K=eB2r22mK = \frac{eB^2r^2}{2m} eV =1.6×10−19×(5×10−4)2(7×10−3)22×9.1×10−31≈1.08= \frac{1.6 \times 10^{-19} \times (5 \times 10^{-4})^2 (7 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.08 eV
  3. ϕ=1.89−1.08≈0.81\phi = 1.89 - 1.08 \approx 0.81 eV
  4. So the answer is (c) 0.82 eV.
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Q178JEE Main 2021 · 25 JUL · SHIFT 2Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular arc of radius 30 cm. The value of B is ___________ ×\times 10−-7 T.

Given hc = 20 ×\times 10−-26 J-m, mass of electron = 9 ×\times 10 −-31 kg

Numerical answer type. Work it out, then open the answer.

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Answer: 125

  1. E=hcλ=20×10−26500×10−9=4×10−19E = \frac{hc}{\lambda} = \frac{20 \times 10^{-26}}{500 \times 10^{-9}} = 4 \times 10^{-19} J =2.5= 2.5 eV.
  2. K=2.5−1.25=1.25K = 2.5 - 1.25 = 1.25 eV =2×10−19= 2 \times 10^{-19} J
  3. p=2mK=2×9×10−31×2×10−19=6×10−25p = \sqrt{2mK} = \sqrt{2 \times 9 \times 10^{-31} \times 2 \times 10^{-19}} = 6 \times 10^{-25} kg m/s
  4. B=per=6×10−251.6×10−19×0.3=1.25×10−5B = \frac{p}{er} = \frac{6 \times 10^{-25}}{1.6 \times 10^{-19} \times 0.3} = 1.25 \times 10^{-5} T =125×10−7= 125 \times 10^{-7} T
  5. So the answer is 125.
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Q179MHT CET 2020 · 19 OCT · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

If the maximum kinetic energy of emitted electrons in photoelectric effect is 3.2×10−19 J3.2 \times 10^{-19}\text{ J} and the work-function for metal is 6.63×10−19 J6.63 \times 10^{-19}\text{ J}, then stopping potential and threshold wavelength respectively are

[Planck's constant, h=6.63×10−34 Jh = 6.63 \times 10^{-34}\text{ J}-s]

[Velocity of light, c=3×108 msc = 3 \times 10^{8}\,\frac{\text{m}}{\text{s}} ]

[Charge on electron =1.6×10−19 C= 1.6 \times 10^{-19}\text{ C} ]

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Answer: (c)

  1. V0=Kmaxe=3.2×10−191.6×10−19=2 VV_0 = \frac{K_{max}}{e} = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2\text{ V}
  2. λ0=hcϕ=6.63×10−34×3×1086.63×10−19=3×10−7 m\lambda_0 = \frac{hc}{\phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6.63 \times 10^{-19}} = 3 \times 10^{-7}\text{ m}
  3. λ0=3000 A˚\lambda_0 = 3000\,\mathring{A}
  4. So the answer is (c) 2 V, 3000 A˚\mathring{A}.
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Q180MHT CET 2020 · 16 OCT · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The light of wavelength λ\lambda incident on the surface of metal having work function ϕ\phi emits the electrons. The maximum velocity of electrons emitted is [ c=c = velocity of light, h=h = Planck's constant, m=m = mass of electron]

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Answer: (b)

  1. 12mvmax2=hcλ−ϕ=hc−λϕλ\frac{1}{2}mv_{max}^2 = \frac{hc}{\lambda} - \phi = \frac{hc - \lambda\phi}{\lambda}
  2. vmax=[2(hc−λϕ)mλ]1/2v_{max} = \left[\frac{2(hc - \lambda\phi)}{m\lambda}\right]^{1/2}
  3. So the answer is (b).
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Q181MHT CET 2020 · 16 OCT · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The maximum velocity of the photoelectron emitted by the metal surface is vv. Charge and mass of the photoelectron is denoted by ee and mm, respectively. The stopping potential in volt is

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Answer: (a)

  1. eV0=12mv2eV_0 = \frac{1}{2}mv^2
  2. V0=mv22e=v22(e/m)V_0 = \frac{mv^2}{2e} = \frac{v^2}{2(e/m)}
  3. So the answer is (a) v22(em)\frac{v^2}{2\left(\frac{e}{m}\right)}.
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Q182MHT CET 2020 · 16 OCT · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Energy of the incident photon on the metal surface is 3W3W and then 5W5W, where WW is the work function for that metal. The ratio of velocities of emitted photoelectrons is

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Answer: (c)

  1. v∝E−Wv \propto \sqrt{E - W}
  2. K1=3W−W=2WK_1 = 3W - W = 2W, K2=5W−W=4WK_2 = 5W - W = 4W
  3. v1v2=2W4W=12\frac{v_1}{v_2} = \sqrt{\frac{2W}{4W}} = \frac{1}{\sqrt{2}}
  4. So the answer is (c) 1:21 : \sqrt{2}.
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Q183JEE Main 2020 · 5 SEP · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

The surface of a metal is illuminated alternately with photons of energies E1 = 4 eV and E2 = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is _____.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. vmax∝E−ϕv_{max} \propto \sqrt{E - \phi}, so v12v22=4−ϕ2.5−ϕ=22=4\frac{v_1^2}{v_2^2} = \frac{4 - \phi}{2.5 - \phi} = 2^2 = 4.
  2. 4−ϕ=10−4ϕ⇒3ϕ=64 - \phi = 10 - 4\phi \Rightarrow 3\phi = 6
  3. Work function ϕ=2\phi = 2 eV.
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Q184JEE Main 2020 · 2 SEP · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When radiation of wavelength λ\lambda is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ\lambda, the stopping potential is V4\frac{V}{4}. If the threshold wavelength for the metallic surface is nλ\lambda then value of n will be __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 9

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=eVA - \phi = eV and A3−ϕ=eV4\frac{A}{3} - \phi = \frac{eV}{4}
  3. A−ϕ=4A3−4ϕ⇒3ϕ=A3⇒ϕ=A9A - \phi = \frac{4A}{3} - 4\phi \Rightarrow 3\phi = \frac{A}{3} \Rightarrow \phi = \frac{A}{9}
  4. λ0=9λ\lambda_0 = 9\lambda
  5. So n=9n = 9.
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Q185JEE Main 2020 · 3 SEP · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:

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Answer: (c)

  1. Photon energies (hc=1240hc = 1240 eV nm): 1240500=2.48\frac{1240}{500} = 2.48 eV, 1240200=6.2\frac{1240}{200} = 6.2 eV.
  2. 3(2.48−ϕ)=6.2−ϕ⇒7.44−3ϕ=6.2−ϕ3(2.48 - \phi) = 6.2 - \phi \Rightarrow 7.44 - 3\phi = 6.2 - \phi
  3. 2ϕ=1.24⇒ϕ≈0.622\phi = 1.24 \Rightarrow \phi \approx 0.62 eV
  4. So the answer is (c) 0.61 eV.
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Q186JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

Radiation, with wavelength 6561 A˚\text{\AA} falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to:

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Answer: (c)

  1. Photon energy =124006561=1.89= \frac{12400}{6561} = 1.89 eV.
  2. K=eB2r22mK = \frac{eB^2r^2}{2m} eV =1.6×10−19×(3×10−4)2(10−2)22×9.1×10−31≈0.79= \frac{1.6 \times 10^{-19} \times (3 \times 10^{-4})^2 (10^{-2})^2}{2 \times 9.1 \times 10^{-31}} \approx 0.79 eV
  3. ϕ=1.89−0.79=1.10\phi = 1.89 - 0.79 = 1.10 eV
  4. So the answer is (c) 1.1 eV.
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Q187JEE Main 2020 · 8 JAN · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA\lambda_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB\lambda_B = 2λA\lambda_A, then the work function of metal B is:

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Answer: (b)

  1. λ=h2mT⇒λ∝1T\lambda = \frac{h}{\sqrt{2mT}} \Rightarrow \lambda \propto \frac{1}{\sqrt{T}}; λB=2λA⇒TB=TA4\lambda_B = 2\lambda_A \Rightarrow T_B = \frac{T_A}{4}.
  2. TA−1.5=TA4⇒TA=2T_A - 1.5 = \frac{T_A}{4} \Rightarrow T_A = 2 eV, TB=0.5T_B = 0.5 eV
  3. ϕB=4.5−0.5=4\phi_B = 4.5 - 0.5 = 4 eV
  4. So the answer is (b) 4 eV.
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Q188NEET 2019Easy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work function of a photosensitive material is 4.0 eV4.0\text{ eV}. The longest wavelength of light that can cause photon emission from the substance is (approximately)

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Answer: (a)

  1. Longest wavelength = threshold: λ0=hcϕ=1240 eV nm4.0 eV\lambda_0 = \dfrac{hc}{\phi} = \dfrac{1240\text{ eV nm}}{4.0\text{ eV}}
  2. λ0=310 nm\lambda_0 = 310\text{ nm}
  3. So the answer is (a) 310 nm.
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Q189MHT CET 2019 · 2 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The maximum velocity of the photoelectron emitted by the metal surface is ' vv '. Charge and mass of the photoelectron is denoted by ' ee ' and ' mm ' respectively. The stopping potential in volt is

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Answer: (b)

  1. eV0=12mv2eV_0 = \frac{1}{2}mv^2
  2. V0=mv22e=v22(e/m)V_0 = \frac{mv^2}{2e} = \frac{v^2}{2(e/m)}
  3. So the answer is (b) v22(em)\frac{v^2}{2\left(\frac{e}{m}\right)}.
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Q190JEE Main 2019 · 10 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be: Given E (in eV) = 1237/λ\lambda (in nm)

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Answer: (d)

  1. Kmax=1237λ−1237λ0K_{max} = \frac{1237}{\lambda} - \frac{1237}{\lambda_0} (in eV, λ\lambda in nm)
  2. Kmax=1237(1260−1380)=1237×12098800K_{max} = 1237\left(\frac{1}{260} - \frac{1}{380}\right) = 1237 \times \frac{120}{98800}
  3. Kmax≈1.5 eVK_{max} \approx 1.5\text{ eV}, so the answer is (d) 1.5 eV.
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Q191MHT CET 2019 · 2 MAY · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When photons of energy hνh\nu fall on a metal plate of work function ' W0W_0 ', photoelectrons of maximum kinetic energy ' KK ' are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

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Answer: (b)

  1. K=hν−W0K = h\nu - W_0.
  2. New: K′=2hν−W0=hν+(hν−W0)=K+hνK' = 2h\nu - W_0 = h\nu + (h\nu - W_0) = K + h\nu
  3. So the answer is (b) K+hνK + h\nu.
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Q192JEE Main 2019 · 9 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The electric field of light wave is given as

E⃗=10−3cos⁡(2πx5×10−7−2π×6×1014t)x^ NC\vec{E} = 10^{-3}\cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14}t\right)\hat{x}\,\frac{\text{N}}{\text{C}}

This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is:

Given, E (in eV) = 12375/λ\lambda(inÅ)

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Answer: (b)

  1. From the wave, k=2πλk = \frac{2\pi}{\lambda} with λ=5×10−7\lambda = 5 \times 10^{-7} m =5000= 5000 Å.
  2. E=123755000=2.475E = \frac{12375}{5000} = 2.475 eV
  3. eV0=E−ϕ=2.475−2=0.475eV_0 = E - \phi = 2.475 - 2 = 0.475 eV
  4. So the answer is (b) 0.48 V.
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Q193MHT CET 2019 · 3 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When certain metal surface is illuminated with a light of wavelength λ\lambda, the stopping potential is VV, When the same surface is illuminated by light of wavelength 2λ2\lambda, the stopping potential is (V3)\left(\frac{V}{3}\right). The threshold wavelength for the surface is

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Answer: (c)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=eVA - \phi = eV and A2−ϕ=eV3\frac{A}{2} - \phi = \frac{eV}{3}
  3. A−ϕ=3A2−3ϕ⇒ϕ=A4A - \phi = \frac{3A}{2} - 3\phi \Rightarrow \phi = \frac{A}{4}
  4. λ0=4λ\lambda_0 = 4\lambda
  5. So the answer is (c) 4λ4\lambda.
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Q194JEE Main 2019 · 12 JAN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is – V0. The threshold frequency for photoelectric emission is:

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Answer: (c)

  1. Einstein: hν=ϕ+e∣Vs∣h\nu = \phi + e|V_s|
  2. hν−ϕ=eV02h\nu - \phi = \frac{eV_0}{2} and hν2−ϕ=eV0\frac{h\nu}{2} - \phi = eV_0
  3. hν2−ϕ=2(hν−ϕ)⇒ϕ=3hν2\frac{h\nu}{2} - \phi = 2(h\nu - \phi) \Rightarrow \phi = \frac{3h\nu}{2}
  4. ν0=3ν2\nu_0 = \frac{3\nu}{2}
  5. So the answer is (c) 3ν2\frac{3\nu}{2}.
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Q195JEE Main 2019 · 11 JAN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: ( hce\frac{hc}{e} = 1240 nm eV)

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Answer: (b)

  1. eVs=hcλ−ϕ⇒ΔV=hce(1λ1−1λ2)eV_s = \frac{hc}{\lambda} - \phi \Rightarrow \Delta V = \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)
  2. ΔV=1240(1300−1400)=12401200\Delta V = 1240\left(\frac{1}{300} - \frac{1}{400}\right) = \frac{1240}{1200}
  3. ΔV=1.03 V\Delta V = 1.03\text{ V}
  4. So the answer is (b) 1.0 V.
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Q196JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 ×\times 107)ct + sin(6.28 ×\times 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons?

(Take c = 3 ×\times 108 ms−1{}^{-1}, h = 6.6 ×\times 10−-34J-s)

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Answer: (d)

  1. ω=2πν\omega = 2\pi\nu: ν1=3.14×107×3×1082π=1.5×1015\nu_1 = \frac{3.14 \times 10^7 \times 3 \times 10^8}{2\pi} = 1.5 \times 10^{15} Hz and ν2=3×1015\nu_2 = 3 \times 10^{15} Hz.
  2. Maximum KE comes from the higher frequency: hν2=6.6×10−34×3×1015=1.98×10−18h\nu_2 = 6.6 \times 10^{-34} \times 3 \times 10^{15} = 1.98 \times 10^{-18} J ≈12.4\approx 12.4 eV.
  3. Kmax=12.4−4.7≈7.7K_{max} = 12.4 - 4.7 \approx 7.7 eV
  4. So the answer is (d) 7.72 eV.
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Q197JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Surface of certain metal is first illuminated with light of wavelength λ\lambda1 = 350 nm and then, by light of wavelength λ\lambda2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to:

(Energy of photon = 1240λ(in nm)\frac{1240}{\lambda(\text{in nm})}eV)

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Answer: (a)

  1. E1=1240350=3.54E_1 = \frac{1240}{350} = 3.54 eV, E2=1240540=2.30E_2 = \frac{1240}{540} = 2.30 eV.
  2. v1=2v2⇒K1=4K2v_1 = 2v_2 \Rightarrow K_1 = 4K_2: 3.54−ϕ=4(2.30−ϕ)3.54 - \phi = 4(2.30 - \phi)
  3. 3ϕ=9.18−3.54=5.64⇒ϕ≈1.883\phi = 9.18 - 3.54 = 5.64 \Rightarrow \phi \approx 1.88 eV
  4. So the answer is (a) 1.8.
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Q198NEET 2018Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

When the light of frequency 2v0v_0 (where v0v_0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5v0v_0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is

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Answer: (a)

  1. 12mv2=hν−hν0\frac{1}{2}mv^2 = h\nu - h\nu_0
  2. At 2ν02\nu_0: K1=hν0K_1 = h\nu_0; at 5ν05\nu_0: K2=4hν0K_2 = 4h\nu_0
  3. v1v2=14=12\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}
  4. So the answer is (a) 1 : 2.
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Q199NEET 2017 · (One or more correct)Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The photoelectric threshold wavelength of silver is 3250 ×\times 10−-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 ×\times 10−-10 m is {Given h = 4.14 ×\times 10−-15 eV s and c = 3 ×\times 108 m s−1{}^{-1} ]

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Answer: (a)

  1. hc=4.14×10−15×3×108=1.242×10−6hc = 4.14 \times 10^{-15} \times 3 \times 10^{8} = 1.242 \times 10^{-6} eV m
  2. E=1.242×10−62536×10−10=4.90E = \frac{1.242 \times 10^{-6}}{2536 \times 10^{-10}} = 4.90 eV, ϕ=1.242×10−63250×10−10=3.82\phi = \frac{1.242 \times 10^{-6}}{3250 \times 10^{-10}} = 3.82 eV
  3. K=1.08K = 1.08 eV =1.72×10−19= 1.72 \times 10^{-19} J
  4. v=2×1.72×10−199.1×10−31≈6.1×105v = \sqrt{\frac{2 \times 1.72 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 6.1 \times 10^{5} m/s
  5. So the answer is (a) ≈0.6×106\approx 0.6 \times 10^{6} m/s.
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Q200JEE Main 2017 · 8 APR · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be:

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Answer: (c)

  1. 12mv2=hn−ϕ\frac{1}{2}mv^2 = hn - \phi
  2. 12mv′2=3hn−ϕ=3(hn−ϕ)+2ϕ=32mv2+2ϕ\frac{1}{2}mv'^2 = 3hn - \phi = 3(hn - \phi) + 2\phi = \frac{3}{2}mv^2 + 2\phi
  3. So v′2>3v2v'^2 > 3v^2, i.e. v′>3 vv' > \sqrt{3}\,v.
  4. So the answer is (c).
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Q201NEET 2016 · PHASE 2Medium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

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Answer: (d)

  1. ϕ=E−Kmax=5−2=3\phi = E - K_{max} = 5 - 2 = 3 eV
  2. For 6 eV photons: Kmax=6−3=3K_{max} = 6 - 3 = 3 eV
  3. To stop them the anode must be 3 V below the cathode: VA=−3V_A = -3 V
  4. So the answer is (d) −3-3 V.
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Q202NEET 2016 · PHASE 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

When a metallic surface is illuminated with radiation of wavelength λ\lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 λ\lambda, the stopping potential is V4\frac{V}{4}. The threshold wavelength for the metallic surface is

Show answer and solution

Answer: (b)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=eVA - \phi = eV and A2−ϕ=eV4\frac{A}{2} - \phi = \frac{eV}{4}
  3. A−ϕ=4(A2−ϕ)=2A−4ϕ⇒ϕ=A3A - \phi = 4\left(\frac{A}{2} - \phi\right) = 2A - 4\phi \Rightarrow \phi = \frac{A}{3}
  4. λ0=hcϕ=3λ\lambda_0 = \frac{hc}{\phi} = 3\lambda
  5. So the answer is (b) 3λ3\lambda.
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Q203JEE Main 2016 · 10 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A photoelectric surface is illuminated successively by monochromatic light of wavelengths λ\lambda and λ2.\frac{\lambda}{2}. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is:

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Answer: (b)

  1. K=hcλ−ϕK = \frac{hc}{\lambda} - \phi; write A=hcλA = \frac{hc}{\lambda}.
  2. First: K1=A−ϕK_1 = A - \phi; second (λ/2\lambda/2): K2=2A−ϕK_2 = 2A - \phi
  3. 2A−ϕ=3(A−ϕ)⇒2ϕ=A2A - \phi = 3(A - \phi) \Rightarrow 2\phi = A
  4. So the answer is (b) ϕ=hc2λ\phi = \frac{hc}{2\lambda}.
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Q204JEE Main 2016Medium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Radiation of wavelength λ,\lambda, is incident on a photocell. The fastest emitted electron has speed v.v. If the wavelength is changed to 3λ4,\frac{3\lambda}{4}, the speed of the fastest emitted electron will be:

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Answer: (c)

  1. 12mv2=hcλ−ϕ\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi
  2. At 3λ4\frac{3\lambda}{4}: 12mv′2=43⋅hcλ−ϕ=43(hcλ−ϕ)+ϕ3\frac{1}{2}mv'^2 = \frac{4}{3}\cdot\frac{hc}{\lambda} - \phi = \frac{4}{3}\left(\frac{hc}{\lambda} - \phi\right) + \frac{\phi}{3}
  3. So 12mv′2>43⋅12mv2⇒v′>v(43)1/2\frac{1}{2}mv'^2 > \frac{4}{3}\cdot\frac{1}{2}mv^2 \Rightarrow v' > v\left(\frac{4}{3}\right)^{1/2}
  4. So the answer is (c) >v(43)1/2> v\left(\frac{4}{3}\right)^{1/2}.
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Q205JEE Main 2016 · 9 APR · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

When photons of wavelength λ1\lambda_1 are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ2\lambda_2 are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength λ3\lambda_3 is used then find the stopping potential for this case:

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Answer: (c)

  1. eV=hcλ1−ϕeV = \frac{hc}{\lambda_1} - \phi and 3eV=hcλ2−ϕ3eV = \frac{hc}{\lambda_2} - \phi.
  2. Subtract: eV=hc2(1λ2−1λ1)eV = \frac{hc}{2}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right), so ϕ=hcλ1−eV=hc(32λ1−12λ2)\phi = \frac{hc}{\lambda_1} - eV = hc\left(\frac{3}{2\lambda_1} - \frac{1}{2\lambda_2}\right).
  3. V3=1e(hcλ3−ϕ)=hce[1λ3+12λ2−32λ1]V_3 = \frac{1}{e}\left(\frac{hc}{\lambda_3} - \phi\right) = \frac{hc}{e}\left[\frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1}\right]
  4. So the answer is (c).
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Q206NEET 2015 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A photoelectric surface is illuminated successively by monochromatic light of wavelength λ\lambda and λ/2\lambda/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is

(h = Planck's constant, C = speed of light)

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Answer: (c)

  1. K=hcλ−ϕK = \frac{hc}{\lambda} - \phi; write A=hcλA = \frac{hc}{\lambda}.
  2. First: K1=A−ϕK_1 = A - \phi; second (λ/2\lambda/2): K2=2A−ϕK_2 = 2A - \phi
  3. 2A−ϕ=3(A−ϕ)⇒2ϕ=A2A - \phi = 3(A - \phi) \Rightarrow 2\phi = A
  4. So the answer is (c) ϕ=hc2λ\phi = \frac{hc}{2\lambda}.
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Q207NEET 2015 · AIPMT · CANCELLED PAPERMedium

Photoelectric Effect: Einstein's Equation · Stopping potential with two wavelengths

A certain metallic surface is illuminated with monochromatic light of wavelength, λ.\lambda. The stopping potential for photo-electric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2 λ\lambda, the stopping potential is V0. The threshold wavelength for this surface for photo- electric effect is

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Answer: (d)

  1. Einstein: hcλ=ϕ+eVs\frac{hc}{\lambda} = \phi + eV_s; write A=hcλA = \frac{hc}{\lambda}.
  2. A−ϕ=3eV0A - \phi = 3eV_0 and A2−ϕ=eV0\frac{A}{2} - \phi = eV_0
  3. Subtract: A2=2eV0⇒A=4eV0, ϕ=eV0=A4\frac{A}{2} = 2eV_0 \Rightarrow A = 4eV_0,\ \phi = eV_0 = \frac{A}{4}
  4. λ0=hcϕ=4λ\lambda_0 = \frac{hc}{\phi} = 4\lambda
  5. So the answer is (d) 4λ4\lambda.
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Q208NEET 2015 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

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Answer: (a)

  1. Photon energy =1240500=2.48= \frac{1240}{500} = 2.48 eV, so Kmax=2.48−2.28=0.20K_{max} = 2.48 - 2.28 = 0.20 eV.
  2. λ=h2mK=1.2270.20\lambda = \frac{h}{\sqrt{2mK}} = \frac{1.227}{\sqrt{0.20}} nm ≈2.8×10−9\approx 2.8 \times 10^{-9} m
  3. Electrons with less than maximum KE have larger λ\lambda, so λ≥2.8×10−9\lambda \geq 2.8 \times 10^{-9} m.
  4. So the answer is (a).
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Q209NEET 2014 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When the energy of the incident radiation is increased nby 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is

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Answer: (b)

  1. E=K+ϕE = K + \phi: E=0.5+ϕE = 0.5 + \phi and 1.2E=0.8+ϕ1.2E = 0.8 + \phi.
  2. 0.6+1.2ϕ=0.8+ϕ⇒0.2ϕ=0.20.6 + 1.2\phi = 0.8 + \phi \Rightarrow 0.2\phi = 0.2
  3. So the answer is (b) ϕ=1.0\phi = 1.0 eV.
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Q210NEET 2013Easy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

For photoelectric emission from certain metal the cutoff frequency is vv. If radiation of frequency 2ν\nu impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

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Answer: (a)

  1. 12mvmax2=h(2ν)−hν=hν\frac{1}{2}mv_{max}^2 = h(2\nu) - h\nu = h\nu
  2. vmax=2hνmv_{max} = \sqrt{\frac{2h\nu}{m}}
  3. So the answer is (a) 2hνm\sqrt{\frac{2h\nu}{m}}.
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Q211JEE Main 2013 · 25 APR · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

A copper ball of radius 1 cm and work function 4.47 eV is irradiated with ultraviolet radiation of wavelength 2500 Å. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is:

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Answer: (a)

  1. Photon energy =124002500=4.96= \frac{12400}{2500} = 4.96 eV, so the ball charges up to V=4.96−4.47=0.49V = 4.96 - 4.47 = 0.49 V.
  2. Q=Vrk=0.49×0.019×109≈5.4×10−13Q = \frac{Vr}{k} = \frac{0.49 \times 0.01}{9 \times 10^9} \approx 5.4 \times 10^{-13} C
  3. So the answer is (a) 5.5×10−135.5 \times 10^{-13} C.
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Q212NEET 2012 · AIPMT · MAINSEasy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is

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Answer: (b)

  1. Einstein's equation: 12mvmax2=E−ϕ\frac{1}{2}mv_{max}^2 = E - \phi, so vmax∝E−ϕv_{max} \propto \sqrt{E - \phi}.
  2. K1=1−0.5=0.5K_1 = 1 - 0.5 = 0.5 eV, K2=2.5−0.5=2K_2 = 2.5 - 0.5 = 2 eV
  3. v1v2=0.52=12\frac{v_1}{v_2} = \sqrt{\frac{0.5}{2}} = \frac{1}{2}
  4. So the answer is (b) 1:21:2.
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Q213NEET 2012 · AIPMT · PRELIMSMedium

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is

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Answer: (c)

  1. n=2→1n = 2 \to 1: E=13.6(1−14)=10.2E = 13.6\left(1 - \frac{1}{4}\right) = 10.2 eV.
  2. ϕ=10.2−3.57=6.63\phi = 10.2 - 3.57 = 6.63 eV =6.63×1.6×10−19= 6.63 \times 1.6 \times 10^{-19} J
  3. ν0=ϕh=6.63×1.6×10−196.63×10−34=1.6×1015\nu_0 = \frac{\phi}{h} = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.6 \times 10^{15} Hz
  4. So the answer is (c).
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Q214NEET 2011 · AIPMT · PRELIMSEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

Photoelectric emission occurs only when the incident light has more than a certain minimum

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Answer: (d)

  1. Each photon must have hν≥ϕh\nu \ge \phi; intensity or power only changes the number of photons.
  2. So emission needs a minimum frequency (threshold frequency ν0=ϕ/h\nu_0 = \phi/h).
  3. So the answer is (d) frequency.
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Q215NEET 2011 · AIPMT · MAINSEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The threshold frequency for a photosensitive metal is 3.3 ×\times 1014 Hz. If light of frequency 8.2 ×\times 1014 Hz is incident on this metal, the cut- off voltage for the photoelectron emission is nearly

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Answer: (b)

  1. eV0=h(ν−ν0)eV_0 = h(\nu - \nu_0)
  2. V0=6.63×10−34×(8.2−3.3)×10141.6×10−19=3.25×10−191.6×10−19V_0 = \frac{6.63 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}} = \frac{3.25 \times 10^{-19}}{1.6 \times 10^{-19}}
  3. V0≈2.0 VV_0 \approx 2.0\text{ V}, so the answer is (b) 2 V.
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Q216NEET 2011 · AIPMT · PRELIMSEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is

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Answer: (c)

  1. eV0=KmaxeV_0 = K_{max}; the work function is not needed.
  2. Kmax=0.5 eV⇒V0=0.5 VK_{max} = 0.5\text{ eV} \Rightarrow V_0 = 0.5\text{ V}
  3. So the answer is (c) 0.5 V.
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Q217NEET 2011 · AIPMT · PRELIMSEasy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be

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Answer: (b)

  1. 12mvmax2=E−ϕ\frac{1}{2}mv_{max}^2 = E - \phi, so vmax∝E−ϕv_{max} \propto \sqrt{E - \phi}.
  2. K1=1−0.5=0.5K_1 = 1 - 0.5 = 0.5 eV, K2=2.5−0.5=2K_2 = 2.5 - 0.5 = 2 eV
  3. v1v2=0.52=12\frac{v_1}{v_2} = \sqrt{\frac{0.5}{2}} = \frac{1}{2}
  4. So the answer is (b) 1:21:2.
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Q218NEET 2010 · AIPMT · PRELIMSEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work functions 5.01 eV, when ultraviolet light of 200 nm falls on it, must be

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Answer: (b)

  1. E=1240200=6.2 eVE = \frac{1240}{200} = 6.2\text{ eV}
  2. Kmax=6.2−5.01=1.19≈1.2 eVK_{max} = 6.2 - 5.01 = 1.19 \approx 1.2\text{ eV}, so ∣V0∣=1.2 V|V_0| = 1.2\text{ V}.
  3. The collector must be negative with respect to the plate to repel the electrons.
  4. So the answer is (b) −1.2-1.2 V.
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Q219NEET 2010 · AIPMT · MAINSMedium

Photoelectric Effect: Einstein's Equation · Photoelectrons in fields, hydrogen spectra & charged spheres

The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in nth state En = −13.6n2 eV\frac{-13.6}{n^2}\text{ eV})

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Answer: (d)

  1. n=3→1n = 3 \to 1: E=13.6(1−19)=12.09E = 13.6\left(1 - \frac{1}{9}\right) = 12.09 eV.
  2. eV0=12.09−5.1≈7eV_0 = 12.09 - 5.1 \approx 7 eV
  3. So the answer is (d) 7 V.
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Q220JEE Main 2009 · AIEEEEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The surface of a metal is illuminated with the light of 400 nm.400\text{ nm.} The kinetic energy of the ejected photoelectrons was found to be 1.68 eV.1.68\text{ eV.} The work function of the metal is: (hc=1240 eV.nm)(hc = 1240\text{ eV.nm})

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Answer: (a)

  1. E=1240400=3.1 eVE = \frac{1240}{400} = 3.1\text{ eV}
  2. ϕ=E−Kmax=3.10−1.68=1.42≈1.41 eV\phi = E - K_{max} = 3.10 - 1.68 = 1.42 \approx 1.41\text{ eV}
  3. So the answer is (a) 1.41 eV1.41\text{ eV}.
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Q221NEET 2008 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the

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Answer: (c)

  1. E=ϕ+eV0=6.2+5=11.2E = \phi + eV_0 = 6.2 + 5 = 11.2 eV
  2. λ=124011.2\lambda = \frac{1240}{11.2} nm ≈111\approx 111 nm
  3. This is below 400 nm and above the X-ray range, i.e. ultraviolet.
  4. So the answer is (c) Ultraviolet region.
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Q222JEE Main 2006 · AIEEEEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV6.2\text{ eV} and the stopping potential for a radiation incident on this surface is 5 V.5\text{ V.} The incident radiation lies in

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Answer: (a)

  1. E=ϕ+eV0=6.2+5=11.2 eVE = \phi + eV_0 = 6.2 + 5 = 11.2\text{ eV}
  2. λ=124011.2≈111 nm\lambda = \frac{1240}{11.2} \approx 111\text{ nm}
  3. This is below 400 nm and far above X-ray wavelengths (below about 10 nm), so it is ultra-violet.
  4. So the answer is (a) ultra-violet region.
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Q223NEET 2006 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Changing frequency or wavelength: KE ratios

When photons of energy hν\nu fall on an aluminimum plate (of work function E0), photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

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Answer: (a)

  1. K=hν−E0K = h\nu - E_0.
  2. New: K′=2hν−E0=hν+(hν−E0)=K+hνK' = 2h\nu - E_0 = h\nu + (h\nu - E_0) = K + h\nu
  3. So the answer is (a) K+hνK + h\nu.
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Q224NEET 2005 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work functions for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein's equation the metals which will emit photoelectrons for a radiation of wavelength 4100 A˚\mathring{A} is/are

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Answer: (b)

  1. Photon energy E=12400 eV A˚4100 A˚≈3.02 eVE = \dfrac{12400\text{ eV Å}}{4100\text{ Å}} \approx 3.02\text{ eV}
  2. 3.02>1.923.02 > 1.92 (A) and >2.0> 2.0 (B), but <5< 5 (C).
  3. So the answer is (b) A and B only.
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Q225NEET 2005 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

A photosensitive metallic surface has work function, hvv0. If photons of energy 2hv02hv_0 fall on this surface, the electrons come out with a maximum velocity of 4 ×\times 106 m/s. When the photon energy is increased to 5 hvv0, then maximum velocity of photoelectrons will be

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Answer: (c)

  1. 12mv2=E−hν0\frac{1}{2}mv^2 = E - h\nu_0
  2. K1=2hν0−hν0=hν0K_1 = 2h\nu_0 - h\nu_0 = h\nu_0, K2=5hν0−hν0=4hν0K_2 = 5h\nu_0 - h\nu_0 = 4h\nu_0
  3. v2=v14=8×106 m/sv_2 = v_1\sqrt{4} = 8\times10^{6}\text{ m/s}
  4. So the answer is (c) 8×1068\times10^{6} m/s.
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Q226JEE Main 2004 · AIEEEEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The work function of a substance is 4.0 eV.4.0\text{ eV}. The longest wavelength of light that can cause photo- electron emission from this substance is approximately.

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Answer: (a)

  1. Longest wavelength = threshold wavelength: λ0=hcϕ\lambda_0 = \frac{hc}{\phi}
  2. λ0=1240 eV nm4.0 eV=310 nm\lambda_0 = \frac{1240\text{ eV nm}}{4.0\text{ eV}} = 310\text{ nm}
  3. So the answer is (a) 310 nm310\text{ nm}.
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Q227JEE Main 2003 · AIEEEEasy

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

Two identical photo-cathodes receive light of frequencies f1f_1 and f2f_2. If the velocities of the photo electrons (of mass mm) coming out are respectively v1v_1 and v2,v_2, then

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Answer: (a)

  1. Same cathode, same ϕ\phi: 12mv12=hf1−ϕ\frac{1}{2}mv_1^2 = hf_1 - \phi and 12mv22=hf2−ϕ\frac{1}{2}mv_2^2 = hf_2 - \phi.
  2. Subtract: 12m(v12−v22)=h(f1−f2)\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)
  3. v12−v22=2hm(f1−f2)v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)
  4. So the answer is (a).
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Q228NEET 2002 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

When ultraviolet rays incident on metal plate then photoelectric effect does not occur, it occurs by incidence of

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Answer: (b)

  1. If UV photons do not have enough energy, a higher-frequency radiation is needed.
  2. X-rays have higher frequency than UV; infrared, radio and microwaves are lower.
  3. So the answer is (b) X-rays.
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Q229JEE Main 2002 · AIEEEEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

Sodium and copper have work functions 2.3 eV2.3\text{ eV} and 4.5 eV4.5\text{ eV} respectively. Then the ratio of the wavelengths is nearest to

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Answer: (c)

  1. Threshold wavelength λ0=hcϕ\lambda_0 = \frac{hc}{\phi}, so λ0∝1ϕ\lambda_0 \propto \frac{1}{\phi}.
  2. λNaλCu=ϕCuϕNa=4.52.3≈1.96≈2\frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{\phi_{Cu}}{\phi_{Na}} = \frac{4.5}{2.3} \approx 1.96 \approx 2
  3. So the answer is (c) 2:12 : 1.
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Q230NEET 1999 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The photoelectric work function for a metal surface is 4.125 eV4.125\text{ eV}. The cut-off wavelength for this surface is

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Answer: (b)

  1. λ0=hcϕ≈12375 eV A˚4.125 eV\lambda_0 = \dfrac{hc}{\phi} \approx \dfrac{12375\text{ eV Å}}{4.125\text{ eV}}
  2. λ0=3000 A˚\lambda_0 = 3000\text{ Å}
  3. So the answer is (b) 3000 Å.
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Q231NEET 1998 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

Light of wavelength 50005000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV1.9\text{ eV}. The kinetic energy of the photoelectron emitted will be

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Answer: (a)

  1. E=12400λ(A˚) eV=124005000=2.48 eVE = \frac{12400}{\lambda(\mathring{A})}\text{ eV} = \frac{12400}{5000} = 2.48\text{ eV}
  2. Kmax=E−ϕ=2.48−1.9=0.58 eVK_{max} = E - \phi = 2.48 - 1.9 = 0.58\text{ eV}
  3. So the answer is (a) 0.58 eV0.58\text{ eV}.
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Q232NEET 1998 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Maximum speed of photoelectrons

In a photo-emissive cell, with exciting wavelength λ\lambda, the fastest electron has speed vv. If the exciting wavelength is changed to 3λ/43\lambda/4, the speed of the fastest emitted electron will be

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Answer: (d)

  1. 12mv2=hcλ−ϕ\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi
  2. At 3λ4\frac{3\lambda}{4}: 12mv′2=43⋅hcλ−ϕ=43(hcλ−ϕ)+ϕ3\frac{1}{2}mv'^2 = \frac{4}{3}\cdot\frac{hc}{\lambda} - \phi = \frac{4}{3}\left(\frac{hc}{\lambda} - \phi\right) + \frac{\phi}{3}
  3. So 12mv′2>43⋅12mv2⇒v′>v(43)1/2\frac{1}{2}mv'^2 > \frac{4}{3}\cdot\frac{1}{2}mv^2 \Rightarrow v' > v\left(\frac{4}{3}\right)^{1/2}
  4. So the answer is (d) greater than v(4/3)1/2v(4/3)^{1/2}.
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Q233NEET 1993 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

When light of wavelength 300 nm (nanometer) falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?

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Answer: (b)

  1. Taking each wavelength as threshold, ϕ=hcλ0∝1λ0\phi = \dfrac{hc}{\lambda_0} \propto \dfrac{1}{\lambda_0}
  2. ϕ1ϕ2=600300=2\dfrac{\phi_1}{\phi_2} = \dfrac{600}{300} = 2
  3. So the answer is (b) 2:12:1.
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Q234NEET 1991 · AIPMTMedium

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

Photoelectric work function of a metal is 1 eV. Light of wavelength λ\lambda = 3000 Å falls on it. The photo electrons come out with a maximum velocity

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Answer: (d)

  1. E=124003000≈4.13E = \frac{12400}{3000} \approx 4.13 eV, so Kmax=4.13−1=3.13K_{max} = 4.13 - 1 = 3.13 eV ≈5.0×10−19\approx 5.0 \times 10^{-19} J
  2. v=2Km=2×5.0×10−199.1×10−31≈1.05×106v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 5.0 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 1.05 \times 10^{6} m/s
  3. So the answer is (d) 10610^{6} m/s.
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Q235NEET 1989 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Einstein's equation: direct numericals

Ultraviolet radiations of 6.2 eV falls on an aluminium surface. Kinetic energy of fastest electron emitted is (work function = 4.2 eV)

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Answer: (b)

  1. Kmax=E−ϕ=6.2−4.2=2 eVK_{max} = E - \phi = 6.2 - 4.2 = 2\text{ eV}
  2. Kmax=2×1.6×10−19=3.2×10−19 JK_{max} = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\text{ J}
  3. So the answer is (b) 3.2×10−19 J3.2 \times 10^{-19}\text{ J}.
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Q236NEET 1988 · AIPMTEasy

Photoelectric Effect: Einstein's Equation · Work function, threshold frequency & wavelength

The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of 5000 Å. Its work function is

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Answer: (a)

  1. ϕ=hcλ0=6.63×10−34×3×1085000×10−10\phi = \dfrac{hc}{\lambda_0} = \dfrac{6.63\times10^{-34}\times3\times10^{8}}{5000\times10^{-10}}
  2. ϕ≈3.98×10−19 J≈4×10−19 J\phi \approx 3.98\times10^{-19}\text{ J} \approx 4\times10^{-19}\text{ J}
  3. So the answer is (a) 4×10−19 J4\times10^{-19}\text{ J}.
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Q237MHT CET 2026 · 17 APR · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In photoelectric effect, keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased,

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Answer: (a)

  1. Higher intensity means more photons per second, so more photoelectrons per second.
  2. Maximum KE depends only on frequency, which is fixed.
  3. So the answer is (a) the photoelectric current increases.
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Q238MHT CET 2026 · 13 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Which one of the four graphs showing lines P, Q, R and S between maximum kinetic energy (E) and intensity of incident radiation (I) is correct?

Figure

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Answer: (a)

  1. Emax=hν−ϕE_{max} = h\nu - \phi does not depend on intensity II.
  2. So the E–I graph is a horizontal line above the I-axis, which is line P.
  3. So the answer is (a) P.
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Q239MHT CET 2026 · 13 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

A point source of light is used in a photoelectric effect. If the source is moved farther from the emitting metal, then the stopping potential

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Answer: (c)

  1. Moving a point source away reduces intensity (I∝1/r2I \propto 1/r^2) but not frequency.
  2. eV0=hν−ϕeV_0 = h\nu - \phi does not depend on intensity.
  3. So the answer is (c) will remain constant.
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Q240MHT CET 2026 · 15 APR · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

In photoelectric effect experiment, if the frequency of incident radiation (vv) is increased, keeping all other factors constant, the stopping potential, ( v>v0v > v_0) (v0v_0 = threshold frequency)

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Answer: (c)

  1. eV0=hν−ϕ⇒V0=heν−ϕeeV_0 = h\nu - \phi \Rightarrow V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}.
  2. V0V_0 increases linearly as ν\nu increases.
  3. So the answer is (c) increases.
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Q241MHT CET 2026 · 19 MAY · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

In an experiment on photoelectric effect, the stopping potential V0V_0 is plotted against the frequency ν\nu of the incident light. Considering ee as the electronic charge, the work function of the photoelectric surface is given by

Figure

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Answer: (c)

  1. V0=heν−ϕeV_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}: a straight line with slope h/eh/e and V0V_0-intercept −ϕ/e-\phi/e.
  2. The intercept on the V0V_0 axis is OBOB, so OB=ϕ/eOB = \phi/e (in volt), while OAOA is the threshold frequency.
  3. Hence ϕ=e×OB\phi = e \times OB in joules.
  4. So the answer is (c).
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Q242MHT CET 2026 · 19 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

According to Einstein's photoelectric equation, the graph of the kinetic energy of the emitted photoelectrons versus the frequency of incident radiation gives a straight line whose slope

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Answer: (c)

  1. Kmax=hν−ϕK_{max} = h\nu - \phi, so the slope of the KmaxK_{max} vs ν\nu line is hh.
  2. hh is a universal constant: it does not depend on the metal or on intensity.
  3. So the answer is (c).
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Q243MHT CET 2026 · 16 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

Using Einstein's photoelectric equation, the graphical representation between the kinetic energy (E) of emitted Photoelectrons and the frequency of incident radiation (ν\nu) is show correctly in figure

Figure

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Answer: (c)

  1. E=hν−ϕE = h\nu - \phi: a straight line of positive slope hh.
  2. E=0E = 0 at ν=ν0=ϕ/h>0\nu = \nu_0 = \phi/h > 0, so the line starts on the ν\nu axis at the threshold frequency, not at the origin.
  3. Graph C shows this.
  4. So the answer is (c) C.
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Q244JEE Main 2026 · 4 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The graph shows variation of stopping potential VoV_\mathrm{o} with the frequency vv of the incident radiation for three photosensitive metals X1,X2X_1, X_2 and X3X_3. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?

Figure

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Answer: (a)

  1. eV0=hν−hν0eV_0 = h\nu - h\nu_0; the axis intercept is the threshold frequency.
  2. Thresholds: X1=1.0X_1 = 1.0, X2=1.5X_2 = 1.5, X3=2.0X_3 = 2.0 (in 101410^{14} Hz), so X1X_1 has the smallest work function.
  3. Same wavelength means same hνh\nu, so KEmax=hν−ϕKE_{max} = h\nu - \phi is largest for the smallest ϕ\phi.
  4. So the answer is (a) X1X_1.
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Q245MHT CET 2026 · 20 MAY · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Light of incident frequency three times the threshold frequency is incident on a photo sensitive material. If the incident frequency is made (13)rd\left(\frac{1}{3}\right)^{rd} and intensity is doubled then photoelectric current will

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Answer: (a)

  1. Initially ν=3ν0\nu = 3\nu_0; new frequency ν′=3ν03=ν0\nu' = \frac{3\nu_0}{3} = \nu_0.
  2. At the threshold, KEmax=hν0−ϕ=0KE_{max} = h\nu_0 - \phi = 0: electrons are not emitted with any energy, so no photocurrent flows.
  3. Increasing intensity cannot help when frequency is not above threshold.
  4. So the answer is (a) be zero.
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Q246NEET 2026 · RE-NEETMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time tt, then variation of the photocurrent II and magnitude of the stopping potential ∣V∣|V| with time is best represented by:

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Answer: (b)

  1. Photocurrent ∝\propto number of photons per second ∝\propto power, so II falls linearly with tt.
  2. Stopping potential e∣V∣=hν−ϕe|V| = h\nu - \phi depends only on frequency, which is unchanged, so ∣V∣|V| stays constant.
  3. So the answer is (b).
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Q247MHT CET 2025 · 22 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

From photoelectric effect experiment, select the correct statement.

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Answer: (d)

  1. (a) False: wave theory cannot explain threshold frequency or instant emission.
  2. (b) False: maximum KE depends on frequency, not intensity.
  3. (c) False: stopping potential depends on frequency and work function.
  4. (d) True: more photons per second give more electrons, so saturation current rises with intensity.
  5. So the answer is (d).
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Q248MHT CET 2025 · 19 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In a photoelectric experiment, if the intensity of incident light is doubled and the frequency is kept slightly greater than threshold frequency, then the saturation photoelectric current

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Answer: (c)

  1. Above threshold, saturation current ∝\propto number of photons per second ∝\propto intensity.
  2. Doubling intensity doubles the current.
  3. So the answer is (c) is doubled.
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Q249JEE Main 2025 · 4 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R\mathbf{R}

Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases.

Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency.

In the light of the above statements, choose the correct answer from the options given below

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Answer: (b)

  1. A: eV0=hν−ϕeV_0 = h\nu - \phi, so stopping potential does not depend on intensity. A is false.
  2. R: more intensity means more photons per second, so more electrons per second (for ν>ν0\nu > \nu_0). R is true.
  3. So the answer is (b) A is false but R is true.
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Q250JEE Main 2025 · 29 JAN · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.

Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Answer: (b)

  1. A: a sufficiently negative (retarding) potential stops even the fastest electrons, so the photocurrent becomes zero. A is true.
  2. R: V0=heν−ϕeV_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}, which is linear in ν\nu. R is true.
  3. R only describes how V0V_0 varies with frequency; it does not explain why emission can be suppressed.
  4. So the answer is (b).
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Q251JEE Main 2025 · 29 JAN · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

In an experiment with photoelectric effect, the stopping potential,

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Answer: (a)

  1. eV0=Kmax⇒V0=1eKmaxeV_0 = K_{max} \Rightarrow V_0 = \dfrac{1}{e}K_{max}.
  2. V0V_0 does not depend on intensity, and it decreases as wavelength increases (Kmax=hc/λ−ϕK_{max} = hc/\lambda - \phi).
  3. So the answer is (a).
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Q252JEE Main 2025 · 24 JAN · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

In photoelectric effect, the stopping potential (V0) v/s(\mathrm{V}_0)\text{ v/s} frequency (v)(v) curve is plotted. ( h is the Planck's constant and ϕ0\phi_0 is work function of metal)

(A) V0 v/s v\mathrm{V}_0\text{ v/s } v is linear.

(B) The slope of V0 v/s v\mathrm{V}_0\text{ v/s } v curve =ϕ0h= \frac{\phi_0}{\mathrm{h}}

(C) h constant is related to the slope of V0 v/s v\mathrm{V}_0\text{ v/s } v line.

(D) The value of electric charge of electron is not required to determine h using the V0 v/s v\mathrm{V}_0\text{ v/s } v curve. (E) The work function can be estimated without knowing the value of hh.

Choose the correct answer from the options given below:

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Answer: (a)

  1. V0=heν−ϕ0eV_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}.
  2. (A) True: V0V_0 is linear in ν\nu. (B) False: slope is h/eh/e, not ϕ0/h\phi_0/h. (C) True: slope =h/e= h/e.
  3. (D) False: h=e×slopeh = e \times \text{slope}, so ee is needed.
  4. (E) True: the V0V_0-axis intercept is −ϕ0/e-\phi_0/e, giving ϕ0\phi_0 in eV directly without hh.
  5. So the answer is (a) (A), (C) and (E) only.
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Q253MHT CET 2025 · 26 APR · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Light of incident frequency 3 times the threshold frequency is incident on a photosensitive material. If the incident frequency is made (14)\left(\frac{1}{4}\right)th and intensity is tripled then the photoelectric current will

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Answer: (d)

  1. ν=3ν0\nu = 3\nu_0; new frequency ν′=3ν04=0.75ν0<ν0\nu' = \frac{3\nu_0}{4} = 0.75\nu_0 < \nu_0.
  2. Below threshold no photoelectrons are emitted, whatever the intensity.
  3. So the answer is (d) be zero.
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Q254NEET 2025Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?

FigureFigure
FigureFigure
Show answer and solution

Answer: (c)

  1. Above threshold, photocurrent ∝\propto intensity (number of photons): graph A is correct, B is wrong.
  2. Photocurrent does not increase with frequency; frequency only changes the kinetic energy. So C and D are wrong.
  3. So the answer is (c) A only.
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Q255MHT CET 2025 · 25 APR · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

If the frequency of incident light in a photoelectric experiment is doubled, then stopping potential will

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Answer: (c)

  1. eV0=hν−ϕeV_0 = h\nu - \phi.
  2. New: eV0′=2hν−ϕ=2(hν−ϕ)+ϕ=2eV0+ϕeV_0' = 2h\nu - \phi = 2(h\nu - \phi) + \phi = 2eV_0 + \phi.
  3. V0′=2V0+ϕ/e>2V0V_0' = 2V_0 + \phi/e > 2V_0.
  4. So the answer is (c) become more than double.
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Q256MHT CET 2025 · 26 APR · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

The graph shows the variation of photocurrent with anode potential for four different radiations. Let Ia,Ib,Ic\mathrm{I_a}, \mathrm{I_b}, \mathrm{I_c} and Id\mathrm{I_d} are intensities and fa,fb,fc\mathrm{f_a}, \mathrm{f_b}, \mathrm{f_c} and fd\mathrm{f_d} be the frequencies for the curves a,b,c\mathrm{a, b, c} and d respectively, then

Figure

Show answer and solution

Answer: (a)

  1. Stopping potential (where the curve meets the axis) depends on frequency; saturation current depends on intensity.
  2. Stopping potential: aa smallest, bb and cc equal, dd largest, so fa<fb=fc<fdf_a < f_b = f_c < f_d.
  3. Saturation: cc and dd reach the same level, so Ic=IdI_c = I_d.
  4. So the answer is (a) fb>faf_b > f_a, fb=fcf_b = f_c, Ic=IdI_c = I_d.
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Q257MHT CET 2025 · 23 APR · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

Graph shows variation of stopping potential with frequency of incident radiation on a metal plate. The value of Planck's constant is [ e=\mathrm{e} = charge on photoelectron]

Figure

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Answer: (c)

  1. eV=hν−ϕeV = h\nu - \phi, so slope of VV vs ν\nu =he= \frac{h}{e}.
  2. From points C and D: slope =V2−V1ν2−ν1= \frac{V_2 - V_1}{\nu_2 - \nu_1}.
  3. h=e(V2−V1)ν2−ν1h = \frac{e(V_2 - V_1)}{\nu_2 - \nu_1}
  4. So the answer is (c).
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Q258MHT CET 2025 · 23 APR · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

For two different photosensitive materials having work function ϕ\phi and 2ϕ2\phi respectively, are illuminated with light of sufficient energy to emit electrons. If the graph of stopping potential versus frequency is drawn, for these two different photosensitive materials, the ratio of slope of graph for these two materials is

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Answer: (a)

  1. V0=heν−ϕeV_0 = \frac{h}{e}\nu - \frac{\phi}{e}; slope =h/e= h/e for any material.
  2. Work function changes only the intercept.
  3. So the answer is (a) 1:11 : 1.
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Q259MHT CET 2024 · 10 MAY · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In photoelectric effect, the photocurrent

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Answer: (c)

  1. Each photon ejects at most one electron, so photocurrent ∝\propto number of photons per second ∝\propto intensity.
  2. Frequency (above ν0\nu_0) only changes the energy of each electron, Kmax=hν−ϕK_{max} = h\nu - \phi, not their number.
  3. So the answer is (c): photocurrent depends only on intensity.
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Q260MHT CET 2024 · 2 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

The number of photoelectrons emitted for light of frequency vv (higher than the threshold frequency (v0)(v_0) is proportional to

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Answer: (b)

  1. One photon ejects one electron, so number of photoelectrons ∝\propto number of photons falling per second.
  2. For a fixed frequency, number of photons per second ∝\propto intensity II.
  3. So the answer is (b) intensity of light.
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Q261MHT CET 2024 · 16 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, then the stopping potential will

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Answer: (c)

  1. Moving the source away only lowers intensity (I∝1/r2I \propto 1/r^2); frequency is the same.
  2. V0=(hν−ϕ)/eV_0 = (h\nu - \phi)/e is independent of intensity.
  3. So the answer is (c) remain constant.
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Q262MHT CET 2024 · 11 MAY · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

If the frequency of incident radiation (ν)(\nu) is increased, keeping other factors constant, the stopping potential ( ν>ν0\nu > \nu_0, threshold frequency)

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Answer: (a)

  1. V0=heν−ϕeV_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}, a straight line of positive slope h/eh/e.
  2. Increasing ν\nu increases V0V_0.
  3. So the answer is (a) increases.
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Q263JEE Main 2024 · 5 APR · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

Which of the following statement is not true about stopping potential (V0)(\mathrm{V}_0)?

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Answer: (c)

  1. eV0=hν−ϕeV_0 = h\nu - \phi: V0V_0 depends on frequency (a true) and on the emitter's work function (d true).
  2. V0=Kmax/eV_0 = K_{max}/e (b true).
  3. V0V_0 does not depend on intensity, so (c) is the false statement.
  4. So the answer is (c).
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Q264MHT CET 2024 · 9 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

The figure shows the variation of photocurrent with anode potential for four different radiations. Let fa,fb,fcf_a, f_b, f_c and fdf_d be the frequencies for the curves a,b,ca, b, c and dd respectively

Figure

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Answer: (b)

  1. Stopping potential V0=(hν−ϕ)/eV_0 = (h\nu - \phi)/e increases with frequency for the same metal.
  2. From the graph, the stopping potentials are in the order V01<V02<V03<V04V_0^1 < V_0^2 < V_0^3 < V_0^4 for curves a,b,c,da, b, c, d.
  3. All curves have the same saturation current, so only the frequency differs.
  4. So fa<fb<fc<fdf_a < f_b < f_c < f_d: the answer is (b).
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Q265JEE Main 2024 · 4 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light (I1<I2)(\mathrm{I_1} < \mathrm{I_2}) of same wavelengths:

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Answer: (c)

  1. Same wavelength (frequency) means the same stopping potential, so both curves meet the V axis at the same −V0-V_0.
  2. Saturation current ∝\propto intensity, so the I2I_2 curve is higher since I2>I1I_2 > I_1.
  3. Only graph (c) shows a common −V0-V_0 with I2I_2 above I1I_1.
  4. So the answer is (c).
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Q266MHT CET 2024 · 15 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces A and B. The graph shows that the work function of A is

Figure

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Answer: (b)

  1. V0=he(ν−ν0)V_0 = \dfrac{h}{e}(\nu - \nu_0); each line meets the ν\nu axis at its threshold frequency ν0\nu_0.
  2. From the graph, ν0A<ν0B\nu_{0A} < \nu_{0B}.
  3. ϕ=hν0\phi = h\nu_0, so ϕA<ϕB\phi_A < \phi_B.
  4. So the answer is (b) smaller than that of B.
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Q267MHT CET 2024 · 11 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

In case of photoelectric effect, the graph of measured stopping potential (V0)(\mathrm{V}_0) against frequency ' vv ' of incident light is a straight line. The slope of this line multiplied by the charge of electron (e) gives

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Answer: (b)

  1. V0=heν−ϕeV_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}, so slope =h/e= h/e.
  2. Slope ×e=h\times e = h.
  3. So the answer is (b) the Planck's constant.
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Q268MHT CET 2024 · 9 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

Using Einstein's photoelectric equation, the graph between kinetic energy of emitted photoelectrons and the frequency of incident radiation is shown correctly by graph

Figure

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Answer: (a)

  1. E=hν−ϕE = h\nu - \phi: a straight line with positive slope hh.
  2. E=0E = 0 at ν=ν0\nu = \nu_0, and no electrons are emitted below ν0\nu_0, so the line starts from a point on the ν\nu axis.
  3. Graph 1 shows this.
  4. So the answer is (a) 1.
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Q269MHT CET 2024 · 3 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The graph of stopping potential ' Vs\mathrm{V_s} ' against frequency ' vv ' of incident radiation is plotted for two different metals ' X ' and ' Y ' as shown in graph. ' ϕx\phi_x ' and ' ϕy\phi_y ' are work functions of ' xx ' and ' YY ' respectively then

Figure

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Answer: (b)

  1. Vs=he(ν−ν0)V_s = \dfrac{h}{e}(\nu - \nu_0); each line cuts the ν\nu axis at its threshold frequency.
  2. From the graph, X cuts at ν0\nu_0 and Y at ν0′>ν0\nu_0' > \nu_0.
  3. ϕ=hν0\phi = h\nu_0, so ϕx<ϕy\phi_x < \phi_y.
  4. So the answer is (b).
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Q270JEE Main 2024 · 5 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

Given below are two statements:

Figure

Statement I: Figure shows the variation of stopping potential with frequency (v)(v) for the two photosensitive materials M1M_1 and M2M_2. The slope gives value of he\frac{h}{e}, where hh is Planck's constant, e is the charge of electron.

Statement II: M2\mathrm{M}_2 will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (a)

  1. V0=heν−ϕeV_0 = \frac{h}{e}\nu - \frac{\phi}{e}, so the slope of each line is h/eh/e: Statement I is correct.
  2. M2M_2 has the higher threshold frequency, so larger work function ϕ\phi.
  3. For the same ν\nu, KEmax=hν−ϕKE_{max} = h\nu - \phi is smaller for M2M_2: Statement II is incorrect.
  4. So the answer is (a).
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Q271JEE Main 2024 · 30 JAN · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

For the photoelectric effect, the maximum kinetic energy (Ek)(E_k) of the photoelectrons is plotted against the frequency (v)(v) of the incident photons as shown in figure. The slope of the graph gives

Figure

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Answer: (a)

  1. Ek=hν−ϕE_k = h\nu - \phi, so dEkdν=h\frac{dE_k}{d\nu} = h.
  2. Slope tan⁡θ=h\tan\theta = h.
  3. So the answer is (a) Planck's constant.
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Q272JEE Main 2024 · 31 JAN · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:

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Answer: (c)

  1. ν=1.5ν0\nu = 1.5\nu_0; halved: ν′=0.75ν0<ν0\nu' = 0.75\nu_0 < \nu_0.
  2. Below threshold no electrons are emitted, even if intensity is doubled.
  3. So the answer is (c) zero.
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Q273MHT CET 2024 · 16 MAY · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

The figure shows the variation of photocurrent with anode potential for four different radiations. Let Ia,Ib,Ic\mathrm{I_a}, \mathrm{I_b}, \mathrm{I_c} and Id\mathrm{I_d} be the intensities for the curves a,b,ca, b, c and dd respectively [fa,fb,fc[f_a, f_b, f_c and fdf_d are frequencies respectively]

Figure

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Answer: (c)

  1. All four curves meet the axis at the same retarding potential, so the frequencies are equal: fa=fb=fc=fdf_a = f_b = f_c = f_d.
  2. Saturation currents rise from aa to dd, and saturation current ∝\propto intensity: Ia<Ib<Ic<IdI_a < I_b < I_c < I_d.
  3. So the answer is (c).
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Q274MHT CET 2023 · 10 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

The maximum kinetic energy of the photoelectrons varies

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Answer: (d)

  1. Einstein's equation: Kmax=hν−ϕ0K_{max} = h\nu - \phi_0.
  2. This is a straight line in ν\nu; intensity does not appear in it.
  3. So the answer is (d): varies linearly with frequency and is independent of intensity.
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Q275NEET 2023Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

The maximum kinetic energy of the emitted photoelectrons in photoelectric effect is independent of:

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Answer: (b)

  1. Kmax=hν−ϕ=hcλ−ϕK_{max} = h\nu - \phi = \dfrac{hc}{\lambda} - \phi.
  2. It depends on ϕ\phi, ν\nu and λ\lambda, but intensity does not appear.
  3. So the answer is (b) intensity of incident radiation.
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Q276JEE Main 2023 · 8 APR · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In photo electric effect

A. The photocurrent is proportional to the intensity of the incident radiation

B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.

C. Max. K.E with which photoelectrons are emitted depends on the frequency of incident light. D. The emission of photoelectrons require a minimum threshold intensity of incident radiation. E. Max. K.E of the photoelectrons is independent of the frequency of the incident light. Choose the correct answer from the options given below:

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Answer: (d)

  1. A true: photocurrent ∝\propto intensity.
  2. B false and C true: Kmax=hν−ϕK_{max} = h\nu - \phi depends on frequency, not intensity.
  3. D false: there is a threshold frequency, not a threshold intensity. E false (contradicts C).
  4. So the answer is (d) A and C only.
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Q277JEE Main 2023 · 24 JAN · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

From the photoelectric effect experiment, following observations are made. Identify which of these are correct.

A. The stopping potential depends only on the work function of the metal.

B. The saturation current increases as the intensity of incident light increases.

C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light. D. Photoelectric effect can be explained using wave theory of light.

Choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

  1. A false: V0=(hν−ϕ)/eV_0 = (h\nu - \phi)/e depends on frequency as well as work function.
  2. B true: saturation current ∝\propto intensity.
  3. C false: KmaxK_{max} is independent of intensity. D false: wave theory cannot explain threshold frequency or instant emission.
  4. So the answer is (d) B only.
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Q278JEE Main 2023 · 25 JAN · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

Given below are two statements:

Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source.

Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.

In the light of above statements, choose the most appropriate answer from the options given below

Show answer and solution

Answer: (c)

  1. Statement I: eV0=hν−ϕeV_0 = h\nu - \phi has no intensity (power) term, so it is correct.
  2. Statement II: Kmax=hcλ−ϕK_{max} = \dfrac{hc}{\lambda} - \phi depends on wavelength, so it is correct.
  3. So the answer is (c) both statements are correct.
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Q279JEE Main 2023 · 6 APR · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The work functions of Aluminium and Gold are 4.1 eV4.1\text{ eV} and 5.1 eV5.1\text{ eV} respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

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Answer: (c)

  1. V0=heν−ϕeV_0 = \frac{h}{e}\nu - \frac{\phi}{e}, so slope =h/e= h/e for every metal.
  2. The work function only shifts the line, it does not change the slope.
  3. Ratio of slopes =1= 1.
  4. So the answer is (c) 1.
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Q280MHT CET 2023 · 13 MAY · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photocurrent becomes

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Answer: (d)

  1. ν=1.5ν0\nu = 1.5\nu_0; halved: ν′=0.75ν0<ν0\nu' = 0.75\nu_0 < \nu_0.
  2. Below threshold no emission, even with double intensity.
  3. So the answer is (d) zero.
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Q281JEE Main 2023 · 13 APR · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Given below are two statements:

Statement I: Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.

Statement II: Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.

In the light of above statements, choose the correct answer form the options given below

Show answer and solution

Answer: (b)

  1. Statement I: UV has the highest frequency of the three, so it is most effective. True.
  2. Statement II: KEmax=hν−ϕKE_{max} = h\nu - \phi increases linearly with ν\nu, not inversely. False.
  3. So the answer is (b).
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Q282JEE Main 2023 · 10 APR · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The variation of stopping potential (V0)(\mathrm{V}_0) as a function of the frequency (v)(v) of the incident light for a metal is shown in figure. The work function of the surface is

Figure

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Answer: (d)

  1. The line cuts the ν\nu-axis at ν0=5×1014\nu_0 = 5 \times 10^{14} Hz.
  2. ϕ=hν0=6.63×10−34×5×1014=3.32×10−19\phi = h\nu_0 = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.32 \times 10^{-19} J
  3. ϕ=3.32×10−191.6×10−19≈2.07\phi = \frac{3.32 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.07 eV
  4. So the answer is (d) 2.07 eV.
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Q283JEE Main 2022 · 27 JUL · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

With reference to the observations in photo-electric effect, identify the correct statements from below:

(A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.

(B) The value of saturation current increases on moving the source of light away from the metal surface.

(C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light.

(D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.

(E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves.

Choose the correct answer from the options given below:

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Answer: (b)

  1. (A) True: 12mvmax2=hν−ϕ\frac{1}{2}mv_{max}^2 = h\nu - \phi, so vmax2v_{max}^2 is linear in ν\nu.
  2. (B) False: moving the source away lowers intensity, so saturation current decreases. (C) False: KmaxK_{max} does not depend on source power.
  3. (D) False: the photon (particle) picture explains instant emission. (E) True: wave theory cannot explain a threshold.
  4. So the answer is (b) (A) and (E) only.
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Q284MHT CET 2021 · 24 SEP · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In photoelectric experiment keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased,

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Answer: (c)

  1. More intensity at fixed frequency means more photons per second, so more electrons per second.
  2. Kmax=hν−ϕK_{max} = h\nu - \phi is unchanged because ν\nu is unchanged.
  3. So the answer is (c) photoelectric current increases.
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Q285MHT CET 2021 · 23 SEP · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In photoelectric effect, the photo current

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Answer: (a)

  1. Photocurrent ∝\propto number of photons per second ∝\propto intensity.
  2. Frequency above threshold only sets the KE of each electron, Kmax=hν−ϕK_{max} = h\nu - \phi.
  3. So the answer is (a).
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Q286JEE Main 2021 · 27 AUG · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

In a photoelectric experiment, increasing the intensity of incident light:

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Answer: (d)

  1. Intensity I=nhνI = n h\nu per unit area per second; at fixed ν\nu, more intensity means more photons nn.
  2. Frequency is unchanged, so Kmax=hν−ϕK_{max} = h\nu - \phi is unchanged.
  3. So the answer is (d).
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Q287MHT CET 2021 · 21 SEP · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

On a photosensitive surface, if the intensity of incident radiation is increased, the stopping potential

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Answer: (d)

  1. eV0=hν−ϕeV_0 = h\nu - \phi contains no intensity term.
  2. Higher intensity only gives more electrons, each with the same KmaxK_{max}.
  3. So the answer is (d) remains unchanged.
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Q288JEE Main 2021 · 16 MAR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation:

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Answer: (b)

  1. eV0=hν−ϕeV_0 = h\nu - \phi.
  2. Of the properties of the radiation, only frequency enters; phase, amplitude and intensity do not.
  3. So the answer is (b) frequency.
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Q289MHT CET 2021 · 21 SEP · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Light of frequency two times the threshold frequency is incident on photosensitive material. If the incident frequency is made (13)\left(\frac{1}{3}\right)rd and intensity is doubled, then the photoelectric current will

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Answer: (c)

  1. ν=2ν0\nu = 2\nu_0; new frequency ν′=2ν03<ν0\nu' = \frac{2\nu_0}{3} < \nu_0.
  2. Below threshold no emission, regardless of intensity.
  3. So the answer is (c) be zero.
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Q290MHT CET 2021 · 24 SEP · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

In a photoelectric experiment, a graph of maximum kinetic energy (KE(\mathrm{KE}max )) against the frequency of incident radiation (v) is plotted. If A\mathrm{A} and B\mathrm{B} are the intercepts on the X\mathrm{X} and Y\mathrm{Y} axis respectively then the Planck's constant is given by

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Answer: (b)

A-BKEνslope = h
  1. KEmax=hν−ϕKE_{max} = h\nu - \phi.
  2. X-intercept (KE=0KE = 0): A=ν0=ϕ/hA = \nu_0 = \phi/h. Y-intercept magnitude: B=ϕB = \phi.
  3. Slope h=ϕν0=BAh = \frac{\phi}{\nu_0} = \frac{B}{A}.
  4. So the answer is (b) BA\frac{B}{A}.
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Q291MHT CET 2020 · 19 OCT · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The graph of stopping potential VsV_s against frequency vv of incident radiation is plotted for two different metals PP and QQ as shown in the graph. ϕp\phi_p and ϕQ\phi_Q are work-functions of PP and QQ respectively, then

Figure

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Answer: (b)

  1. eVs=hν−ϕeV_s = h\nu - \phi, so the intercept on the ν\nu-axis is the threshold frequency ν0=ϕ/h\nu_0 = \phi/h.
  2. Line PP cuts the axis at a lower frequency than line QQ: ν0P<ν0Q\nu_{0P} < \nu_{0Q}.
  3. ϕ=hν0⇒ϕP<ϕQ\phi = h\nu_0 \Rightarrow \phi_P < \phi_Q
  4. So the answer is (b) ϕP<ϕQ\phi_P < \phi_Q.
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Q292MHT CET 2020 · 16 OCT · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The graph of kinetic energy against the frequency vv of incident light is as shown in the figure. The slope of the graph and intercept on XX-axis respectively are

Figure

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Answer: (b)

  1. Einstein's equation: KE=hν−ϕ0KE = h\nu - \phi_0, a straight line y=mx+cy = mx + c.
  2. Slope =h= h (Planck's constant).
  3. At KE=0KE = 0: ν=ϕ0/h=ν0\nu = \phi_0/h = \nu_0, the threshold frequency.
  4. So the answer is (b) Planck's constant, threshold frequency.
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Q293NEET 2020 · PHASE 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

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Answer: (c)

  1. ν=1.5ν0\nu = 1.5\nu_0; halved: ν′=0.75ν0<ν0\nu' = 0.75\nu_0 < \nu_0.
  2. Below threshold no photoelectrons, whatever the intensity.
  3. So the answer is (c) zero.
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Q294JEE Main 2020 · 4 SEP · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

In a photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:

Figure

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Answer: (b)

  1. eV=hcλ−ϕeV = \frac{hc}{\lambda} - \phi: slope hc/ehc/e and intercept depend only on constants and the metal.
  2. Intensity changes only the number of photoelectrons (current), not VV.
  3. So the answer is (b) the graph does not change.
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Q295JEE Main 2020 · 4 SEP · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

Given figure shows few data points in a photoelectric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is:

(Planck's constant h = 6.62 × 10–34 J.s)

Figure

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Answer: (b)

  1. Minimum energy = work function =hν0= h\nu_0, where ν0\nu_0 is where Vstop=0V_{stop} = 0.
  2. The data point B (5.5,0)(5.5, 0) gives ν0=5.5×1014\nu_0 = 5.5 \times 10^{14} Hz.
  3. ϕ=6.62×10−34×5.5×1014=3.64×10−19\phi = 6.62 \times 10^{-34} \times 5.5 \times 10^{14} = 3.64 \times 10^{-19} J
  4. ϕ=3.64×10−191.6×10−19≈2.27\phi = \frac{3.64 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.27 eV
  5. So the answer is (b) 2.27 eV.
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Q296JEE Main 2020 · 7 JAN · SHIFT 1Difficult

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A beam of electromagnetic radiation of intensity 6.4 × 10–5 W/cm2 is comprised of wavelength, λ\lambda = 310 nm. It falls normally on a metal (work function ϕ\phi = 2eV) of surface area of 1 cm2. If one in 103 photons ejects an electron, total number of electrons ejected in 1 s is 10x. (hc = 1240 eVnm, 1eV = 1.6 × 10–19 J), then x is _____.

Numerical answer type. Work it out, then open the answer.

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Answer: 11

  1. Power on 1 cm2^2: P=6.4×10−5P = 6.4\times10^{-5} W.
  2. Photon energy E=1240310=4E = \dfrac{1240}{310} = 4 eV =6.4×10−19= 6.4\times10^{-19} J (greater than ϕ\phi, so emission occurs).
  3. Photons per second =6.4×10−56.4×10−19=1014= \dfrac{6.4\times10^{-5}}{6.4\times10^{-19}} = 10^{14}.
  4. Electrons per second =1014103=1011= \dfrac{10^{14}}{10^{3}} = 10^{11}, so x=11x = 11.
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Q297MHT CET 2019 · 2 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be

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Answer: (d)

  1. Kmax=hν−ϕK_{max} = h\nu - \phi depends only on frequency and work function.
  2. Reducing intensity to I/4I/4 only reduces the number of photons per second.
  3. So the answer is (d) unchanged.
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Q298MHT CET 2019 · 3 MAY · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

The stopping potential of the photoelectrons, from a photo cell is

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Answer: (b)

  1. V0=heν−ϕeV_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}: V0V_0 increases linearly with frequency and does not depend on intensity.
  2. Of the given options, only (b) relates V0V_0 to frequency in the right direction (strictly it is linear, not exactly proportional).
  3. So the answer is (b).
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Q299JEE Main 2019 · 12 APR · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

The stopping potential V0 (in volt) as a function of frequency (vv) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be: (Given: Planck’s constant (h) = 6.63 × 10–34 Js, electron charges e = 1.6 × 10–19 C)

Figure

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Answer: (d)

  1. From the graph, V0=0V_0 = 0 at ν0=4×1014\nu_0 = 4 \times 10^{14} Hz.
  2. ϕ=hν0=6.63×10−34×4×1014=2.65×10−19\phi = h\nu_0 = 6.63 \times 10^{-34} \times 4 \times 10^{14} = 2.65 \times 10^{-19} J
  3. ϕ=2.65×10−191.6×10−19≈1.66\phi = \frac{2.65 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.66 eV
  4. So the answer is (d) 1.66 eV.
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Q300JEE Main 2019 · 10 JAN · SHIFT 2Difficult

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A metal plate of area 1 ×\times 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be

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Answer: (c)

  1. Power on plate P=IA=16×10−3×10−4=1.6×10−6P = IA = 16\times10^{-3}\times10^{-4} = 1.6\times10^{-6} W.
  2. Photon energy E=10E = 10 eV =1.6×10−18= 1.6\times10^{-18} J, so photons per second =1.6×10−61.6×10−18=1012= \dfrac{1.6\times10^{-6}}{1.6\times10^{-18}} = 10^{12}.
  3. Only 10% eject electrons: n=1011n = 10^{11} per second.
  4. Kmax=E−ϕ=10−5=5K_{max} = E - \phi = 10 - 5 = 5 eV.
  5. So the answer is (c) 101110^{11} and 5 eV.
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Q301JEE Main 2014 · 12 APR · SHIFT 1Difficult

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A beam of light has two wavelengths 4972 A˚\text{\AA} and 6216 A˚\text{\AA} with a total intensity of 3.6 ×\times 10−-3 Wm−2{}^{-2} equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2s is approximately:

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Answer: (b)

  1. Photon energies: E=12400λ(A˚)E = \dfrac{12400}{\lambda(\text{Å})} eV gives 124004972≈2.49\dfrac{12400}{4972} \approx 2.49 eV and 124006216≈1.99\dfrac{12400}{6216} \approx 1.99 eV.
  2. Only the 4972 Å photons exceed ϕ=2.3\phi = 2.3 eV, so only they eject electrons.
  3. Power of that half: P=1.8×10−3×10−4=1.8×10−7P = 1.8\times10^{-3}\times10^{-4} = 1.8\times10^{-7} W.
  4. Electrons per second =1.8×10−72.49×1.6×10−19≈4.5×1011= \dfrac{1.8\times10^{-7}}{2.49\times1.6\times10^{-19}} \approx 4.5\times10^{11}; in 2 s ≈9×1011\approx 9\times10^{11}.
  5. So the answer is (b) 9×10119\times10^{11}.
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Q302NEET 2013Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

A source of light is placed at a distance of 50 cm from a photo cell and the stopping potential is found to be V0. If the distance between the light source and photo cell is made 25 cm, the new stopping potential will be:

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Answer: (b)

  1. Halving the distance makes intensity 44 times (I∝1/r2I \propto 1/r^2), but frequency is unchanged.
  2. eV0=hν−ϕeV_0 = h\nu - \phi is independent of intensity.
  3. So the answer is (b) V0V_0.
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Q303JEE Main 2013Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

The anode voltage of a photocell is kept fixed. The wavelength λ\lambda of the light falling on the cathode is gradually changed. The plate current II of the photocell varies as follows:

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Answer: (d)

  1. Increasing λ\lambda lowers photon energy hc/λhc/\lambda, so electrons are ejected with less energy and the plate current falls.
  2. At the threshold wavelength λ0\lambda_0, emission stops and I=0I = 0; beyond it II stays zero.
  3. So II decreases with λ\lambda and reaches zero at λ0\lambda_0.
  4. So the answer is (d).
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Q304JEE Main 2013 · 25 APR · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

In an experiment on photoelectric effect, a student plots stopping potential V0V_0 against reciprocal of the wavelength λ\lambda of the incident light for two different metals A and B. These are shown in the figure.

Figure

Looking at the graphs, you can most appropriately say that:

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Answer: (d)

  1. eV0=hcλ−ϕeV_0 = \frac{hc}{\lambda} - \phi, so V0=hce⋅1λ−ϕeV_0 = \frac{hc}{e}\cdot\frac{1}{\lambda} - \frac{\phi}{e}.
  2. The slope hc/ehc/e is the same for every metal, so the two lines must be parallel.
  3. The plotted lines have different slopes, which is impossible.
  4. So the answer is (d) the student's data is not correct.
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Q305JEE Main 2011 · AIEEEMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

This question has Statement - 11 and Statement - 22. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement - 11: A metallic surface is irradiated by a monochromatic light of frequency v>v0v > v_0 (the threshold frequency). The maximum kinetic energy and the stopping potential are KmaxK_\mathrm{max} and V0V_0 respectively. If the frequency incident on the surface is doubled, both the KmaxK_\mathrm{max} and V0V_0 are also doubled.

Statement - 22: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

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Answer: (c)

  1. Kmax=hν−ϕK_{max} = h\nu - \phi and eV0=hν−ϕeV_0 = h\nu - \phi.
  2. Doubling ν\nu: K′=2hν−ϕ=2Kmax+ϕ≠2KmaxK' = 2h\nu - \phi = 2K_{max} + \phi \neq 2K_{max}. Statement 1 is false.
  3. Both KmaxK_{max} and V0V_0 are linear (not proportional) in ν\nu. Statement 2 is true.
  4. So the answer is (c).
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Q306NEET 2010 · AIPMT · MAINSEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

When monochromatic radiation of intensity II falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2II, the number of emitted electrons and their maximum kinetic energy are respectively

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Answer: (b)

  1. Number of photoelectrons ∝\propto intensity: I→2II \to 2I gives N→2NN \to 2N.
  2. Kmax=hν−ϕK_{max} = h\nu - \phi is independent of intensity: TT stays TT.
  3. So the answer is (b) 2N2N and TT.
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Q307JEE Main 2010 · AIEEEMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

Statement - 11: When ultraviolet light is incident on a photocell, its stopping potential is V0V_0 and the maximum kinetic energy of the photoelectrons is KmaxK_\mathrm{max}. When the ultraviolet light is replaced by XX-rays, both V0V_0 and KmaxK_\mathrm{max} increase.

Statement - 22: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

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Answer: (b)

  1. X-rays have higher frequency than UV, so Kmax=hν−ϕK_{max} = h\nu - \phi and V0V_0 both increase. Statement 1 is true.
  2. The spread of speeds from zero to maximum occurs even for monochromatic light, because electrons lose different amounts of energy coming from different depths and bound states. Statement 2 is false.
  3. So the answer is (d).
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Q308NEET 2009 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

The number of photo electrons emitted for light of a frequency vv (higher than the threshold frequency v0v_0) is proportional to

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Answer: (b)

  1. One photon ejects one electron, so number of photoelectrons ∝\propto photons per second.
  2. At fixed frequency, photons per second ∝\propto intensity.
  3. So the answer is (b) intensity of light.
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Q309NEET 2009 · AIPMTMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement?

Figure

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Answer: (a)

  1. Same stopping potential means same frequency; different saturation current means different intensity.
  2. Curves aa and bb share the stopping potential but saturate at different currents.
  3. Curves bb and cc have different stopping potentials (different frequencies) but the same saturation current (same intensity).
  4. So the answer is (a).
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Q310NEET 2007 · AIPMTMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · What stopping potential depends on

A 5 watt source emits monochromatic light of wavelength 5000 A˚\text{\AA}. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of

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Answer: (d)

  1. Intensity of a point source ∝1r2\propto \frac{1}{r^2}; number of photoelectrons ∝\propto intensity.
  2. r:0.5→1.0r: 0.5 \to 1.0 m (doubled) ⇒\Rightarrow intensity becomes 14\frac{1}{4}.
  3. So the answer is (d) 4.
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Q311NEET 2006 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

A photocell employs photoelectric effect to convert

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Answer: (c)

  1. In a photocell the frequency is fixed (above threshold) and the photocurrent ∝\propto intensity.
  2. So a change in illumination appears as a change in current; the work function is a fixed property of the cathode.
  3. So the answer is (c).
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Q312JEE Main 2006 · AIEEEEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

The time taken by a photoelectron to come out after the photon strikes is approximately

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Answer: (b)

  1. Photoemission is practically instantaneous: a photon gives all its energy to one electron in a single collision.
  2. The observed time lag is about 10−910^{-9} to 10−10 s10^{-10}\ s.
  3. So the answer is (b) 10−10 s10^{-10}\ s.
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Q313JEE Main 2006 · AIEEEMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: photocurrent vs anode potential

The anode voltage of a photocell is kept fixed. The wavelength λ\lambda of the light falling on the cathode is gradually changed. The plate current I\mathrm{I} of the photocell varies as follows

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Answer: (b)

  1. As λ\lambda increases, photon energy hc/λhc/\lambda decreases and the plate current falls.
  2. At the threshold wavelength λ0\lambda_0 the current becomes zero.
  3. So the answer is (b), the curve decreasing to zero at λ0\lambda_0.
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Q314JEE Main 2005 · AIEEEEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A photocell is illuminated by a small bright source placed 1 m1\ m away. When the same source of light is placed 12 m\frac{1}{2}\ m away, the number of electrons emitted by photo-cathode would

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Answer: (a)

  1. I∝1r2I \propto \frac{1}{r^2} and number of photoelectrons ∝I\propto I.
  2. r:1→12r: 1 \to \frac{1}{2} m ⇒I′=4I\Rightarrow I' = 4I.
  3. So the answer is (a) increases by a factor of 4.
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Q315NEET 2004 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

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Answer: (d)

  1. KEmax=hν−ϕKE_{max} = h\nu - \phi: a straight line with positive slope hh.
  2. KE=0KE = 0 for ν≤ν0\nu \le \nu_0, so the line starts on the frequency axis at ν0>0\nu_0 > 0, not at the origin.
  3. So the answer is (d), the straight line starting at a non-zero frequency.
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Q316JEE Main 2004 · AIEEEEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal VsVs the frequency of the incident radiation gives a straight line whose slope

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Answer: (d)

  1. KEmax=hν−ϕKE_{max} = h\nu - \phi, so slope =h= h, a universal constant.
  2. It does not depend on the metal (only the intercept does) or on intensity.
  3. So the answer is (d).
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Q317NEET 2003 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m then

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Answer: (d)

  1. Intensity of a point source I∝1r2I \propto \frac{1}{r^2}; number of photoelectrons ∝I\propto I.
  2. r:1→2r: 1 \to 2 m ⇒I′=I/4\Rightarrow I' = I/4.
  3. Energy of each electron depends only on frequency, so it is unchanged.
  4. So the answer is (d) number of electrons becomes a quarter.
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Q318NEET 2001 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Photocurrent numericals & source distance

A photo-cell is illuminated by a source of light, which is placed at a distance d from the cell. If the distance become d/2, then number of electrons emitted per second, will be

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Answer: (b)

  1. I∝1d2I \propto \frac{1}{d^2} and electrons emitted per second ∝I\propto I.
  2. d→d/2⇒I′=4Id \to d/2 \Rightarrow I' = 4I.
  3. So the answer is (b) four times.
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Q319NEET 1999 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

When intensity of incident light increases

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Answer: (a)

  1. Higher intensity means more photons per second, so more electrons per second: photocurrent increases.
  2. Kmax=hν−ϕK_{max} = h\nu - \phi does not change with intensity.
  3. So the answer is (a) photo-current increases.
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Q320NEET 1997 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Which of the following statement is correct?

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Answer: (a)

  1. (a) True: photocurrent ∝\propto intensity.
  2. (b) False: stopping potential V0=(hν−ϕ)/eV_0 = (h\nu - \phi)/e does not depend on intensity.
  3. (c) False: saturation current does not increase with frequency. (d) False: current saturates, it is not proportional to voltage.
  4. So the answer is (a).
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Q321NEET 1996 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Graphs: stopping potential & KE vs frequency

According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

Show answer and solution

Answer: (d)

  1. KEmax=hν−ϕKE_{max} = h\nu - \phi: a straight line with positive slope hh.
  2. No emission below ν0\nu_0, so the line starts on the frequency axis at ν0>0\nu_0 > 0.
  3. So the answer is (d), the straight line starting at a non-zero frequency.
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Q322NEET 1993 · AIPMTEasy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

Number of ejected photoelectrons increases with increase

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Answer: (a)

  1. Number of ejected electrons ∝\propto number of photons per second ∝\propto intensity.
  2. Changing wavelength or frequency changes only the KE of electrons, not their number.
  3. So the answer is (a) in intensity of light.
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Q323NEET 1992 · AIPMTMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs · Effect of intensity & frequency

The cathode of a photoelectric cell is changed such that the work function changes from W1W_1 to W2W_2 (W2>W1W_2 > W_1 ). If the current before and after changes are I1I_1 and I2I_2, all other conditions remaining unchanged, then (assuming hν>W2h\nu > W_2)

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Answer: (a)

  1. Photocurrent ∝\propto number of photons per second (intensity), as long as hν>Wh\nu > W.
  2. Changing WW changes only the kinetic energy of the electrons, not their number.
  3. So the answer is (a) I1=I2I_1 = I_2.
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Q324MHT CET 2026 · 14 MAY · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

Two particles A and B of masses m1m_1 and m2m_2 (m1>m2m_1 > m_2) have the same de-Broglie wavelength, then

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Answer: (a)

  1. λ=hp\lambda = \dfrac{h}{p}, so equal λ\lambda means equal momentum pp, whatever the masses.
  2. Energy E=p22mE = \dfrac{p^2}{2m} then differs: the heavier A has less energy.
  3. So the answer is (a) their momenta are the same.
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Q325MHT CET 2026 · 14 MAY · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de-Broglie wavelength λ\lambda of a particle

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Answer: (c)

  1. de Broglie relation: λ=hp\lambda = \dfrac{h}{p}, so λ∝1p\lambda \propto \dfrac{1}{p}.
  2. So the answer is (c) inversely proportional to momentum.
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Q326MHT CET 2026 · 17 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The kinetic energy of a free electron increases to 3 times the previous K.E. The ratio of new de- Broglie wavelength to previous de-Broglie wavelength is

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Answer: (a)

  1. λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}}, so λ∝1K\lambda \propto \dfrac{1}{\sqrt{K}}.
  2. K′=3K⇒λ′λ=K3K=13K' = 3K \Rightarrow \dfrac{\lambda'}{\lambda} = \sqrt{\dfrac{K}{3K}} = \dfrac{1}{\sqrt{3}}.
  3. So the answer is (a) 13\dfrac{1}{\sqrt{3}}.
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Q327MHT CET 2026 · 20 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The ratio of de-Broglie wavelength of an α\alpha-particle and proton accelerated from rest by the same potential is 1m\frac{1}{\sqrt{m}}. The value of mm is

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Answer: (b)

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}, so for the same VV, λ∝1mq\lambda \propto \dfrac{1}{\sqrt{mq}}.
  2. λαλp=mpe4mp⋅2e=18\dfrac{\lambda_\alpha}{\lambda_p} = \sqrt{\dfrac{m_p e}{4m_p\cdot 2e}} = \dfrac{1}{\sqrt{8}}.
  3. So m=8m = 8; the answer is (b) 8.
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Q328MHT CET 2026 · 20 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The proton and α\alpha - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and α\alpha - particle is (mass of α\alpha-particle is 44 times mass of proton, charge of α\alpha-particle is 22 times charge of proton)

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Answer: (d)

  1. λ=h2mqV⇒λ∝1mq\lambda = \dfrac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mq}} for the same VV.
  2. λpλα=mαqαmpqp=4×2=22\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4\times2} = 2\sqrt{2}.
  3. So the answer is (d) 222\sqrt{2}.
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Q329MHT CET 2026 · 18 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

If the kinetic energy of a particle is increased to 16 times, the percentage change in the de Broglie wavelength of a particle is

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Answer: (c)

  1. λ=h2mK∝1K\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{K}}.
  2. K′=16K⇒λ′=λ4K' = 16K \Rightarrow \lambda' = \dfrac{\lambda}{4}
  3. Change =λ−λ/4λ×100=75%= \dfrac{\lambda - \lambda/4}{\lambda} \times 100 = 75\% (decrease).
  4. So the answer is (c) 75%75\%.
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Q330JEE Main 2026 · 5 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

An electron is travelling with a velocity vv in free space and when it enters a medium, its velocity is reduced by 20%20\%. The de Broglie wavelength of electron in the medium is αλ0\alpha\lambda_0, where λ0\lambda_0 is its de Broglie wavelength in free space. The value of α\alpha is ______.

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Answer: (c)

  1. λ=hmv∝1v\lambda = \dfrac{h}{mv} \propto \dfrac{1}{v}.
  2. In the medium v′=0.8vv' = 0.8v, so λ=λ00.8=1.25λ0\lambda = \dfrac{\lambda_0}{0.8} = 1.25\lambda_0.
  3. So α=1.25\alpha = 1.25, answer (c).
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Q331MHT CET 2026 · 13 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

An electron accelerated through a potential difference V1V_1 has a de-Broglie wavelength of λ\lambda. When the potential is changed to V2V_2, its de-Broglie wavelength increases to 2λ2\lambda. The value of (V1/V2)(V_1/V_2) is equal to

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Answer: (d)

  1. λ=h2meV⇒λ∝1V\lambda = \dfrac{h}{\sqrt{2meV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{V}}, i.e. V∝1λ2V \propto \dfrac{1}{\lambda^2}.
  2. V1V2=(λ2λ1)2=(2λλ)2=4\dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2 = \left(\dfrac{2\lambda}{\lambda}\right)^2 = 4
  3. So the answer is (d) 4:14 : 1.
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Q332JEE Main 2026 · 6 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

The de Broglie wavelength for an electron accelerated through the potential difference of V1V_1 volt is λ1\lambda_1. When the potential difference is changed to V2V_2 volt, the associated de Broglie wavelength is increased by 50%50\%. If (V1/V2)=(9/α)(V_1/V_2) = (9/\alpha), then the value of α\alpha is ______。

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

  1. λ∝1V⇒V1V2=(λ2λ1)2\lambda \propto \dfrac{1}{\sqrt{V}} \Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2.
  2. λ2=1.5λ1⇒V1V2=94\lambda_2 = 1.5\lambda_1 \Rightarrow \dfrac{V_1}{V_2} = \dfrac{9}{4}
  3. So α=4\alpha = 4.
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Q333JEE Main 2026 · 21 JAN · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

A particle having electric charge 3×10−19 C3 \times 10^{-19}\text{ C} and mass 6×10−27 kg6 \times 10^{-27}\text{ kg} is accelerated by applying an electric potential of 1.21 V.

Wavelength of the matter wave associated with the particle is α×10−12 m\alpha \times 10^{-12}\text{ m}. The value of α\alpha is ______. (Take Planck's constant =6.6×10−34 J s= 6.6 \times 10^{-34}\text{ J s})

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 10

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}
  2. 2mqV=2×6×10−27×3×10−19×1.21=43.56×10−462mqV = 2 \times 6 \times 10^{-27} \times 3 \times 10^{-19} \times 1.21 = 43.56 \times 10^{-46}, so 2mqV=6.6×10−23\sqrt{2mqV} = 6.6 \times 10^{-23}
  3. λ=6.6×10−346.6×10−23=10−11 m=10×10−12 m\lambda = \dfrac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} = 10^{-11}\text{ m} = 10 \times 10^{-12}\text{ m}
  4. So α=10\alpha = 10.
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Q334MHT CET 2026 · 17 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The ratio of the accelerating potentials required to accelerate (i) an α\alpha-particle and (ii) a proton to have the same de Broglie wavelength associated with them is

(mass of α\alpha - particle = 6.4×10−276.4 \times 10^{-27} kg, mass of proton = 1.6×10−271.6 \times 10^{-27} kg)

Show answer and solution

Answer: (a)

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}; same λ\lambda needs the same mqVmqV.
  2. VαVp=mpqpmαqα=1.6×10−27×e6.4×10−27×2e=18\dfrac{V_\alpha}{V_p} = \dfrac{m_p q_p}{m_\alpha q_\alpha} = \dfrac{1.6 \times 10^{-27} \times e}{6.4 \times 10^{-27} \times 2e} = \dfrac{1}{8}
  3. So the answer is (a) 1:81 : 8.
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Q335JEE Main 2026 · 4 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The de Broglie wavelength associated with an electron accelerated through a potential difference V is λe\lambda_\mathrm{e} and the de Broglie wavelength associated with a proton accelerated through the same potential difference is λp\lambda_\mathrm{p}. If their corresponding masses are mem_\mathrm{e} and mpm_\mathrm{p}, respectively, then the ratio of their de Broglie wavelengths (λeλp)\left(\frac{\lambda_e}{\lambda_p}\right) is ______.

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Answer: (a)

  1. λ=h2meV\lambda = \dfrac{h}{\sqrt{2meV}}; same charge magnitude and same VV, so λ∝1m\lambda \propto \dfrac{1}{\sqrt{m}}.
  2. λeλp=mpme\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}}
  3. So the answer is (a) mpme\sqrt{\frac{m_p}{m_e}}.
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Q336JEE Main 2026 · 28 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The ratio of de Broglie wavelength of a deuteron with kinetic energy EE to that of an alpha particle with kinetic energy 2E2E, is n:1n : 1. The value of nn is ______.

(Assume mass of proton == mass of neutron)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. λ=h2mK⇒λ∝1mK\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mK}}.
  2. Deuteron: m=2mpm = 2m_p, K=EK = E, so mK=2mpEmK = 2m_pE. Alpha: m=4mpm = 4m_p, K=2EK = 2E, so mK=8mpEmK = 8m_pE.
  3. λdλα=82=2\dfrac{\lambda_d}{\lambda_\alpha} = \sqrt{\dfrac{8}{2}} = 2
  4. So n=2n = 2.
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Q337MHT CET 2026 · 12 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperatures 27∘C27^\circ C and 127∘C127^\circ C respectively is (mHe=2mH2m_{He} = 2m_{H_2})

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Answer: (b)

  1. Thermal de Broglie wavelength: λ=h3mkT⇒λ∝1mT\lambda = \dfrac{h}{\sqrt{3mkT}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mT}}.
  2. TH2=300 KT_{H_2} = 300\text{ K}, THe=400 KT_{He} = 400\text{ K}, mHe=2mH2m_{He} = 2m_{H_2}.
  3. λH2λHe=mHeTHemH2TH2=2×400300=83\dfrac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\dfrac{m_{He}T_{He}}{m_{H_2}T_{H_2}}} = \sqrt{\dfrac{2 \times 400}{300}} = \sqrt{\dfrac{8}{3}}
  4. So the answer is (b) 83\sqrt{\frac{8}{3}}.
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Q338JEE Main 2026 · 23 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The de Broglie wavelength of an oxygen molecule at 27∘C27^\circ\text{C} is x×10−12 mx \times 10^{-12}\text{ m}. The value of xx is (take Planck's constant =6.63×10−34 J s= 6.63 \times 10^{-34}\text{ J s}, Boltzmann constant =1.38×10−23 J/K= 1.38 \times 10^{-23}\text{ J/K}, mass of oxygen molecule =5.31×10−26 kg= 5.31 \times 10^{-26}\text{ kg})

Show answer and solution

Answer: (d)

  1. λ=h3mkT\lambda = \dfrac{h}{\sqrt{3mkT}} with T=300 KT = 300\text{ K}.
  2. 3mkT=3×5.31×10−26×1.38×10−23×300≈6.60×10−463mkT = 3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300 \approx 6.60 \times 10^{-46}, so 3mkT≈2.57×10−23\sqrt{3mkT} \approx 2.57 \times 10^{-23}
  3. λ=6.63×10−342.57×10−23≈2.58×10−11 m≈26×10−12 m\lambda = \dfrac{6.63 \times 10^{-34}}{2.57 \times 10^{-23}} \approx 2.58 \times 10^{-11}\text{ m} \approx 26 \times 10^{-12}\text{ m}
  4. So the answer is (d) 26.
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Q339MHT CET 2026 · 20 MAY · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The wavelength 'λ\lambda' of photon and the de-Broglie wavelength of an electron have same value. The ratio of kinetic energy of electron to the energy of a photon is (m=mass of electron, h = Planck's constant, c = velocity of light)

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Answer: (a)

  1. Same λ\lambda means same momentum p=h/λp = h/\lambda.
  2. Electron: K=p22m=h22mλ2K = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}. Photon: E=hcλE = \dfrac{hc}{\lambda}.
  3. KE=h22mλ2⋅λhc=h2λmc\dfrac{K}{E} = \dfrac{h^2}{2m\lambda^2} \cdot \dfrac{\lambda}{hc} = \dfrac{h}{2\lambda mc}
  4. So the answer is (a).
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Q340MHT CET 2026 · 13 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The energy of a photon is equal to the kinetic energy of a proton. If λ1\lambda_1 is the de-Broglie wavelength of a proton, λ2\lambda_2 is the wavelength associated with the photon and if the energy of the photon is EE then (λ1/λ2)(\lambda_1/\lambda_2) is proportional to

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Answer: (b)

  1. Proton: λ1=h2mE\lambda_1 = \dfrac{h}{\sqrt{2mE}}. Photon: λ2=hcE\lambda_2 = \dfrac{hc}{E}.
  2. λ1λ2=Ec2mE=Ec2m\dfrac{\lambda_1}{\lambda_2} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{\sqrt{E}}{c\sqrt{2m}}
  3. So λ1/λ2∝E\lambda_1/\lambda_2 \propto \sqrt{E}: the answer is (b).
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Q341MHT CET 2026 · 12 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electromagnetic wave of wavelength λ\lambda is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have de-Broglie wavelength λ1\lambda_1, then

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Answer: (b)

  1. Work function ≈0\approx 0, so the photon energy becomes the electron's KE: hcλ=p22m\dfrac{hc}{\lambda} = \dfrac{p^2}{2m}.
  2. With p=h/λ1p = h/\lambda_1: hcλ=h22mλ12\dfrac{hc}{\lambda} = \dfrac{h^2}{2m\lambda_1^2}
  3. λ=2mchλ12⇒λ∝λ12\lambda = \dfrac{2mc}{h}\lambda_1^2 \Rightarrow \lambda \propto \lambda_1^2
  4. So the answer is (b).
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Q342MHT CET 2026 · 11 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

According to de-Broglie hypothesis, the ratio of the wave-length of a photon and that of an electron having same energy 'E' is (m = mass of electron, c = velocity of light)

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Answer: (c)

  1. Photon: λph=hcE\lambda_{ph} = \dfrac{hc}{E}. Electron: λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}.
  2. λphλe=c2mEE=c2mE\dfrac{\lambda_{ph}}{\lambda_e} = \dfrac{c\sqrt{2mE}}{E} = c\sqrt{\dfrac{2m}{E}}
  3. So the answer is (c).
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Q343MHT CET 2026 · 11 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

A photon and an electron have equal energy E. The ratio of λ\lambda(electron) to λ\lambda(photon) is proportional to

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Answer: (b)

  1. λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}, λph=hcE\lambda_{ph} = \dfrac{hc}{E}.
  2. λeλph=Ec2mE=Ec2m∝E\dfrac{\lambda_e}{\lambda_{ph}} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{\sqrt{E}}{c\sqrt{2m}} \propto \sqrt{E}
  3. So the answer is (b) E:1\sqrt{E} : 1.
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Q344NEET 2026 · RE-NEETMedium

Matter Waves: de Broglie Wavelength · Photon vs electron

A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron pep_e to that of photon pPh,pepPhp_{Ph}, \frac{p_e}{p_{Ph}} is:

[Take speed of light =3×108 ms−1= 3 \times 10^{8}\text{ ms}^{-1}, charge of electron =−1.6×10−19 C= -1.6 \times 10^{-19}\text{ C} and mass of electron =9×10−31 kg= 9 \times 10^{-31}\text{ kg} ]

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Answer: (d)

  1. E=20 eV=20×1.6×10−19=3.2×10−18 JE = 20\text{ eV} = 20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-18}\text{ J}.
  2. Electron: pe=2mE=2×9×10−31×3.2×10−18=2.4×10−24 kg m/sp_e = \sqrt{2mE} = \sqrt{2 \times 9 \times 10^{-31} \times 3.2 \times 10^{-18}} = 2.4 \times 10^{-24}\text{ kg m/s}
  3. Photon: pPh=Ec=3.2×10−183×108≈1.067×10−26 kg m/sp_{Ph} = \dfrac{E}{c} = \dfrac{3.2 \times 10^{-18}}{3 \times 10^8} \approx 1.067 \times 10^{-26}\text{ kg m/s}
  4. pepPh=2.4×10−241.067×10−26=225\dfrac{p_e}{p_{Ph}} = \dfrac{2.4 \times 10^{-24}}{1.067 \times 10^{-26}} = 225
  5. So the answer is (d) 225.
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Q345MHT CET 2026 · 12 MAY · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron is moving with initial velocity v⃗=ak^\vec{v} = a\hat{k} in a magnetic field B⃗=bj^\vec{B} = b\hat{j}. Its de-Broglie wavelength will

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Answer: (d)

  1. v⃗\vec v along k^\hat k and B⃗\vec B along j^\hat j: magnetic force F⃗=−e(v⃗×B⃗)\vec F = -e(\vec v\times\vec B) is always perpendicular to v⃗\vec v.
  2. It does no work, so speed and ∣p∣|p| stay constant.
  3. λ=h/p\lambda = h/p stays constant.
  4. So the answer is (d) remain constant.
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Q346MHT CET 2026 · 20 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron is moving with initial velocity v⃗=V0j^\vec{v} = V_0\hat{j} in a magnetic field B⃗=B0i^\vec{B} = B_0\hat{i}. Its de-Broglie wavelength will

Show answer and solution

Answer: (c)

  1. Magnetic force is always perpendicular to velocity, so it does no work.
  2. Speed, hence momentum magnitude p=mvp = mv, is constant.
  3. λ=h/(mv)\lambda = h/(mv) is constant.
  4. So the answer is (c) remains constant.
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Q347MHT CET 2026 · 16 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · Photon vs electron

A photoemissive substance is illuminated with a radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e. The longest wavelength of radiation that can emit photoelectron is λ0\lambda_0. Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)

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Answer: (c)

  1. Einstein: K=hc(1λi−1λ0)K = hc\left(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_0}\right)
  2. λe=h2mK=h2mhc(1λi−1λ0)\lambda_e = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mhc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}}
  3. λe=[h2mc(1λi−1λ0)]1/2\lambda_e = \left[\dfrac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}\right]^{1/2}
  4. So the answer is (c), with the whole bracket in the denominator.
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Q348MHT CET 2026 · 15 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · Photon vs electron

Light of wavelength 'λ\lambda' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential

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Answer: (b)

  1. eV0=hcλ−ϕeV_0 = \dfrac{hc}{\lambda} - \phi
  2. Decreasing λ\lambda raises photon energy hc/λhc/\lambda, so KmaxK_{max} and V0V_0 rise.
  3. So the answer is (b) increases.
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Q349MHT CET 2026 · 15 MAY · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron of mass mm and charge ee initially at rest gets accelerated by a constant electric field EE. The rate of change of de-Broglie wavelength of the electron at time tt is

(hh = Planck's constant)

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Answer: (a)

  1. From rest: p=eEtp = eEt, so λ=heEt\lambda = \dfrac{h}{eEt}.
  2. dλdt=−heEt2\dfrac{d\lambda}{dt} = -\dfrac{h}{eEt^2}
  3. So the answer is (a).
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Q350JEE Main 2026 · 5 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron of mass mm is moving in an electric field E⃗=−2Eoi^ (Eo=\vec{E} = -2E_o\hat{i}\ (E_o = constant >0)> 0), with an initial velocity V⃗=voi^ (vo=\vec{V} = v_o\hat{i}\ (v_o = constant >0)> 0). If λo=h4mvo\lambda_o = \frac{h}{4mv_o}, its de Broglie wavelength at time tt is ______.

( e=e = charge of electron)

Show answer and solution

Answer: (c)

  1. Force on electron =−eE⃗=2eEoi^= -e\vec E = 2eE_o\hat i, along v⃗\vec v: v=vo+2eEomtv = v_o + \dfrac{2eE_o}{m}t.
  2. λ=hmv=hmvo(1+2Eoemtvo)\lambda = \dfrac{h}{mv} = \dfrac{h}{mv_o\left(1+\frac{2E_oe}{m}\frac{t}{v_o}\right)}
  3. hmvo=4λo\dfrac{h}{mv_o} = 4\lambda_o, so λ=4λo1+2Eoemtvo\lambda = \dfrac{4\lambda_o}{1+\frac{2E_oe}{m}\frac{t}{v_o}}
  4. So the answer is (c).
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Q351MHT CET 2025 · 21 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de-Broglie wavelength (λ)(\lambda) of a particle

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Answer: (a)

  1. Impulse equals change in momentum; for a particle starting from rest, impulse =p= p.
  2. λ=hp\lambda = \dfrac{h}{p}, so λ∝1impulse\lambda \propto \dfrac{1}{\text{impulse}}.
  3. So the answer is (a) inversely proportional to impulse.
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Q352JEE Main 2025 · 29 JAN · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

If λ\lambda and KK are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:

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Answer: (d)

  1. λ=h2mK⇒1K=2mh2λ2\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \dfrac{1}{K} = \dfrac{2m}{h^2}\lambda^2.
  2. So 1K∝λ2\dfrac{1}{K} \propto \lambda^2: a parabola through the origin, curving upward.
  3. So the answer is (d).
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Q353MHT CET 2025 · 23 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

Electron beam when accelerated by a voltage of 10 kV, has a de-Broglie wavelength ' λ\lambda '. If the voltage is increased to 20 kV then the deBroglie wavelength associated with the electron beam would be

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Answer: (d)

  1. λ=h2meV⇒λ∝1V\lambda = \dfrac{h}{\sqrt{2meV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{V}}.
  2. λ′λ=1020=12\dfrac{\lambda'}{\lambda} = \sqrt{\dfrac{10}{20}} = \dfrac{1}{\sqrt{2}}.
  3. So the answer is (d) λ2\dfrac{\lambda}{\sqrt{2}}.
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Q354MHT CET 2025 · 19 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

An electron accelerated by a potential difference ' VV ' has de-Broglie wavelength ' λ\lambda '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be

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Answer: (b)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. V′=9V⇒λ′=λ3V' = 9V \Rightarrow \lambda' = \dfrac{\lambda}{3}.
  3. So the answer is (b) λ3\dfrac{\lambda}{3}.
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Q355MHT CET 2025 · 23 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The energy that should be added to an electron to reduce its de-Broglie wavelength from λ\lambda to λ2\frac{\lambda}{2} is nn times the initial energy. The value of ' nn ' is

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Answer: (b)

  1. λ∝1K⇒K∝1λ2\lambda \propto \dfrac{1}{\sqrt{K}} \Rightarrow K \propto \dfrac{1}{\lambda^2}.
  2. λ→λ2⇒K′=4K\lambda \to \dfrac{\lambda}{2} \Rightarrow K' = 4K
  3. Energy added =4K−K=3K= 4K - K = 3K, so n=3n = 3.
  4. So the answer is (b) 3.
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Q356JEE Main 2025 · 23 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

A sub-atomic particle of mass 10−30 kg10^{-30}\text{ kg} is moving with a velocity 2.21×106 m/s2.21 \times 10^{6}\text{ m/s}. Under the matter wave consideration, the particle will behave closely like (h=6.63×10−34 J s)(\mathrm{h} = 6.63 \times 10^{-34}\text{ J s})

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Answer: (a)

  1. λ=hmv=6.63×10−3410−30×2.21×106\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^{6}}
  2. λ=3×10−10 m=3 A˚\lambda = 3 \times 10^{-10}\text{ m} = 3\text{ Å}
  3. Wavelengths of a few Å lie in the X-ray region.
  4. So the answer is (a) X-rays.
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Q357MHT CET 2025 · 22 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The de-Broglie wavelength of a neutron at 27∘C27^\circ\text{C} is ' λ0\lambda_0 '. What will be its wavelength at 927∘C927^\circ\text{C}?

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Answer: (c)

  1. λ=h3mkT⇒λ∝1T\lambda = \dfrac{h}{\sqrt{3mkT}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{T}}.
  2. T1=300 KT_1 = 300\text{ K}, T2=1200 KT_2 = 1200\text{ K}, so T2=4T1T_2 = 4T_1.
  3. λ2=λ04=λ02\lambda_2 = \dfrac{\lambda_0}{\sqrt{4}} = \dfrac{\lambda_0}{2}
  4. So the answer is (c) λ02\frac{\lambda_0}{2}.
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Q358MHT CET 2025 · 26 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The wavelength ' λ\lambda ' of a photon and the deBroglie wavelength of an electron have same value. The ratio of kinetic energy of the electron to the energy of a photon is

( m=\mathrm{m} = mass of electron, c=\mathrm{c} = velocity of light, h=\mathrm{h} = Planck's constant)

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Answer: (c)

  1. Same λ\lambda means same p=h/λp = h/\lambda.
  2. Electron: K=h22mλ2K = \dfrac{h^2}{2m\lambda^2}. Photon: E=hcλE = \dfrac{hc}{\lambda}.
  3. KE=h2λmc\dfrac{K}{E} = \dfrac{h}{2\lambda mc}
  4. So the answer is (c).
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Q359MHT CET 2025 · 20 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

A light of wavelength λ\lambda is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength λ1\lambda_1. Then ratio λ:λ1 2\lambda : \lambda_1^{\,2} is ( h=\mathrm{h} = Planck's constant, c=\mathrm{c} = velocity of light, m=\mathrm{m} = mass of electron)

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Answer: (c)

  1. Negligible work function: hcλ=p22m=h22mλ12\dfrac{hc}{\lambda} = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda_1^2}.
  2. λ=2mchλ12⇒λλ12=2mch\lambda = \dfrac{2mc}{h}\lambda_1^2 \Rightarrow \dfrac{\lambda}{\lambda_1^2} = \dfrac{2mc}{h}
  3. So the answer is (c) 2mc:h2mc : h.
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Q360NEET 2025Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

A photon and an electron (mass mm) have the same energy EE. The ratio ( λ\lambdaphoton /λ/\lambdaelectron) of their de Broglie wavelengths is: ( cc is the speed of light)

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Answer: (a)

  1. λph=hcE\lambda_{ph} = \dfrac{hc}{E}, λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}.
  2. λphλe=c2mEE=c2mE\dfrac{\lambda_{ph}}{\lambda_e} = \dfrac{c\sqrt{2mE}}{E} = c\sqrt{\dfrac{2m}{E}}
  3. So the answer is (a).
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Q361JEE Main 2025 · 28 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

A proton of mass ' mPm_P ' has same energy ' EE ' as that of a photon of wavelength ' λ\lambda '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

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Answer: (d)

  1. Proton: λdB=h2mpE\lambda_{dB} = \dfrac{h}{\sqrt{2m_pE}}. Photon: λ=hcE\lambda = \dfrac{hc}{E}.
  2. λdBλ=Ec2mpE=1cE2mp\dfrac{\lambda_{dB}}{\lambda} = \dfrac{E}{c\sqrt{2m_pE}} = \dfrac{1}{c}\sqrt{\dfrac{E}{2m_p}}
  3. So the answer is (d).
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Q362JEE Main 2025 · 24 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron of mass ' m ' with an initial velocityv⃗=v0i^ (v0>0)\vec{v} = v_0\hat{i}\ (v_0 > 0) enters an electric field E⃗=−Eok^\vec{E} = -E_o\hat{k}. If the initial de Broglie wavelength is λ0\lambda_0, the value after time t would be

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Answer: (c)

  1. Force on electron =−eE⃗=eE0k^= -e\vec E = eE_0\hat k, perpendicular to v⃗0\vec v_0.
  2. vx=v0v_x = v_0, vz=eE0mtv_z = \dfrac{eE_0}{m}t, so v=v01+e2E02t2m2v02v = v_0\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}.
  3. λ=hmv=λ01+e2E02t2m2v02\lambda = \dfrac{h}{mv} = \dfrac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}
  4. So the answer is (c).
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Q363MHT CET 2025 · 22 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie waves & Bohr orbits

When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number =n= n ), the de-Broglie wavelength ( λ\lambda) associated with it is related to nn as

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Answer: (d)

  1. Bohr condition: 2πrn=nλ⇒λ=2πrnn2\pi r_n = n\lambda \Rightarrow \lambda = \dfrac{2\pi r_n}{n}.
  2. rn∝n2r_n \propto n^2, so λ∝n2n=n\lambda \propto \dfrac{n^2}{n} = n (equivalently v∝1/nv \propto 1/n).
  3. So the answer is (d) λ∝n\lambda \propto n.
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Q364NEET 2025Medium

Matter Waves: de Broglie Wavelength · de Broglie waves & Bohr orbits

De-Broglie wavelength of an electron orbiting in the n=2n = 2 state of hydrogen atom is close to (Given Bohr radius =0.052 nm= 0.052\text{ nm})

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Answer: (d)

  1. 2πrn=nλ⇒λ=2πrnn2\pi r_n = n\lambda \Rightarrow \lambda = \dfrac{2\pi r_n}{n}
  2. r2=4a0=4×0.052=0.208r_2 = 4a_0 = 4\times0.052 = 0.208 nm
  3. λ=2π×0.2082≈0.65\lambda = \dfrac{2\pi\times0.208}{2} \approx 0.65 nm
  4. So the answer is (d) 0.67 nm (closest).
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Q365JEE Main 2025 · 22 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie waves & Bohr orbits

An electron in the ground state of the hydrogen atom has the orbital radius of 5.3×10−11 m5.3 \times 10^{-11}\text{ m} while that for the electron in third excited state is 8.48×10−10 m8.48 \times 10^{-10}\text{ m}. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

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Answer: (a)

  1. r4/r1=8.48×10−105.3×10−11=16=n2r_4/r_1 = \dfrac{8.48\times10^{-10}}{5.3\times10^{-11}} = 16 = n^2, so the excited state is n=4n = 4.
  2. λ=2πrnn∝n\lambda = \dfrac{2\pi r_n}{n} \propto n
  3. λgroundλexcited=14\dfrac{\lambda_{ground}}{\lambda_{excited}} = \dfrac{1}{4}, i.e. λexcited=4λground\lambda_{excited} = 4\lambda_{ground}.
  4. The key takes the magnitude 4, option (a).
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Q366MHT CET 2025 · 26 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · Photon vs electron

Let EcE_c and EpE_p represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then Ep/EcE_p/E_c is (speed of electron =C/100= C/100 where C is the velocity of light)

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Answer: (b)

  1. Electron: pe=mc100p_e = m\dfrac{c}{100}, λe=100hmc\lambda_e = \dfrac{100h}{mc}, Ec=12m(c100)2=mc22×104E_c = \dfrac12 m\left(\dfrac{c}{100}\right)^2 = \dfrac{mc^2}{2\times10^4}.
  2. Photon: λp=2λe=200hmc\lambda_p = 2\lambda_e = \dfrac{200h}{mc}, Ep=hcλp=mc2200E_p = \dfrac{hc}{\lambda_p} = \dfrac{mc^2}{200}.
  3. EpEc=2×104200=100\dfrac{E_p}{E_c} = \dfrac{2\times10^4}{200} = 100
  4. So the answer is (b) 10210^2.
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Q367JEE Main 2025 · 7 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · Photon vs electron

A photoemissive substance is illuminated with a radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e. The longest wavelength of radiation that can emit photoelectron is λo\lambda_o. Expression for de-Broglie wavelength is given by:

(m: mass of the electron, h: Planck's constant and c: speed of light)

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Answer: (c)

  1. Einstein: K=hc(1λi−1λo)K = hc\left(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_o}\right)
  2. λe=h2mK=h2mhc(1λi−1λo)\lambda_e = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mhc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}
  3. λe=h2mc(1λi−1λo)\lambda_e = \sqrt{\dfrac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}
  4. So the answer is (c).
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Q368MHT CET 2025 · 25 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron of mass ' mm ' and charge ' ee ' initially at rest gets accelerated by a constant electric field ' E '. The rate of change of de-Broglie wavelength of the electron at time ' tt ' is

(Ignore relativistic effect)( h=h = Planck's constant)

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Answer: (a)

  1. Constant force eEeE from rest: p=eEtp = eEt.
  2. λ=heEt⇒dλdt=−heEt2\lambda = \dfrac{h}{eEt} \Rightarrow \dfrac{d\lambda}{dt} = -\dfrac{h}{eEt^2}
  3. So the answer is (a).
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Q369JEE Main 2025 · 2 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron with mass ' m ' with an initial velocity (t=0) v⃗=v0i^ (v0>0)(t = 0)\ \vec{v} = v_0\hat{i}\ (v_0 > 0) enters a magnetic field B⃗=B0j^\vec{B} = B_0\hat{j}. If the initial de-Broglie wavelength at t=0t = 0 is λ0\lambda_0 then its value after time ' t ' would be:

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Answer: (b)

  1. Magnetic force −e(v⃗×B⃗)-e(\vec v\times\vec B) is perpendicular to v⃗\vec v and does no work.
  2. Speed and momentum magnitude stay constant.
  3. λ=h/(mv)=λ0\lambda = h/(mv) = \lambda_0
  4. So the answer is (b).
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Q370JEE Main 2025 · 8 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron is released from rest near an infinite non-conducting sheet of uniform charge density '−σ-\sigma'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nth power of time. The numerical value of n is _____.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Infinite sheet: uniform field E=σ2ε0E = \dfrac{\sigma}{2\varepsilon_0}, so constant force eEeE.
  2. From rest: p=eEtp = eEt, λ=heEt\lambda = \dfrac{h}{eEt}.
  3. dλdt=−heEt2∝t−2\dfrac{d\lambda}{dt} = -\dfrac{h}{eEt^2} \propto t^{-2}
  4. Answer: n=2n = 2
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Q371MHT CET 2024 · 4 MAY · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The kinetic energy of an electron is increased by 2 times, then the de-Broglie wavelength associated with it changes by a factor.

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Answer: (b)

  1. "Increased by 2 times" means K′=K+2K=3KK' = K + 2K = 3K.
  2. λ∝1K⇒λ′λ=13\lambda \propto \dfrac{1}{\sqrt{K}} \Rightarrow \dfrac{\lambda'}{\lambda} = \dfrac{1}{\sqrt{3}}.
  3. So the answer is (b) 13\dfrac{1}{\sqrt{3}}.
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Q372NEET 2024Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The graph which shows the variation of (1λ2)\left(\frac{1}{\lambda^2}\right) and its kinetic energy, EE is (where λ\lambda is de Broglie wavelength of a free particle):

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Answer: (d)

  1. λ=h2mE⇒1λ2=2mh2E\lambda = \dfrac{h}{\sqrt{2mE}} \Rightarrow \dfrac{1}{\lambda^2} = \dfrac{2m}{h^2}E.
  2. So 1λ2∝E\dfrac{1}{\lambda^2} \propto E: a straight line through the origin.
  3. So the answer is (d).
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Q373MHT CET 2024 · 15 MAY · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

Electrons are accelerated through a potential difference of 16 kV. If the potential difference is increased to 64 kV, then de-Broglie wavelength associated with electron will

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Answer: (b)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. λ′λ=1664=12\dfrac{\lambda'}{\lambda} = \sqrt{\dfrac{16}{64}} = \dfrac{1}{2}.
  3. So the answer is (b) becomes half.
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Q374MHT CET 2024 · 11 MAY · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

If the potential difference used to accelerate electrons is doubled. By what factor does the deBroglie wavelength (λ)(\lambda) associated with the electrons change?

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Answer: (c)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. V′=2V⇒λ′=λ2V' = 2V \Rightarrow \lambda' = \dfrac{\lambda}{\sqrt{2}}, a decrease.
  3. So the answer is (c) λ\lambda is decreased to 12\dfrac{1}{\sqrt{2}} times.
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Q375MHT CET 2024 · 10 MAY · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

If the potential difference used to accelerate electrons is increased four times, by what factor does the de-Broglie wavelength associated with the electrons change?

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Answer: (b)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. V′=4V⇒λ′=λ2V' = 4V \Rightarrow \lambda' = \dfrac{\lambda}{2}.
  3. So the answer is (b) wavelength decreased to half.
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Q376NEET 2024Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An electron and an alpha particle are accelerated by the same potential difference. Let λe\lambda_e and λα\lambda_\alpha denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

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Answer: (a)

  1. For a charge qq accelerated through VV: λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}, so λ∝1mq\lambda \propto \dfrac{1}{\sqrt{mq}} for the same VV.
  2. λeλα=mαqαmeqe=4mp⋅2eme⋅e=8mpme≫1\dfrac{\lambda_e}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_e q_e}} = \sqrt{\dfrac{4m_p \cdot 2e}{m_e \cdot e}} = \sqrt{\dfrac{8m_p}{m_e}} \gg 1
  3. So the answer is (a) λe>λα\lambda_e > \lambda_\alpha.
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Q377JEE Main 2024 · 9 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

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Answer: (b)

  1. Same kinetic energy: λ=h2mK∝1m\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}.
  2. Masses: me<mp<mαm_e < m_p < m_\alpha, so λe>λp>λα\lambda_e > \lambda_p > \lambda_\alpha.
  3. So the answer is (b) λα<λp<λe\lambda_\alpha < \lambda_p < \lambda_e.
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Q378JEE Main 2024 · 8 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton and an electron have the same de Broglie wavelength. If Kp\mathrm{K_p} and Ke\mathrm{K_e} be the kinetic energies of proton and electron respectively, then choose the correct relation:

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Answer: (c)

  1. Same λ=hp\lambda = \dfrac{h}{p} means same momentum pp.
  2. K=p22m∝1mK = \dfrac{p^2}{2m} \propto \dfrac{1}{m}; since mp>mem_p > m_e, Kp<KeK_p < K_e.
  3. So the answer is (c) Kp<KeK_p < K_e.
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Q379JEE Main 2024 · 1 FEB · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The de Broglie wavelengths of a proton and an α\alpha particle are λ\lambda and 2λ2\lambda respectively. The ratio of the velocities of proton and α\alpha particle will be:

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Answer: (a)

  1. λ=hmv⇒v=hmλ\lambda = \dfrac{h}{mv} \Rightarrow v = \dfrac{h}{m\lambda}.
  2. vpvα=mαλαmpλp=4mp⋅2λmp⋅λ=8\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha \lambda_\alpha}{m_p \lambda_p} = \dfrac{4m_p \cdot 2\lambda}{m_p \cdot \lambda} = 8
  3. So the answer is (a) 8:18 : 1.
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Q380JEE Main 2024 · 8 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume h = 6.63 ×10−34 J s, me=9.0×10−31 kg\times 10^{-34}\text{ J s},\ \mathrm{m_e} = 9.0 \times 10^{-31}\text{ kg} and mp=1836\mathrm{m_p} = 1836 times me\mathrm{m_e})

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Answer: (c)

  1. Same λ=h/p\lambda = h/p means same momentum pp.
  2. K=p22m⇒K∝1mK = \dfrac{p^2}{2m} \Rightarrow K \propto \dfrac{1}{m} for the same pp.
  3. KpKe=memp=11836\dfrac{K_p}{K_e} = \dfrac{m_e}{m_p} = \dfrac{1}{1836}
  4. So the answer is (c) 1:18361 : 1836.
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Q381MHT CET 2024 · 16 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

Kinetic energy of a proton is equal to energy EE of a photon. Let ' λ1\lambda_1 ' be the de-Broglie wavelength of proton and ' λ2\lambda_2 ' be the wavelength of photon. If (λ1λ2)∝En\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^n then the value of ' nn ' is

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Answer: (d)

  1. Proton: λ1=h2mE\lambda_1 = \dfrac{h}{\sqrt{2mE}}. Photon: λ2=hcE\lambda_2 = \dfrac{hc}{E}.
  2. λ1λ2=Ec2m∝E1/2\dfrac{\lambda_1}{\lambda_2} = \dfrac{\sqrt{E}}{c\sqrt{2m}} \propto E^{1/2}
  3. So n=0.5n = 0.5: the answer is (d).
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Q382MHT CET 2024 · 3 MAY · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The ratio of the wavelength of a photon of energy ' EE ' to that of the electron of same energy is ( m=m = mass of an electron, c=c = speed of light, h=h = Planck's constant)

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Answer: (d)

  1. λph=hcE\lambda_{ph} = \dfrac{hc}{E}, λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}.
  2. λphλe=c2mEE=c2mE\dfrac{\lambda_{ph}}{\lambda_e} = \dfrac{c\sqrt{2mE}}{E} = c\sqrt{\dfrac{2m}{E}}
  3. So the answer is (d).
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Q383JEE Main 2024 · 29 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25%25\% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

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Answer: (c)

  1. Same λ\lambda means same momentum p=mvp = mv.
  2. Electron: Ke=12mv2=12pvK_e = \frac{1}{2}mv^2 = \frac{1}{2}pv. Photon: E=pcE = pc.
  3. KeE=v2c=0.252=18\dfrac{K_e}{E} = \dfrac{v}{2c} = \dfrac{0.25}{2} = \dfrac{1}{8}
  4. So the answer is (c) 18\frac{1}{8}.
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Q384MHT CET 2024 · 3 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie waves & Bohr orbits

When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

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Answer: (d)

  1. λ∝n\lambda \propto n in a Bohr orbit.
  2. Ground state n=1n=1 to third excited state n=4n=4: λ\lambda becomes 44 times.
  3. So the answer is (d) will increase.
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Q385MHT CET 2023 · 14 MAY · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de-Broglie wavelength (λ)(\lambda) of a particle is related to its kinetic energy (E) as

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Answer: (d)

  1. p=2mEp = \sqrt{2mE}, so λ=h2mE\lambda = \dfrac{h}{\sqrt{2mE}}.
  2. Hence λ∝E−1/2\lambda \propto E^{-1/2}.
  3. So the answer is (d) λ∝E−12\lambda \propto E^{-\frac{1}{2}}.
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Q386JEE Main 2023 · 15 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de Broglie wavelength of an electron having kinetic energy E\mathrm{E} is λ\lambda. If the kinetic energy of electron becomes E4\frac{E}{4}, then its de-Broglie wavelength will be:

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Answer: (b)

  1. λ∝1E\lambda \propto \dfrac{1}{\sqrt{E}}.
  2. E′=E4⇒λ′=λ4=2λE' = \dfrac{E}{4} \Rightarrow \lambda' = \lambda\sqrt{4} = 2\lambda.
  3. So the answer is (b) 2λ2\lambda.
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Q387MHT CET 2023 · 12 MAY · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

When an electron is accelerated through a potential 'VV', the de-Broglie wavelength associated with it is '4λ4\lambda'. When the accelerating potential is increased to 4 V4\ \mathrm{V}, its wavelength will be

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Answer: (d)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. At VV the wavelength is 4λ4\lambda; at 4V4V it is 4λ4=2λ\dfrac{4\lambda}{\sqrt{4}} = 2\lambda.
  3. So the answer is (d) 2λ2\lambda.
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Q388MHT CET 2023 · 10 MAY · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

If the potential difference used to accelerate electrons is doubled, by what factor does the de- Broglie wavelength associated with electrons change?

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Answer: (c)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. V′=2V⇒λ′=λ2V' = 2V \Rightarrow \lambda' = \dfrac{\lambda}{\sqrt{2}}, a decrease.
  3. So the answer is (c) wavelength is decreased to 12\dfrac{1}{\sqrt{2}} times.
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Q389NEET 2023Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:

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Answer: (d)

  1. For an electron, λ=1.227V\lambda = \dfrac{1.227}{\sqrt{V}} nm.
  2. λ=1.22781=1.2279≈0.136\lambda = \dfrac{1.227}{\sqrt{81}} = \dfrac{1.227}{9} \approx 0.136 nm.
  3. So the answer is (d) 0.136 nm.
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Q390JEE Main 2023 · 25 JAN · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de- Broglie wavelength of λ0\lambda_0. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be:

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Answer: (c)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. λ′λ0=2040=12\dfrac{\lambda'}{\lambda_0} = \sqrt{\dfrac{20}{40}} = \dfrac{1}{\sqrt{2}}.
  3. So the answer is (c) λ02\dfrac{\lambda_0}{\sqrt{2}}.
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Q391JEE Main 2023 · 8 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

Proton (P)(\mathrm{P}) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, mp=1849 me\mathrm{m_p} = 1849\, \mathrm{m_e} ):

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Answer: (a)

  1. λ=hp\lambda = \dfrac{h}{p} depends only on momentum, not on mass.
  2. Same λ⇒pP=pe\lambda \Rightarrow p_P = p_e.
  3. So the answer is (a) 1:11 : 1.
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Q392JEE Main 2023 · 29 JAN · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The ratio of de-Broglie wavelength of an α\alpha particle and a proton accelerated from rest by the same potential is 1m\frac{1}{\sqrt{m}}, the value of m is -

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Answer: (c)

  1. λ=h2mqV⇒λ∝1mq\lambda = \dfrac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mq}} for the same VV.
  2. λαλp=mpe4mp⋅2e=18\dfrac{\lambda_\alpha}{\lambda_p} = \sqrt{\dfrac{m_p e}{4m_p \cdot 2e}} = \dfrac{1}{\sqrt{8}}
  3. So m=8m = 8, answer (c).
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Q393JEE Main 2023 · 24 JAN · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An α\alpha-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

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Answer: (d)

  1. Same kinetic energy: λ=h2mK∝1m\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}.
  2. mα>mp>me⇒λα<λp<λem_\alpha > m_p > m_e \Rightarrow \lambda_\alpha < \lambda_p < \lambda_e
  3. So the answer is (d).
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Q394JEE Main 2023 · 10 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ1\lambda_1. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

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Answer: (c)

  1. Thermal kinetic energy ∝T\propto T, so p=3mkTp = \sqrt{3mkT} and λ=h3mkT∝1T\lambda = \dfrac{h}{\sqrt{3mkT}} \propto \dfrac{1}{\sqrt{T}}.
  2. λ2λ1=300600=12\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{300}{600}} = \dfrac{1}{\sqrt{2}}
  3. So the answer is (c) 12λ1\dfrac{1}{\sqrt{2}}\lambda_1.
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Q395MHT CET 2023 · 12 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

Graph shows the variation of de-Broglie wavelength (λ)(\lambda) versus 1V\frac{1}{\sqrt{V}} where 'VV' is the accelerating potential for four particles A, B, C, D carrying same charge but of masses m1,m2,m3,m4\mathrm{m_1, m_2, m_3, m_4}. Which on represents a particle of largest mass?

Figure

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Answer: (a)

  1. λ=h2mq⋅1V\lambda = \dfrac{h}{\sqrt{2mq}} \cdot \dfrac{1}{\sqrt{V}}: a straight line through the origin with slope h2mq\dfrac{h}{\sqrt{2mq}}.
  2. Same qq, so slope ∝1m\propto \dfrac{1}{\sqrt{m}}: the largest mass has the smallest slope.
  3. The line for m1m_1 is the least steep.
  4. So the answer is (a) m1m_1.
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Q396MHT CET 2023 · 10 MAY · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

An electron accelerated through a potential difference 'V1V_1 ' has a de-Broglie wavelength 'λ\lambda'. When the potential is changed to 'V2V_2' its de-Broglie wavelength increases by 50%50\%. The value of (V1V2)\left(\frac{V_1}{V_2}\right) is

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Answer: (b)

  1. λ∝1V⇒V1V2=(λ2λ1)2\lambda \propto \dfrac{1}{\sqrt{V}} \Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2.
  2. A 50%50\% increase gives λ2=1.5λ\lambda_2 = 1.5\lambda.
  3. V1V2=(1.5)2=94\dfrac{V_1}{V_2} = (1.5)^2 = \dfrac{9}{4}
  4. So the answer is (b) 9:49 : 4.
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Q397JEE Main 2023 · 30 JAN · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

An electron accelerated through a potential difference V1V_1 has a de-Broglie wavelength of λ\lambda. When the potential is changed to V2V_2, its de-Broglie wavelength increases by 50%50\%. The value of (V1V2)\left(\frac{V_1}{V_2}\right) is equal to

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Answer: (d)

  1. λ∝1V⇒V1V2=(λ2λ1)2\lambda \propto \dfrac{1}{\sqrt{V}} \Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2.
  2. λ2=1.5λ⇒V1V2=(1.5)2=94\lambda_2 = 1.5\lambda \Rightarrow \dfrac{V_1}{V_2} = (1.5)^2 = \dfrac{9}{4}
  3. So the answer is (d) 94\dfrac{9}{4}.
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Q398JEE Main 2023 · 12 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton and an α\alpha-particle are accelerated from rest by 2 V2\text{ V} and 4 V4\text{ V} potentials, respectively. The ratio of their de-Broglie wavelength is:

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Answer: (a)

  1. λ=h2mqV⇒λ∝1mqV\lambda = \dfrac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mqV}}.
  2. Proton: m⋅e⋅2=2mem \cdot e \cdot 2 = 2me. Alpha: 4m⋅2e⋅4=32me4m \cdot 2e \cdot 4 = 32me.
  3. λpλα=322=4\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{32}{2}} = 4
  4. So the answer is (a) 4:14:1.
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Q399JEE Main 2023 · 11 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy: (Assume mp=me×1849m_p = m_e \times 1849)

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Answer: (a)

  1. λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}}; same KK gives λ∝1m\lambda \propto \dfrac{1}{\sqrt{m}}.
  2. λpλe=memp=11849=143\dfrac{\lambda_p}{\lambda_e} = \sqrt{\dfrac{m_e}{m_p}} = \dfrac{1}{\sqrt{1849}} = \dfrac{1}{43}
  3. So the answer is (a) 1:431:43.
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Q400JEE Main 2023 · 6 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The kinetic energy of an electron, α\alpha-particle and a proton are given as 4 K,2 K4\text{ K}, 2\text{ K} and K\text{K} respectively. The de-Broglie wavelength associated with electron (λe),α(\lambda \mathrm{e}), \alpha-particle (λα)(\lambda\alpha) and the proton (λp)(\lambda p) are as follows:

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Answer: (a)

  1. λ=h2mK⇒λ∝1mK\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mK}}.
  2. Electron: me(4K)m_e(4K), tiny since me≪mpm_e \ll m_p. Proton: mpKm_pK. Alpha: (4mp)(2K)=8mpK(4m_p)(2K) = 8m_pK.
  3. Largest mKmK gives smallest λ\lambda: λα<λp<λe\lambda_\alpha < \lambda_p < \lambda_e.
  4. So the answer is (a).
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Q401JEE Main 2023 · 1 FEB · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of λ\lambda. An alpha particle having certain kinetic energy has the same de-Broglie wavelength λ\lambda. The ratio of kinetic energy of proton and that of alpha particle is:

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Answer: (c)

  1. Same λ\lambda means same momentum pp.
  2. K=p22m⇒KpKα=mαmp=4mpmp=4K = \dfrac{p^2}{2m} \Rightarrow \dfrac{K_p}{K_\alpha} = \dfrac{m_\alpha}{m_p} = \dfrac{4m_p}{m_p} = 4
  3. So the answer is (c) 4:14:1.
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Q402MHT CET 2022 · 11 AUG · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

If the kinetic energy of a free electron doubles, it's de Broglie wavelength (λ\lambda) changes by a factor

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Answer: (a)

  1. λ=h2mK⇒λ∝1K\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{K}}.
  2. K′=2K⇒λ′=λ2K' = 2K \Rightarrow \lambda' = \dfrac{\lambda}{\sqrt{2}}.
  3. So the answer is (a) 12\dfrac{1}{\sqrt{2}}.
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Q403JEE Main 2022 · 28 JUL · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The equation λ=1.227x nm\lambda = \frac{1.227}{x}\text{ nm} can be used to find the de-Brogli wavelength of an electron. In this equation xx stands for:

Where

m=\mathrm{m} = mass of electron

P=\mathrm{P} = momentum of electron

K=\mathrm{K} = Kinetic energy of electron

V=\mathrm{V} = Accelerating potential in volts for electron

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Answer: (d)

  1. For an electron accelerated through VV: λ=h2meV=1.227V\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{1.227}{\sqrt{V}} nm.
  2. So x=Vx = \sqrt{V}.
  3. So the answer is (d) V\sqrt{V}.
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Q404JEE Main 2022 · 30 JUL · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

The de Broglie wavelength associated with an electron, accelerated through a potential difference of 15056 V15056\text{ V}, is:

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Answer: (a)

  1. For an electron, λ=12.27V\lambda = \dfrac{12.27}{\sqrt{V}} Å.
  2. 15056≈122.7\sqrt{15056} \approx 122.7, so λ=12.27122.7=0.1\lambda = \dfrac{12.27}{122.7} = 0.1 Å.
  3. So the answer is (a) 0.1 Å.
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Q405JEE Main 2022 · 29 JUL · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An α\alpha particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

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Answer: (b)

  1. After acceleration, K=qVK = qV and p=2mK=2mqVp = \sqrt{2mK} = \sqrt{2mqV}, so p∝mqp \propto \sqrt{mq}.
  2. pαpp=4mp⋅2emp⋅e=8=22\dfrac{p_\alpha}{p_p} = \sqrt{\dfrac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{8} = 2\sqrt{2}
  3. So the answer is (b) 22:12\sqrt{2} : 1.
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Q406JEE Main 2022 · 27 JUN · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An α\alpha particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths (λα:λC12)(\lambda_\alpha : \lambda_{C12}) is:

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Answer: (b)

  1. Same kinetic energy: λ=h2mK∝1m\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}.
  2. λαλC=mCmα=124=3\dfrac{\lambda_\alpha}{\lambda_C} = \sqrt{\dfrac{m_C}{m_\alpha}} = \sqrt{\dfrac{12}{4}} = \sqrt{3}
  3. So the answer is (b) 3:1\sqrt{3} : 1.
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Q407JEE Main 2022 · 25 JUN · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton, a neutron, an electron and an α\alpha particle have same energy. If λ\lambdap, λ\lambdan, λ\lambdae and λα\lambda_\alpha are the de Broglie's wavelengths of proton, neutron, electron and α\alpha particle respectively, then choose the correct relation from the following:

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Answer: (b)

  1. Same energy: λ=h2mK∝1m\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}.
  2. Masses: me<mp<mn<mαm_e < m_p < m_n < m_\alpha (the neutron is slightly heavier than the proton).
  3. So λα<λn<λp<λe\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e, answer (b).
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Q408JEE Main 2022 · 25 JUL · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1: 2\sqrt{2}. Then the ratio of Vp to Vd will be:

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Answer: (d)

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}; proton and deuteron have the same charge, md=2mpm_d = 2m_p.
  2. λpλd=mdVdmpVp=2VdVp=12\dfrac{\lambda_p}{\lambda_d} = \sqrt{\dfrac{m_dV_d}{m_pV_p}} = \sqrt{\dfrac{2V_d}{V_p}} = \dfrac{1}{\sqrt{2}}
  3. 2VdVp=12⇒VpVd=4\dfrac{2V_d}{V_p} = \dfrac{1}{2} \Rightarrow \dfrac{V_p}{V_d} = 4
  4. So the answer is (d) 4:14:1.
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Q409JEE Main 2022 · 28 JUN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The de Broglie wavelengths for an electron and a photon are λ\lambdae and λ\lambdap respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

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Answer: (a)

  1. Electron: λe=h2mK⇒K=h22mλe2\lambda_e = \dfrac{h}{\sqrt{2mK}} \Rightarrow K = \dfrac{h^2}{2m\lambda_e^2}.
  2. Photon with the same energy: λp=hcK=2mchλe2\lambda_p = \dfrac{hc}{K} = \dfrac{2mc}{h}\lambda_e^2
  3. So λp∝λe2\lambda_p \propto \lambda_e^2: the answer is (a).
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Q410JEE Main 2022 · 26 JUN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?

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Answer: (b)

  1. Same λ\lambda means same momentum: pe=pph=pp_e = p_{ph} = p, so (c) and (d) are wrong.
  2. Electron: Ee=12mv2=12pvE_e = \frac{1}{2}mv^2 = \frac{1}{2}pv. Photon: Eph=pcE_{ph} = pc.
  3. EeEph=v2c\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}
  4. So the answer is (b).
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Q411JEE Main 2022 · 27 JUL · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron (mass mm) with an initial velocity v⃗=v0i^ (v0>0)\vec{v} = v_0\hat{i}\ (v_0 > 0) is moving in an electric field E⃗=−E0i^ (E0>0)\vec{E} = -E_0\hat{i}\ (E_0 > 0) where E0E_0 is constant. If at t=0t = 0 de Broglie wavelength is λ0=hmv0\lambda_0 = \frac{h}{mv_0}, then its de Broglie wavelength after time t is given by

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Answer: (d)

  1. Force on electron =−eE⃗=eE0i^= -e\vec E = eE_0\hat i, along the motion.
  2. v=v0+eE0tmv = v_0 + \dfrac{eE_0t}{m}
  3. λ=hmv=λ01+eE0tmv0\lambda = \dfrac{h}{mv} = \dfrac{\lambda_0}{1+\frac{eE_0t}{mv_0}}
  4. So the answer is (d).
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Q412JEE Main 2021 · 26 AUG · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de-Broglie wavelength of a particle having kinetic energy E is λ\lambda. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

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Answer: (b)

  1. λ∝1E⇒E′=E(λλ′)2\lambda \propto \dfrac{1}{\sqrt{E}} \Rightarrow E' = E\left(\dfrac{\lambda}{\lambda'}\right)^2.
  2. λ′=0.75λ=34λ⇒E′=169E\lambda' = 0.75\lambda = \dfrac{3}{4}\lambda \Rightarrow E' = \dfrac{16}{9}E.
  3. Extra energy =169E−E=79E= \dfrac{16}{9}E - E = \dfrac{7}{9}E.
  4. So the answer is (b) 79E\dfrac{7}{9}E.
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Q413MHT CET 2021 · 23 SEP · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

de-Broglie wavelength associated with an electron accelerated through a potential difference 'V\mathrm{V}' is 'λ\lambda'. When the accelerating potential is increased to '4 V4\ \mathrm{V}', de-Broglie wavelength.

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Answer: (a)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. V′=4V⇒λ′=λ2V' = 4V \Rightarrow \lambda' = \dfrac{\lambda}{2}.
  3. So the answer is (a) reduces to half.
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Q414MHT CET 2021 · 21 SEP · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)

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Answer: (b)

  1. λ∝1mq\lambda \propto \dfrac{1}{\sqrt{mq}} for the same accelerating potential.
  2. λpλα=4m⋅2em⋅e=8=22\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{4m\cdot2e}{m\cdot e}} = \sqrt{8} = 2\sqrt{2}.
  3. So the answer is (b) 22:12\sqrt{2} : 1.
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Q415JEE Main 2021 · 31 AUG · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option:

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Answer: (a)

  1. Same λ=hP\lambda = \dfrac{h}{P} gives Pp=PeP_p = P_e.
  2. K=P22mK = \dfrac{P^2}{2m} and mp>mem_p > m_e, so Kp<KeK_p < K_e.
  3. So the answer is (a) Kp<KeK_p < K_e and Pp=PeP_p = P_e.
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Q416JEE Main 2021 · 25 JUL · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

What should be the order of arrangement of de-Broglie wavelength of electron (λ\lambdae), an α\alpha-particle (λ\lambdaa) and proton (λ\lambdap) given that all have the same kinetic energy?

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Answer: (c)

  1. Same kinetic energy: λ=h2mK∝1m\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}.
  2. me<mp<mα⇒λe>λp>λαm_e < m_p < m_\alpha \Rightarrow \lambda_e > \lambda_p > \lambda_\alpha
  3. So the answer is (c).
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Q417JEE Main 2021 · 22 JUL · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

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Answer: (d)

  1. λ=h2meV\lambda = \dfrac{h}{\sqrt{2meV}}; both have charge ee and the same VV, so λ∝1m\lambda \propto \dfrac{1}{\sqrt{m}}.
  2. λeλp=mpme\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}}
  3. So the answer is (d).
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Q418JEE Main 2021 · 16 MAR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (mp = 1.00727u me = 0.00055u)

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Answer: (d)

  1. Same charge and same VV: λeλp=mpme\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}}.
  2. 1.007270.00055=1831≈42.8≈43\sqrt{\dfrac{1.00727}{0.00055}} = \sqrt{1831} \approx 42.8 \approx 43
  3. So the answer is (d) 43:143 : 1.
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Q419JEE Main 2021 · 25 FEB · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength λelectronλproton\frac{\lambda_{electron}}{\lambda_{proton}} will be:

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Answer: (b)

  1. λ=hmv\lambda = \dfrac{h}{mv}; same speed gives λ∝1m\lambda \propto \dfrac{1}{m}.
  2. λeλp=mpme=1836\dfrac{\lambda_e}{\lambda_p} = \dfrac{m_p}{m_e} = 1836
  3. So the answer is (b) 1836.
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Q420JEE Main 2021 · 24 FEB · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

The de-Broglie wavelength of a proton and α\alpha-particle are equal. The ratio of their velocities is:

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Answer: (c)

  1. Equal λ\lambda means equal momentum: mpvp=mαvαm_p v_p = m_\alpha v_\alpha.
  2. vpvα=mαmp=4\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha}{m_p} = 4
  3. So the answer is (c) 4:14 : 1.
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Q421MHT CET 2021 · 20 SEP · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

What is the additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from 1 nm1\text{ nm} to 0.5 nm0.5\text{ nm}?

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Answer: (d)

  1. K∝1λ2K \propto \dfrac{1}{\lambda^2}.
  2. λ\lambda halves (1 nm→0.5 nm1\text{ nm} \to 0.5\text{ nm}) ⇒K′=4K\Rightarrow K' = 4K.
  3. Additional energy =4K−K=3K= 4K - K = 3K.
  4. So the answer is (d) three times its initial energy.
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Q422JEE Main 2021 · 18 MAR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

A particle is travelling 4 time as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2: 1, the mass of the particle is:

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Answer: (d)

  1. λ=hmv\lambda = \dfrac{h}{mv}, so λλe=mevemv\dfrac{\lambda}{\lambda_e} = \dfrac{m_e v_e}{m v}.
  2. With v=4vev = 4v_e: 2=me4m2 = \dfrac{m_e}{4m}
  3. m=me8m = \dfrac{m_e}{8}, so the answer is (d).
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Q423MHT CET 2021 · 23 SEP · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

According to de-Broglie hypothesis if an electron of mass 'mm' is accelerated by potential difference 'VV', the associated wavelength is 'λ\lambda'. When a proton of mass 'MM' is accelerated through potential difference 9 V9\text{ V}, then the wavelength associated with it is

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Answer: (a)

  1. Electron: λ=h2meV\lambda = \dfrac{h}{\sqrt{2meV}}.
  2. Proton: λp=h2Me(9V)=13mM⋅h2meV\lambda_p = \dfrac{h}{\sqrt{2Me(9V)}} = \dfrac{1}{3}\sqrt{\dfrac{m}{M}} \cdot \dfrac{h}{\sqrt{2meV}}
  3. λp=λ3mM\lambda_p = \dfrac{\lambda}{3}\sqrt{\dfrac{m}{M}}, so the answer is (a).
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Q424JEE Main 2021 · 25 FEB · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An α\alpha particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are λα\lambda_\alpha and λ\lambdap respectively. The ratio λpλα\frac{\lambda_p}{\lambda_\alpha} is:

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Answer: (b)

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}, same VV for both.
  2. λpλα=mαqαmpqp=4×2=8≈2.83\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4 \times 2} = \sqrt{8} \approx 2.83
  3. So the answer is (b) 2.8.
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Q425JEE Main 2021 · 1 SEP · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[me = mass of electron = 9 ×\times 10−-31 kg, h = Planck constant = 6.6 ×\times 10−-34 Js, kB = Boltzmann constant = 1.38 ×\times 10−-23 JK−1{}^{-1}]

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Answer: (a)

  1. λ=h3mkT\lambda = \dfrac{h}{\sqrt{3mkT}} (using K=32kTK = \frac{3}{2}kT).
  2. 3mkT=3×9×10−31×1.38×10−23×300≈1.118×10−503mkT = 3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300 \approx 1.118 \times 10^{-50}, so 3mkT≈1.057×10−25\sqrt{3mkT} \approx 1.057 \times 10^{-25}
  3. λ=6.6×10−341.057×10−25≈6.2×10−9 m\lambda = \dfrac{6.6 \times 10^{-34}}{1.057 \times 10^{-25}} \approx 6.2 \times 10^{-9}\text{ m}
  4. So the answer is (a) 6.26 nm.
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Q426MHT CET 2021 · 23 SEP · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electron of mass 'mm' and a photon have same energy 'EE'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is (c=(c = velocity of light)

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Answer: (c)

  1. λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}, λph=hcE\lambda_{ph} = \dfrac{hc}{E}.
  2. λeλph=Ec2mE=1cE2m\dfrac{\lambda_e}{\lambda_{ph}} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{1}{c}\sqrt{\dfrac{E}{2m}}
  3. So the answer is (c).
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Q427MHT CET 2021 · 22 SEP · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

Kinetic energy of a proton is equal to energy 'EE' of a photon. Let 'λ1\lambda_1' be the de-Broglie wavelength of proton and 'λ2\lambda_2' is the wavelength of photon. If λ1λ2∝En\frac{\lambda_1}{\lambda_2} \propto E^{n}, then the value of 'nn' is

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Answer: (a)

  1. Proton: λ1=h2mE\lambda_1 = \dfrac{h}{\sqrt{2mE}}. Photon: λ2=hcE\lambda_2 = \dfrac{hc}{E}.
  2. λ1λ2=Ec2m∝E1/2\dfrac{\lambda_1}{\lambda_2} = \dfrac{\sqrt{E}}{c\sqrt{2m}} \propto E^{1/2}
  3. So n=12n = \frac{1}{2}: the answer is (a).
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Q428NEET 2021Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electromagnetic wave of wavelength 'λ\lambda' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ\lambdad, then:

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Answer: (d)

  1. Negligible work function: hcλ=p22m\dfrac{hc}{\lambda} = \dfrac{p^2}{2m} with p=hλdp = \dfrac{h}{\lambda_d}.
  2. hcλ=h22mλd2⇒λ=(2mch)λd2\dfrac{hc}{\lambda} = \dfrac{h^2}{2m\lambda_d^2} \Rightarrow \lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2
  3. So the answer is (d).
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Q429JEE Main 2021 · 25 JUL · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electron moving with speed v and a photon moving with speed c, have same de-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is:

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Answer: (c)

  1. Same λ\lambda means same momentum pp.
  2. Electron: K=12pvK = \frac{1}{2}pv. Photon: E=pcE = pc.
  3. KE=v2c\dfrac{K}{E} = \dfrac{v}{2c}
  4. So the answer is (c) v2c\frac{v}{2c}.
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Q430JEE Main 2021 · 17 MAR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is: (c being the velocity of light)

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Answer: (b)

  1. Electron: p=2mEp = \sqrt{2mE}, so λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}.
  2. Photon: E=hcλpE = \dfrac{hc}{\lambda_p}, so λp=hcE\lambda_p = \dfrac{hc}{E}.
  3. λeλp=Ec2mE=1c(E2m)1/2\dfrac{\lambda_e}{\lambda_p} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{1}{c}\left(\dfrac{E}{2m}\right)^{1/2}
  4. So the answer is (b).
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Q431JEE Main 2021 · 27 JUL · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

A particle of mass 9.1 ×\times 10−-31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation of linear momentum 10−-27 kg m/s travels in vacuum. The wavelength of photon is __________ times the wavelength of the particle.

Numerical answer type. Work it out, then open the answer.

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Answer: 910

  1. λ=h/p\lambda = h/p, so λphotonλparticle=pparticlepphoton\dfrac{\lambda_{photon}}{\lambda_{particle}} = \dfrac{p_{particle}}{p_{photon}}.
  2. pparticle=9.1×10−31×106=9.1×10−25p_{particle} = 9.1\times10^{-31}\times10^6 = 9.1\times10^{-25} kg m/s
  3. Ratio =9.1×10−2510−27=910= \dfrac{9.1\times10^{-25}}{10^{-27}} = 910
  4. Answer: 910
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Q432JEE Main 2021 · 25 JUL · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be:

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Answer: (d)

  1. Initial momentum is zero, so the two fragments have equal and opposite momenta: ∣p1∣=∣p2∣|p_1| = |p_2|.
  2. λ=h/p\lambda = h/p depends only on momentum.
  3. λ1:λ2=1:1\lambda_1 : \lambda_2 = 1 : 1
  4. So the answer is (d).
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Q433NEET 2020 · PHASE 2Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly

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Answer: (b)

  1. For an electron, λ=1.227V\lambda = \dfrac{1.227}{\sqrt{V}} nm, where K=eVK = eV, so V=144V = 144 V.
  2. λ=1.22712≈0.102\lambda = \dfrac{1.227}{12} \approx 0.102 nm.
  3. So the answer is (b) 102×10−3102\times10^{-3} nm.
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Q434JEE Main 2020 · 9 JAN · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

A particle moving with kinetic energy E has de Broglie wavelength λ\lambda. If energy Δ\DeltaE is added to its energy, the wavelength become λ\lambda/2. Value of Δ\DeltaE, is:

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Answer: (b)

  1. λ∝1E\lambda \propto \dfrac{1}{\sqrt{E}}, so halving λ\lambda needs E′=4EE' = 4E.
  2. ΔE=4E−E=3E\Delta E = 4E - E = 3E.
  3. So the answer is (b) 3E.
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Q435NEET 2020 · PHASE 1Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

An electron is accelerated from rest through a potential difference of V volt. If the de-Broglie wavelength of the electron is 1.227 ×\times 10-2 nm, the potential difference is:

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Answer: (c)

  1. λ=1.227V\lambda = \dfrac{1.227}{\sqrt{V}} nm ⇒V=1.2271.227×10−2=100\Rightarrow \sqrt{V} = \dfrac{1.227}{1.227\times10^{-2}} = 100.
  2. V=104V = 10^4 V.
  3. So the answer is (c) 10410^4 V.
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Q436JEE Main 2020 · 6 SEP · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An electron, a doubly ionized helium ion (He++) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths λ\lambdae, λ\lambdaHe++ and λ\lambdap is:

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Answer: (c)

  1. He++^{++} is an alpha particle. Same kinetic energy: λ∝1m\lambda \propto \dfrac{1}{\sqrt{m}}.
  2. me<mp<mHem_e < m_p < m_{He}, so λe>λp>λHe++\lambda_e > \lambda_p > \lambda_{He^{++}}.
  3. So the answer is (c).
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Q437JEE Main 2020 · 2 SEP · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.878 ×\times 10–4. The mass of the particle is close to

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Answer: (d)

  1. λλe=mevemv=me5m\dfrac{\lambda}{\lambda_e} = \dfrac{m_e v_e}{m v} = \dfrac{m_e}{5m} (since v=5vev = 5v_e).
  2. m=me5×1.878×10−4=9.1×10−319.39×10−4m = \dfrac{m_e}{5 \times 1.878 \times 10^{-4}} = \dfrac{9.1 \times 10^{-31}}{9.39 \times 10^{-4}}
  3. m≈9.7×10−28 kgm \approx 9.7 \times 10^{-28}\text{ kg}, so the answer is (d).
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Q438JEE Main 2020 · 6 SEP · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to:

(Given: nitrogen molecule weight: 4.64 ×\times 10–26 kg,

Boltzmann constant: 1.38 ×\times 10–23 J/K,

Planck constant: 6.63 ×\times 10–34 J.s)

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Answer: (d)

  1. vrms=3kT/m⇒p=3mkTv_{rms} = \sqrt{3kT/m} \Rightarrow p = \sqrt{3mkT}, λ=h3mkT\lambda = \dfrac{h}{\sqrt{3mkT}}.
  2. 3mkT=3×4.64×10−26×1.38×10−23×400≈7.68×10−463mkT = 3 \times 4.64 \times 10^{-26} \times 1.38 \times 10^{-23} \times 400 \approx 7.68 \times 10^{-46}, so 3mkT≈2.77×10−23\sqrt{3mkT} \approx 2.77 \times 10^{-23}
  3. λ=6.63×10−342.77×10−23≈2.4×10−11 m=0.24 A˚\lambda = \dfrac{6.63 \times 10^{-34}}{2.77 \times 10^{-23}} \approx 2.4 \times 10^{-11}\text{ m} = 0.24\ \mathring{A}
  4. So the answer is (d) 0.24 Å.
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Q439JEE Main 2020 · 7 JAN · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)

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Answer: (b)

  1. Electron (non-relativistic, few eV): λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}.
  2. Photon: λp=hcE\lambda_p = \dfrac{hc}{E}.
  3. λeλp=Ec2mE=1c(E2m)1/2\dfrac{\lambda_e}{\lambda_p} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{1}{c}\left(\dfrac{E}{2m}\right)^{1/2}
  4. So the answer is (b).
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Q440JEE Main 2020 · 8 JAN · SHIFT 2Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron (mass m) with initial velocity v⃗=v0i^+v0j^\vec{v} = v_0\hat{i} + v_0\hat{j} is in an electric field E⃗=−E0k^\vec{E} = -E_0\hat{k}. If λ0\lambda_0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by:

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Answer: (c)

  1. Initial speed 2v0\sqrt2 v_0, so λ0=h2mv0\lambda_0 = \dfrac{h}{\sqrt2 mv_0}.
  2. Force eE0k^eE_0\hat k adds vz=eE0tmv_z = \dfrac{eE_0t}{m}; v2=2v02+e2E02t2m2v^2 = 2v_0^2 + \dfrac{e^2E_0^2t^2}{m^2}.
  3. λ=hmv=λ01+e2E02t22m2v02\lambda = \dfrac{h}{mv} = \dfrac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{2m^2v_0^2}}}
  4. So the answer is (c).
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Q441JEE Main 2020 · 9 JAN · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is:

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Answer: (c)

  1. From rest under constant force ∣e∣E|e|E: p=∣e∣Etp = |e|Et.
  2. λ=h∣e∣Et⇒dλdt=−h∣e∣Et2\lambda = \dfrac{h}{|e|Et} \Rightarrow \dfrac{d\lambda}{dt} = \dfrac{-h}{|e|Et^2}
  3. So the answer is (c).
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Q442JEE Main 2020 · 4 SEP · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

Particle A of mass mA = m2\frac{m}{2} moving along the x-axis with velocity v0 collides elastically with another particle B at rest having mass mB = m3\frac{m}{3}. If both particles move along the x-axis after the collision, the change Δλ\Delta\lambda in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength (λ\lambda0) before collision is:

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Answer: (d)

  1. Elastic, B at rest: vA′=mA−mBmA+mBv0=m/2−m/3m/2+m/3v0=v05v_A' = \dfrac{m_A - m_B}{m_A + m_B}v_0 = \dfrac{m/2 - m/3}{m/2 + m/3}v_0 = \dfrac{v_0}{5}
  2. λA′=hmAv0/5=5λ0\lambda_A' = \dfrac{h}{m_A v_0/5} = 5\lambda_0
  3. Δλ=5λ0−λ0=4λ0\Delta\lambda = 5\lambda_0 - \lambda_0 = 4\lambda_0
  4. So the answer is (d).
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Q443NEET 2019Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is (nearly): (me = 9 ×\times 10–31 kg)

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Answer: (a)

  1. λ=1.227V\lambda = \dfrac{1.227}{\sqrt{V}} nm =1.227104= \dfrac{1.227}{\sqrt{10^4}} nm =0.01227= 0.01227 nm.
  2. λ≈12.2×10−12\lambda \approx 12.2\times10^{-12} m.
  3. So the answer is (a) 12.2×10−1212.2\times10^{-12} m.
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Q444NEET 2019Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A proton and an α\alpha-particle are accelerated from rest to the same energy. The de Broglie wavelengths λp\lambda_p and λα\lambda_\alpha are in the ratio,

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Answer: (b)

  1. Same kinetic energy KK: λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}}, so λ∝1m\lambda \propto \dfrac{1}{\sqrt{m}}.
  2. λpλα=mαmp=4=2\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha}{m_p}} = \sqrt{4} = 2
  3. So the answer is (b) 2:12 : 1.
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Q445JEE Main 2019 · 12 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths λAλB\frac{\lambda_A}{\lambda_B} is close to:

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Answer: (c)

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}, same charge qq.
  2. λAλB=mBVBmAVA=4m×2500m×50=200\dfrac{\lambda_A}{\lambda_B} = \sqrt{\dfrac{m_BV_B}{m_AV_A}} = \sqrt{\dfrac{4m \times 2500}{m \times 50}} = \sqrt{200}
  3. 200≈14.14\sqrt{200} \approx 14.14
  4. So the answer is (c) 14.14.
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Q446JEE Main 2019 · 11 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

If the de Broglie wavelength of an electron is equal to 10–3 times the wavelength of a photon of frequency 6 ×\times 1014 Hz, then the speed of electron is equal to: (Speed of light = 3 ×\times 108 m/s, Planck's constant = 6.63 ×\times 10–34 J.s, Mass of electron = 9.1 ×\times 10–31 kg)

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Answer: (b)

  1. Photon: λp=cf=3×1086×1014=5×10−7\lambda_p = \dfrac{c}{f} = \dfrac{3\times10^8}{6\times10^{14}} = 5\times10^{-7} m.
  2. Electron: λe=10−3λp=5×10−10\lambda_e = 10^{-3}\lambda_p = 5\times10^{-10} m.
  3. v=hmλe=6.63×10−349.1×10−31×5×10−10≈1.45×106v = \dfrac{h}{m\lambda_e} = \dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times5\times10^{-10}} \approx 1.45\times10^6 m/s
  4. So the answer is (b).
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Q447JEE Main 2019 · 9 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths 'λ\lambdax' and 'λ\lambday' respectively. If x and y were moving in opposite directions, then the de- Broglie wavelength of 'P' is :-

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Answer: (b)

  1. Opposite directions: p=∣px−py∣=h∣1λx−1λy∣p = |p_x - p_y| = h\left|\dfrac{1}{\lambda_x} - \dfrac{1}{\lambda_y}\right|
  2. λP=hp=λxλy∣λx−λy∣\lambda_P = \dfrac{h}{p} = \dfrac{\lambda_x\lambda_y}{|\lambda_x - \lambda_y|}
  3. So the answer is (b).
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Q448JEE Main 2019 · 8 APR · SHIFT 2Difficult

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

A nucleus A, with a finite de-broglie wavelength λ\lambdaA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths λ\lambdaB and λ\lambdaC of B and C are respectively:

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Answer: (d)

  1. Let A have mass MM, speed vv; B and C have mass M/2M/2, velocities vBv_B and −vB/2-v_B/2.
  2. Momentum: Mv=M2vB−M2⋅vB2=M4vB⇒vB=4vMv = \dfrac{M}{2}v_B - \dfrac{M}{2}\cdot\dfrac{v_B}{2} = \dfrac{M}{4}v_B \Rightarrow v_B = 4v
  3. pB=2Mv=2pA⇒λB=λA2p_B = 2Mv = 2p_A \Rightarrow \lambda_B = \dfrac{\lambda_A}{2}; pC=M2(2v)=Mv⇒λC=λAp_C = \dfrac{M}{2}(2v) = Mv \Rightarrow \lambda_C = \lambda_A
  4. So the answer is (d).
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Q449JEE Main 2019 · 8 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

Two particles move at right angle to each other. Their de-Broglie wavelengths are λ1\lambda_1 and λ2\lambda_2 respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength λ\lambda of the final particle, is given by:

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Answer: (b)

p₁ = h/λ₁p₂p = h/λ
  1. Momentum is conserved; p⃗1⊥p⃗2\vec p_1 \perp \vec p_2, so p2=p12+p22p^2 = p_1^2 + p_2^2.
  2. With p=h/λp = h/\lambda: 1λ2=1λ12+1λ22\dfrac{1}{\lambda^2} = \dfrac{1}{\lambda_1^2} + \dfrac{1}{\lambda_2^2}
  3. So the answer is (b).
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Q450JEE Main 2018 · 15 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

If the de Broglie wavelengths associated with a proton and an α\alpha-particle are equal, then the ratio of velocities of the proton and the α\alpha-particle will be:

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Answer: (a)

  1. Equal λ=hmv\lambda = \dfrac{h}{mv} means mpvp=mαvαm_p v_p = m_\alpha v_\alpha.
  2. vpvα=mαmp=4\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha}{m_p} = 4
  3. So the answer is (a) 4:14 : 1.
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Q451NEET 2018Medium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An electron of mass m with an initial velocityv⃗=v0i^\vec{v} = v_0\hat{i} (v0 > 0) enters an electric field E⃗=−E⃗0i^\vec{E} = -\vec{E}_0\hat{i} (E0 = constant > 0) at t = 0. If λ\lambda0 is its de-Broglie wavelength initially, then its de- Broglie wavelength at time t is

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Answer: (a)

  1. Force on electron F⃗=−eE⃗=+eE0i^\vec F = -e\vec E = +eE_0\hat i, along v⃗\vec v: it speeds up.
  2. v=v0+eE0mtv = v_0 + \dfrac{eE_0}{m}t
  3. λ=hmv=hmv0(1+eE0mv0t)=λ01+eE0mv0t\lambda = \dfrac{h}{mv} = \dfrac{h}{mv_0\left(1+\frac{eE_0}{mv_0}t\right)} = \dfrac{\lambda_0}{1+\frac{eE_0}{mv_0}t}
  4. So the answer is (a).
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Q452JEE Main 2018 · 16 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · de Broglie waves & Bohr orbits

The de-Broglie wavelength (λ\lambdaB) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λ\lambdaG) by:

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Answer: (b)

  1. λ=2πrnn∝n\lambda = \dfrac{2\pi r_n}{n} \propto n
  2. Second excited state: n=3n = 3; ground state: n=1n = 1.
  3. λB=3λG\lambda_B = 3\lambda_G
  4. So the answer is (b).
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Q453JEE Main 2018 · 15 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are λ1\lambda_1 and λ2,\lambda_2, their de Broglie wavelength in the frame of reference attached to their center of masses:

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Answer: (b)

  1. Equal masses: v⃗CM=v⃗1+v⃗22\vec v_{CM} = \dfrac{\vec v_1 + \vec v_2}{2}, so each electron's velocity relative to CM is v⃗1−v⃗22\dfrac{\vec v_1 - \vec v_2}{2} in magnitude.
  2. v⃗1⊥v⃗2\vec v_1 \perp \vec v_2: pCM=12p12+p22=h21λ12+1λ22p_{CM} = \dfrac{1}{2}\sqrt{p_1^2 + p_2^2} = \dfrac{h}{2}\sqrt{\dfrac{1}{\lambda_1^2} + \dfrac{1}{\lambda_2^2}}
  3. λCM=hpCM=2λ1λ2λ12+λ22\lambda_{CM} = \dfrac{h}{p_{CM}} = \dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}
  4. So the answer is (b).
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Q454NEET 2017Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (kelvin) and mass m, is

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Answer: (a)

  1. In thermal equilibrium, K=32kTK = \frac{3}{2}kT.
  2. p=2mK=3mkTp = \sqrt{2mK} = \sqrt{3mkT}
  3. λ=hp=h3mkT\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{3mkT}}
  4. So the answer is (a).
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Q455JEE Main 2017Medium

Matter Waves: de Broglie Wavelength · Collisions, decay & fission

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA\lambda_A to λB\lambda_B after the collision is:

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Answer: (b)

  1. Elastic head-on, B at rest: vA=m−m/2m+m/2v=v3v_A = \dfrac{m - m/2}{m + m/2}v = \dfrac{v}{3}, vB=2mm+m/2v=4v3v_B = \dfrac{2m}{m+m/2}v = \dfrac{4v}{3}.
  2. pA=mv3p_A = \dfrac{mv}{3}, pB=m2⋅4v3=2mv3p_B = \dfrac{m}{2}\cdot\dfrac{4v}{3} = \dfrac{2mv}{3}
  3. λAλB=pBpA=2\dfrac{\lambda_A}{\lambda_B} = \dfrac{p_B}{p_A} = 2
  4. So the answer is (b).
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Q456NEET 2016 · PHASE 1Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is

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Answer: (c)

  1. Electron: λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}. Photon: λph=hcE\lambda_{ph} = \dfrac{hc}{E}.
  2. λeλph=Ec2mE=1c(E2m)1/2\dfrac{\lambda_e}{\lambda_{ph}} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{1}{c}\left(\dfrac{E}{2m}\right)^{1/2}
  3. So the answer is (c).
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Q457NEET 2015 · AIPMT · CANCELLED PAPEREasy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

Which of the following figures represent the variation of particle momentum and the associated de- Broglie wavelength?

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Answer: (d)

  1. p=hλp = \dfrac{h}{\lambda}, so pλ=p\lambda = constant.
  2. The pp vs λ\lambda graph is a rectangular hyperbola (p falls as λ\lambda rises, never touching the axes).
  3. So the answer is (d).
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Q458JEE Main 2015 · 10 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to (∣e∣|e| = 1.6 ×\times 10−-19 C, me = 9.1 ×\times 10−-31 kg, h = 6.6 ×\times 10−-34 Js):

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Answer: (c)

  1. λ=h2meV\lambda = \dfrac{h}{\sqrt{2meV}}.
  2. 2×9.1×10−31×1.6×10−19×50=1.456×10−47≈3.82×10−24\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times50} = \sqrt{1.456\times10^{-47}} \approx 3.82\times10^{-24} kg m/s.
  3. λ=6.6×10−343.82×10−24≈1.7×10−10\lambda = \dfrac{6.6\times10^{-34}}{3.82\times10^{-24}} \approx 1.7\times10^{-10} m.
  4. So the answer is (c) 1.7 Å.
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Q459JEE Main 2014 · 9 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

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Answer: (d)

  1. λ=hmv\lambda = \dfrac{h}{mv}: a large mass gives an extremely small wavelength.
  2. A dust particle is far heavier than an electron, proton or α\alpha-particle, so its λ\lambda is too small to detect by diffraction.
  3. So the answer is (d) a dust particle.
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Q460NEET 2014 · AIPMTMedium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is

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Answer: (b)

  1. λ∝1K\lambda \propto \dfrac{1}{\sqrt{K}}.
  2. K′=16K⇒λ′=λ4K' = 16K \Rightarrow \lambda' = \dfrac{\lambda}{4}, a decrease of 34=75%\dfrac{3}{4} = 75\%.
  3. So the answer is (b) 75.
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Q461JEE Main 2013 · 23 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are λe\lambda_e and λp\lambda_p for electrons and protons respectively. The ratio of λeλp\frac{\lambda_e}{\lambda_p} is given by: (given mem_e is mass of electron and mpm_p is mass of proton).

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Answer: (d)

  1. λ=h2meV\lambda = \dfrac{h}{\sqrt{2meV}}, same charge magnitude.
  2. λeλp=mp(4V)meV=2mpme\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p(4V)}{m_eV}} = 2\sqrt{\dfrac{m_p}{m_e}}
  3. So the answer is (d).
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Q462NEET 2013Medium

Matter Waves: de Broglie Wavelength · Thermal neutrons & gas molecules

The de-broglie wavelength of neutrons in thermal equilibrium at temperature T is

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Answer: (d)

  1. λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}} with mn=1.675×10−27 kgm_n = 1.675 \times 10^{-27}\text{ kg}.
  2. Taking K≈kTK \approx kT: 2mnk=2×1.675×10−27×1.38×10−23≈2.15×10−25\sqrt{2m_nk} = \sqrt{2 \times 1.675 \times 10^{-27} \times 1.38 \times 10^{-23}} \approx 2.15 \times 10^{-25}
  3. λ=6.63×10−342.15×10−25T≈3.08×10−9T m=30.8T A˚\lambda = \dfrac{6.63 \times 10^{-34}}{2.15 \times 10^{-25}\sqrt{T}} \approx \dfrac{3.08 \times 10^{-9}}{\sqrt{T}}\text{ m} = \dfrac{30.8}{\sqrt{T}}\ \mathring{A}
  4. (With K=32kTK = \frac{3}{2}kT the value is 25.2/T25.2/\sqrt{T} Å, not listed, so the options use K=kTK = kT.)
  5. So the answer is (d) 30.8T A˚\frac{30.8}{\sqrt{T}}\ \mathring{A}.
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Q463NEET 2013Medium

Matter Waves: de Broglie Wavelength · Photon vs electron

The wavelength λ\lambdae of an electron and λ\lambdap of a photon of same energy E are related by

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Answer: (c)

  1. Electron: λe=h2mE⇒E=h22mλe2\lambda_e = \dfrac{h}{\sqrt{2mE}} \Rightarrow E = \dfrac{h^2}{2m\lambda_e^2}.
  2. Photon: λp=hcE=2mchλe2\lambda_p = \dfrac{hc}{E} = \dfrac{2mc}{h}\lambda_e^2
  3. So λp∝λe2\lambda_p \propto \lambda_e^2: the answer is (c).
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Q464JEE Main 2013 · 22 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength · de Broglie waves & Bohr orbits

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nthn^{th} orbital will therefore be proportional to:

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Answer: (c)

  1. Condition: 2πr=nλ=nhp2\pi r = n\lambda = \dfrac{nh}{p}.
  2. In field BB: p=qBrp = qBr.
  3. 2πr=nhqBr⇒r2=nh2πqB⇒r∝n1/22\pi r = \dfrac{nh}{qBr} \Rightarrow r^2 = \dfrac{nh}{2\pi qB} \Rightarrow r \propto n^{1/2}
  4. So the answer is (c).
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Q465NEET 2012 · AIPMT · MAINSMedium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

If the momentum of an electron is changed by P, then the de Broglie wavelength associated with changes by 0.5%. The initial momentum of electron will be

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Answer: (a)

  1. λ=hp⇒Δλλ=Δpp\lambda = \dfrac{h}{p} \Rightarrow \dfrac{\Delta\lambda}{\lambda} = \dfrac{\Delta p}{p} (in magnitude).
  2. Pp0=0.5%=1200\dfrac{P}{p_0} = 0.5\% = \dfrac{1}{200}
  3. p0=200Pp_0 = 200P, so the answer is (a).
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Q466NEET 2012 · AIPMT · PRELIMSMedium

Matter Waves: de Broglie Wavelength · de Broglie wavelength in electric & magnetic fields

An α\alpha-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de Broglie wavelength associated with the particle will be

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Answer: (d)

  1. Circular motion: r=mvqB⇒p=qBrr = \dfrac{mv}{qB} \Rightarrow p = qBr, so λ=hqBr\lambda = \dfrac{h}{qBr}.
  2. q=2e=3.2×10−19q = 2e = 3.2\times10^{-19} C, qBr=3.2×10−19×0.25×0.83×10−2=6.64×10−22qBr = 3.2\times10^{-19}\times0.25\times0.83\times10^{-2} = 6.64\times10^{-22}
  3. λ=6.63×10−346.64×10−22≈10−12\lambda = \dfrac{6.63\times10^{-34}}{6.64\times10^{-22}} \approx 10^{-12} m =0.01= 0.01 Å
  4. So the answer is (d) 0.01 Å.
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Q467NEET 2011 · AIPMT · PRELIMSEasy

Matter Waves: de Broglie Wavelength · Particle accelerated through a potential

Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would

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Answer: (b)

  1. λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.
  2. λ′λ=25100=12\dfrac{\lambda'}{\lambda} = \sqrt{\dfrac{25}{100}} = \dfrac{1}{2}.
  3. So the answer is (b) decrease by 2 times.
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Q468NEET 2008 · AIPMTMedium

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 ×\times 106 m s−1{}^{-1}. The velocity of the particle is

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Answer: (c)

  1. Same λ=hmv\lambda = \dfrac{h}{mv} means same momentum: mv=mevem v = m_e v_e.
  2. m=1 mg=10−6 kgm = 1\text{ mg} = 10^{-6}\text{ kg}
  3. v=9.1×10−31×3×10610−6=2.7×10−18 m s−1v = \dfrac{9.1 \times 10^{-31} \times 3 \times 10^{6}}{10^{-6}} = 2.7 \times 10^{-18}\text{ m s}^{-1}
  4. So the answer is (c).
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Q469JEE Main 2005 · AIEEEEasy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor

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Answer: (d)

  1. λ=h2mK⇒λ∝1K\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{K}}.
  2. K′=2K⇒λ′=λ2K' = 2K \Rightarrow \lambda' = \dfrac{\lambda}{\sqrt{2}}.
  3. So the answer is (d) 12\dfrac{1}{\sqrt{2}}.
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Q470NEET 2002 · AIPMTEasy

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

If particles are moving with same velocity, then which has maximum de Broglie wavelength?

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Answer: (d)

  1. λ=hmv\lambda = \dfrac{h}{mv}; for the same vv, λ∝1m\lambda \propto \dfrac{1}{m}.
  2. The β\beta-particle (electron) has the smallest mass, so the largest λ\lambda.
  3. So the answer is (d) β\beta-particle.
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Q471NEET 1996 · AIPMTEasy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

An electron beam has a kinetic energy equal to 100 eV. Find its wavelength associated with a beam, if mass of electron = 9.1×10−319.1 \times 10^{-31} kg and 1 eV = 1.6×10−191.6 \times 10^{-19} J/eV.

(Planck's constant = 6.6×10−346.6 \times 10^{-34} Js)

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Answer: (c)

  1. λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}}, with K=100×1.6×10−19=1.6×10−17K = 100\times1.6\times10^{-19} = 1.6\times10^{-17} J.
  2. 2×9.1×10−31×1.6×10−17≈5.4×10−24\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-17}} \approx 5.4\times10^{-24} kg m/s.
  3. λ=6.6×10−345.4×10−24≈1.2×10−10\lambda = \dfrac{6.6\times10^{-34}}{5.4\times10^{-24}} \approx 1.2\times10^{-10} m.
  4. So the answer is (c) 1.2 Å.
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Q472NEET 1995 · AIPMTEasy

Matter Waves: de Broglie Wavelength · Photon vs electron

If we consider electrons and photons of same wavelength, then they will have same

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Answer: (a)

  1. For both photons and electrons, λ=hp\lambda = \dfrac{h}{p}.
  2. Same λ\lambda gives the same momentum p=hλp = \dfrac{h}{\lambda}; energy and speed differ.
  3. So the answer is (a) momentum.
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Q473NEET 1995 · AIPMTMedium

Matter Waves: de Broglie Wavelength · Comparing electron, proton & alpha

An electron of mass mm, when accelerated through a potential difference VV, has de Broglie wavelength λ\lambda. The de Broglie wavelength associated with a proton of mass MM accelerated through the same potential difference, will be

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Answer: (d)

  1. For a charge qq accelerated through VV: λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}.
  2. Electron and proton have the same ∣q∣|q| and the same VV, so λ∝1m\lambda \propto \dfrac{1}{\sqrt{m}}.
  3. λpλ=mM⇒λp=λmM\dfrac{\lambda_p}{\lambda} = \sqrt{\dfrac{m}{M}} \Rightarrow \lambda_p = \lambda\sqrt{\dfrac{m}{M}}
  4. So the answer is (d) λmM\lambda\sqrt{\frac{m}{M}}.
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Q474NEET 1989 · AIPMTEasy

Matter Waves: de Broglie Wavelength · de Broglie wavelength, momentum & KE

The de Broglie wave corresponding to a particle of mass mm and velocity vv has a wavelength associated with it

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Answer: (a)

  1. de Broglie relation: λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{mv}.
  2. So the answer is (a) hmv\dfrac{h}{mv}.
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Q475JEE Main 2023 · 31 JAN · SHIFT 1EasyOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R

Assertion A: The beam of electrons show wave nature and exhibit interference and diffraction. Reason R: Davisson Germer Experimentally verified the wave nature of electrons. In the light of the above statements, choose the most appropriate answer from the options given below:

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Answer: (b)

  1. A: electron beams do show interference and diffraction (wave nature) — true.
  2. R: Davisson–Germer electron diffraction experimentally established this — true and it is the basis for A.
  3. So the answer is (b).
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Q476NEET 2023Easy

Davisson-Germer, Uncertainty, Compton & X-Rays · X-rays & cut-off wavelength

The minimum wavelength of XX-rays produced by an electron accelerated through a potential difference of VV volts is proportional to:

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Answer: (a)

  1. The whole kinetic energy eVeV goes into one photon for the shortest wavelength: eV=hcλmineV = \frac{hc}{\lambda_{min}}.
  2. λmin=hceV⇒λmin∝1V\lambda_{min} = \frac{hc}{eV} \Rightarrow \lambda_{min} \propto \frac{1}{V}
  3. So the answer is (a) 1V\frac{1}{V}.
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Q477JEE Main 2022 · 25 JUN · SHIFT 1EasyOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Given below are two statements:

Statement I: Davisson-Germer experiment establishes the wave nature of electrons. Statement II: If electrons have wave nature, they can interfere and show diffraction. In the light of the above statements choose the correct answer from the option given below:

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Answer: (a)

  1. Statement I: Davisson–Germer electron diffraction established the wave nature of electrons — true.
  2. Statement II: any wave can interfere and diffract, so electrons with wave nature can too — true.
  3. So the answer is (a).
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Q478JEE Main 2021 · 26 FEB · SHIFT 1Easy

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Given below are two statements: one is labeled as Assertion A and the other is labelled as Reason R.

Assertion A: An electron microscope can achieve better resolving power than an optical microscope.

Reason R: The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (c)

  1. Resolving power improves as wavelength decreases.
  2. Electrons accelerated through kV have λ∼10−11\lambda \sim 10^{-11} m, far below visible light (∼5×10−7\sim 5\times10^{-7} m).
  3. So A is true and R correctly explains it: answer (c).
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Q479JEE Main 2021 · 24 FEB · SHIFT 2Easy

Davisson-Germer, Uncertainty, Compton & X-Rays · X-rays & cut-off wavelength

An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:

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Answer: (c)

  1. λmin=hceV\lambda_{min} = \frac{hc}{eV} with hc=1240 eV nmhc = 1240\text{ eV nm}.
  2. eV=1.24×106 eVeV = 1.24 \times 10^{6}\text{ eV}
  3. λmin=12401.24×106=10−3 nm\lambda_{min} = \frac{1240}{1.24 \times 10^{6}} = 10^{-3}\text{ nm}
  4. So the answer is (c) 10−3 nm10^{-3}\text{ nm}.
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Q480JEE Main 2021 · 18 MAR · SHIFT 2MediumOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

The speed of electrons in a scanning electron microscope is 1 ×\times 107 ms-1. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:

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Answer: (b)

  1. Resolving power ∝1λ\propto \frac{1}{\lambda} and λ=hmv\lambda = \frac{h}{mv}, so resolving power ∝mv\propto mv.
  2. Same speed: RPpRPe=mpme=1837\frac{RP_p}{RP_e} = \frac{m_p}{m_e} = 1837
  3. So the answer is (b) 1837.
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Q481JEE Main 2021 · 31 AUG · SHIFT 2Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · Uncertainty principle & electron diffraction

Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases is the same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-

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Answer: (c)

  1. Uncertainty principle: Δx≈h4πΔp\Delta x \approx \frac{h}{4\pi \Delta p}, take Δp∼p\Delta p \sim p.
  2. Same temperature: p=3mkT⇒p∝mp = \sqrt{3mkT} \Rightarrow p \propto \sqrt{m}
  3. Δx∝1m⇒ΔxeΔxp=mpme\Delta x \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{\Delta x_e}{\Delta x_p} = \sqrt{\frac{m_p}{m_e}}
  4. So the answer is (c).
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Q482JEE Main 2021 · 20 JUL · SHIFT 2Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · X-rays & cut-off wavelength

An electron having de-Broglie wavelength λ\lambda is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:

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Answer: (c)

  1. Kinetic energy of electron: E=h22mλ2E = \frac{h^2}{2m\lambda^2}
  2. All of it can go into one photon: hcλ0=h22mλ2\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}
  3. λ0=2mcλ2h\lambda_0 = \frac{2mc\lambda^2}{h}
  4. So the answer is (c).
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Q483NEET 2020 · PHASE 2Easy

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

The wave nature of electrons was experimentally verified by,

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Answer: (a)

  1. de Broglie proposed matter waves; Davisson and Germer confirmed them by electron diffraction from a nickel crystal (1927).
  2. So the answer is (a) Davisson and Germer.
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Q484JEE Main 2020 · 5 SEP · SHIFT 1DifficultOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

A beam of electrons of energy E scatters from a target having atomic spacing of 1 A˚\text{\AA}. The first maximum intensity occurs at θ\theta = 60o. Then E (in eV) is ______.

(Planck constant h = 6.64 × 10–34 Js,

1 eV = 1.6 × 10–19 J, electron

mass m = 9.1 × 10–31 kg)

Numerical answer type. Work it out, then open the answer.

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Answer: 50.47

  1. Bragg condition for first maximum: 2dsin⁡θ=λ2d\sin\theta = \lambda
  2. λ=2×1×sin⁡60∘=3 A˚=1.732×10−10 m\lambda = 2 \times 1 \times \sin 60^\circ = \sqrt{3}\text{ \AA} = 1.732 \times 10^{-10}\text{ m}
  3. E=h22mλ2=(6.64×10−34)22×9.1×10−31×3×10−20=8.07×10−18 JE = \frac{h^2}{2m\lambda^2} = \frac{(6.64 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 3 \times 10^{-20}} = 8.07 \times 10^{-18}\text{ J}
  4. E=8.07×10−181.6×10−19≈50.47 eVE = \frac{8.07 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 50.47\text{ eV}
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Q485JEE Main 2019 · 10 JAN · SHIFT 1Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -

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Answer: (a)

  1. Required wavelength λ=7.5×10−12 m\lambda = 7.5 \times 10^{-12}\text{ m}, and λ=hp\lambda = \frac{h}{p}, E=p22mE = \frac{p^2}{2m}.
  2. p=6.63×10−347.5×10−12=8.84×10−23 kg m/sp = \frac{6.63 \times 10^{-34}}{7.5 \times 10^{-12}} = 8.84 \times 10^{-23}\text{ kg m/s}
  3. E=(8.84×10−23)22×9.1×10−31≈4.3×10−15 JE = \frac{(8.84 \times 10^{-23})^2}{2 \times 9.1 \times 10^{-31}} \approx 4.3 \times 10^{-15}\text{ J}
  4. E=4.3×10−151.6×10−19≈2.7×104 eV≈25 keVE = \frac{4.3 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 2.7 \times 10^{4}\text{ eV} \approx 25\text{ keV}
  5. So the answer is (a) 25 keV.
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Q486JEE Main 2017Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · X-rays & cut-off wavelength

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X– rays. It produces continuous as well as characteristic X-rays. If λ\lambdamin is the smallest possible wavelength of X-ray in the spectrum, the variation of logλ\lambdamin with log V is correctly represented in:

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Answer: (b)

  1. λmin=hceV\lambda_{min} = \frac{hc}{eV}
  2. Taking logs: log⁡λmin=log⁡hce−log⁡V\log \lambda_{min} = \log\frac{hc}{e} - \log V
  3. This is a straight line with slope −1-1 and positive intercept: falling line.
  4. So the answer is (b).
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Q487NEET 2016 · PHASE 2Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · X-rays & cut-off wavelength

Electrons of mass m with de-Broglie wavelength λ\lambda fall on the target in an X-ray tube. The cutoff wavelength (λ\lambda0) of the emitted X-ray is

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Answer: (a)

  1. Electron kinetic energy: E=p22m=h22mλ2E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}
  2. Cut-off: λ0=hcE=hc⋅2mλ2h2\lambda_0 = \frac{hc}{E} = hc \cdot \frac{2m\lambda^2}{h^2}
  3. λ0=2mcλ2h\lambda_0 = \frac{2mc\lambda^2}{h}
  4. So the answer is (a).
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Q488JEE Main 2015EasyOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Match List - I\text{I} (Fundamental Experiment) with List - II\text{II} (its conclusion) and select the correct option

from the choices given below the list:

Figure

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Answer: (a)

  1. Franck–Hertz: electrons lose energy only in fixed amounts, showing discrete energy levels of atoms: A – (ii).
  2. Photoelectric effect: light acts as photons, showing particle nature of light: B – (i).
  3. Davisson–Germer: electron diffraction, showing wave nature of electrons: C – (iii).
  4. So the answer is (a).
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Q489JEE Main 2015 · 10 APR · SHIFT 1Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · Uncertainty principle & electron diffraction

A parallel beam of electrons travelling in x-direction falls on a slit of width d (see figure). If after passing the slit, an electron acquires momentum py in the y-direction then for a majority of electrons passing through the slit (h is Planck's constant):

Figure

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Answer: (c)

  1. Most electrons land inside the central maximum, where sin⁡θ<λd\sin\theta < \frac{\lambda}{d}.
  2. py=psin⁡θ<pλdp_y = p\sin\theta < p\frac{\lambda}{d}, and pλ=hp\lambda = h.
  3. ∣py∣<hd⇒∣py∣d<h|p_y| < \frac{h}{d} \Rightarrow |p_y|d < h
  4. So the answer is (c).
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Q490JEE Main 2014 · 19 APR · SHIFT 1Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Match List - I (Experiment performed) with List - II (Phenomena discovered/associated) and select

the correct option from the options given below the lists:

List - IList - II
(a) Davisson and Germer Experiment(i) Wave nature of electrons
(b) Millikan’s oil drop experiment(ii) Charge of an electron
(c) Rutherford experiment(iii) Quantisation of energy levels
(d) Franck - Hertz experiment(iv) Existence of nucleus
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Answer: (b)

  1. Davisson and Germer: electron diffraction, so wave nature of electrons, (a)-(i).
  2. Millikan oil drop: charge of the electron, (b)-(ii).
  3. Rutherford alpha scattering: existence of the nucleus, (c)-(iv).
  4. Franck-Hertz: discrete energy losses, so quantised energy levels, (d)-(iii).
  5. So the answer is (b).
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Q491JEE Main 2014 · 11 APR · SHIFT 1Difficult

Davisson-Germer, Uncertainty, Compton & X-Rays · Compton effect

A photon of wavelength λ\lambda is scattered from an electron, which was at rest. The wavelength shift Δλ\Delta\lambda is three times of λ\lambda and the angle of scattering θ\theta is 60o. The angle at which the electron recoiled is ϕ\phi. The value of tan ϕ\phi is: (electron speed is much smaller than the speed of light)

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Answer: (c)

h/λh/4λ60°φelectron
  1. Δλ=3λ\Delta\lambda = 3\lambda, so scattered wavelength λ′=4λ\lambda' = 4\lambda and its momentum is h4λ\frac{h}{4\lambda}.
  2. Momentum along x: px=hλ−h4λcos⁡60∘=78hλp_x = \frac{h}{\lambda} - \frac{h}{4\lambda}\cos 60^\circ = \frac{7}{8}\frac{h}{\lambda}
  3. Momentum along y: py=h4λsin⁡60∘=38hλp_y = \frac{h}{4\lambda}\sin 60^\circ = \frac{\sqrt{3}}{8}\frac{h}{\lambda}
  4. tan⁡ϕ=pypx=37≈0.25\tan\phi = \frac{p_y}{p_x} = \frac{\sqrt{3}}{7} \approx 0.25
  5. So the answer is (c) 0.25.
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Q492NEET 2013Medium

Davisson-Germer, Uncertainty, Compton & X-Rays · Uncertainty principle & electron diffraction

A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?

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Answer: (a)

  1. Angular width of central maximum =2λa= \frac{2\lambda}{a}, where aa is slit width.
  2. λ=hmv\lambda = \frac{h}{mv}, so higher speed gives smaller λ\lambda.
  3. Smaller λ\lambda gives a narrower central maximum.
  4. So the answer is (a): the angular width decreases.
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Q493JEE Main 2012 · AIEEEEasyOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement 1: Davisson - Germer experiment established the wave nature of electrons. Statement 2: If electrons have wave nature, they can interfere and show diffraction.

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Answer: (b)

  1. Statement 1: Davisson and Germer observed electron diffraction from a nickel crystal, which proved the wave nature of electrons. True.
  2. Statement 2: Interference and diffraction are wave properties, so waves of electrons must show them. True.
  3. Their experiment worked exactly because electron waves diffract, so Statement 2 explains Statement 1.
  4. So the answer is (b).
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Q494NEET 2011 · AIPMT · PRELIMSEasyOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by

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Answer: (a)

  1. Electron speed is set by the accelerating voltage: eV=12mv2eV = \dfrac12 mv^2.
  2. Filament current only changes the number of electrons (beam intensity), not their speed.
  3. So the answer is (a) increasing the anode–filament potential difference.
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Q495JEE Main 2008 · AIEEEMediumOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

Figure

Electrons accelerated by potential VV are diffracted from a crystal. If d=1 A˚d = 1\text{ \AA} and i=30∘, Vi = 30^\circ,\ V should be about

(h=6.6×10−34 Js, me=9.1×10−31 kg, e=1.6×10−19 C)\left(h = 6.6 \times 10^{-34}\text{ Js},\ m_e = 9.1 \times 10^{-31}\text{ kg},\ e = 1.6 \times 10^{-19}\text{ C}\right)

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Answer: (b)

  1. From the figure, ii is measured from the normal, so the extra path for the lower plane is 2dcos⁡i2d\cos i.
  2. First-order peak: λ=2dcos⁡30∘=2×1×0.866=1.73 A˚\lambda = 2d\cos 30^\circ = 2 \times 1 \times 0.866 = 1.73\text{ \AA}
  3. For electrons, λ=h2meV=12.27V A˚\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\text{ \AA}
  4. V=12.271.73≈7.1⇒V≈50 V\sqrt{V} = \frac{12.27}{1.73} \approx 7.1 \Rightarrow V \approx 50\text{ V}
  5. So the answer is (b) 50 V.
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Q496JEE Main 2008 · AIEEEMediumOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · Davisson-Germer & electron microscope

Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

Figure

If a strong diffraction peak is observed when electrons are incident at an angle ′i′'i' from the normal to the crystal planes with distance ′d′'d' between them (see figure), de Broglie wavelength λdB\lambda_{dB} of electrons can be calculated by the relationship (nn is an integer)

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Answer: (b)

  1. The angle ii is measured from the normal (see figure), not from the plane.
  2. Path difference between rays from adjacent planes =2dcos⁡i= 2d\cos i.
  3. Strong peak (constructive interference): 2dcos⁡i=nλdB2d\cos i = n\lambda_{dB}
  4. So the answer is (b).
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Q497JEE Main 2008 · AIEEEMedium

Davisson-Germer, Uncertainty, Compton & X-Rays · Uncertainty principle & electron diffraction

In an experiment, electrons are made to pass through a narrow slit of width ′d′'d' comparable to their de Broglie wavelength. They are detected on a screen at a distance ′D′'D' from the slit (see figure).

Figure

Which of the following graphs can be expected to represent the number of electrons ′N′'N' detected as a function of the detector position ′y′ (y=0'y'\ (y = 0 corresponds to the middle of the slit ))

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Answer: (d)

  1. Electrons behave as waves and give a single-slit diffraction pattern on the screen.
  2. This has a broad, tall central maximum at y=0y = 0, spreading beyond the slit width dd, with smaller secondary maxima on both sides.
  3. Graph (d) shows the central peak plus weaker side peaks; (b) is the shadow of a particle beam and (a), (c) lack secondary maxima.
  4. So the answer is (d).
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Q498NEET 1996 · AIPMTEasyOld syllabus

Davisson-Germer, Uncertainty, Compton & X-Rays · X-rays & cut-off wavelength

The minimum wavelength of the X-rays produced by electrons accelerated through a potential difference of VV volts is directly proportional to

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Answer: (b)

  1. Cut-off condition: eV=hcλmineV = \frac{hc}{\lambda_{min}}.
  2. λmin=hceV⇒λmin∝1V\lambda_{min} = \frac{hc}{eV} \Rightarrow \lambda_{min} \propto \frac{1}{V}
  3. So the answer is (b) 1V\frac{1}{V}.
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Q499MHT CET 2024 · 9 MAY · SHIFT 1Medium

Other / Mixed: Atoms & Bohr Links

The gyromagnetic ratio and Bohr magneton are given respectively by [Given e=e = charge on electron, m=m = mass of electron, h=h = Planck's constant]

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Answer: (a)

  1. Gyromagnetic ratio =μL=evr/2mvr=e2m= \frac{\mu}{L} = \frac{evr/2}{mvr} = \frac{e}{2m}
  2. Bohr magneton: μB=e2m⋅h2π=eh4πm\mu_B = \frac{e}{2m} \cdot \frac{h}{2\pi} = \frac{eh}{4\pi m}
  3. So the answer is (a).
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Q500JEE Main 2024 · 6 APR · SHIFT 2Medium

Other / Mixed: Atoms & Bohr Links

In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V10.2\text{ V}. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ________ nm. (Given hc =1245 eV nm, e=1.6×10−19 C= 1245\text{ eV nm},\ e = 1.6 \times 10^{-19}\text{ C}).

Numerical answer type. Work it out, then open the answer.

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Answer: 122

  1. First dip at 10.2 V means the n = 1 to n = 2 excitation needs 10.2 eV.
  2. On returning, the atom emits a photon of 10.2 eV: λ=hcE\lambda = \frac{hc}{E}
  3. λ=124510.2≈122 nm\lambda = \frac{1245}{10.2} \approx 122\text{ nm}
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Q501JEE Main 2023 · 13 APR · SHIFT 2Medium

Other / Mixed: Atoms & Bohr Links

An atom absorbs a photon of wavelength 500 nm500\text{ nm} and emits another photon of wavelength 600 nm600\text{ nm}. The net energy absorbed by the atom in this process is n×10−4 eVn \times 10^{-4}\text{ eV}. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take h=6.6×10−34 Jsh = 6.6 \times 10^{-34}\text{ Js} and c=3×108 m/sc = 3 \times 10^{8}\text{ m/s})

Numerical answer type. Work it out, then open the answer.

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Answer: 4125

  1. hc=6.6×10−34×3×108=1.98×10−25 J m=1237.5 eV nmhc = 6.6 \times 10^{-34} \times 3 \times 10^{8} = 1.98 \times 10^{-25}\text{ J m} = 1237.5\text{ eV nm}
  2. Net energy =hc(1500−1600)=1237.53000 eV= hc\left(\frac{1}{500} - \frac{1}{600}\right) = \frac{1237.5}{3000}\text{ eV}
  3. =0.4125 eV=4125×10−4 eV= 0.4125\text{ eV} = 4125 \times 10^{-4}\text{ eV}
  4. So n=4125n = 4125.
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Q502MHT CET 2021 · 21 SEP · SHIFT 2Medium

Other / Mixed: Atoms & Bohr Links

The wave number of the last line of the Balmer series in the hydrogen spectrum will be (( Rydberg's cons tant,R=107m)\text{tant}, R = \frac{10^{7}}{\text{m}})

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Answer: (d)

  1. Last (series limit) line of Balmer: n1=2n_1 = 2, n2=∞n_2 = \infty.
  2. νˉ=R(122−0)=R4\bar{\nu} = R\left(\frac{1}{2^2} - 0\right) = \frac{R}{4}
  3. νˉ=1074=2.5×106=25×105 m−1\bar{\nu} = \frac{10^7}{4} = 2.5 \times 10^{6} = 25 \times 10^{5}\text{ m}^{-1}
  4. So the answer is (d).
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Q503JEE Main 2021 · 31 AUG · SHIFT 2Difficult

Other / Mixed: Atoms & Bohr Links

A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 ×\times 10−-34 Js)

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Answer: (c)

  1. Energy of first excited state (n=2n = 2) of H: −3.4 eV-3.4\text{ eV}.
  2. Photon energy =2.6+3.4=6.0 eV=9.6×10−19 J= 2.6 + 3.4 = 6.0\text{ eV} = 9.6 \times 10^{-19}\text{ J}
  3. f=Eh=9.6×10−196.6×10−34=1.45×1015 Hzf = \frac{E}{h} = \frac{9.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 1.45 \times 10^{15}\text{ Hz}
  4. =1.45×109 MHz= 1.45 \times 10^{9}\text{ MHz}
  5. So the answer is (c).
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Q504JEE Main 2021 · 27 JUL · SHIFT 2Difficult

Other / Mixed: Atoms & Bohr Links

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000A˚\text{\AA}. What is the maximum kinetic energy of the emitted photoelectron?

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Answer: (b)

  1. Second excited state is n=3n = 3: E3=−13.69=−1.51 eVE_3 = -\frac{13.6}{9} = -1.51\text{ eV}.
  2. Photon energy =3+1.51=4.51 eV= 3 + 1.51 = 4.51\text{ eV}
  3. Work function ϕ=124004000=3.1 eV\phi = \frac{12400}{4000} = 3.1\text{ eV}
  4. Kmax=4.51−3.1=1.41 eVK_{max} = 4.51 - 3.1 = 1.41\text{ eV}
  5. So the answer is (b) 1.41 eV.
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Q505JEE Main 2021 · 26 FEB · SHIFT 2Difficult

Other / Mixed: Atoms & Bohr Links

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be:

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Answer: (d)

  1. Photon energy: E=13.6(1−125)=13.06 eV=2.09×10−18 JE = 13.6\left(1 - \frac{1}{25}\right) = 13.06\text{ eV} = 2.09 \times 10^{-18}\text{ J}
  2. Momentum conservation: Mv=EcMv = \frac{E}{c}
  3. v=2.09×10−181.67×10−27×3×108≈4.17 m/sv = \frac{2.09 \times 10^{-18}}{1.67 \times 10^{-27} \times 3 \times 10^{8}} \approx 4.17\text{ m/s}
  4. So the answer is (d) 4.17 m/s.
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Q506JEE Main 2019 · 12 JAN · SHIFT 2Medium

Other / Mixed: Atoms & Bohr Links

In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

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Answer: (b)

  1. Energy given to a mercury atom =5.6−0.7=4.9 eV= 5.6 - 0.7 = 4.9\text{ eV}.
  2. The atom emits this as a photon: λ=12404.9 nm\lambda = \frac{1240}{4.9}\text{ nm}
  3. λ≈253 nm≈250 nm\lambda \approx 253\text{ nm} \approx 250\text{ nm}
  4. So the answer is (b) 250 nm.
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Q507JEE Main 2002 · AIEEEEasy

Other / Mixed: Atoms & Bohr Links

Formation of covalent bonds in compounds exhibits

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Answer: (a)

  1. A covalent bond forms by overlap of atomic orbitals, i.e. overlap of electron waves.
  2. Orbitals and their overlap describe the wave nature of the electron.
  3. So the answer is (a) wave nature of electron.
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Q508NEET 1998 · AIPMTMedium

Other / Mixed: Atoms & Bohr Links

The 21 cm21\text{ cm} radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

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Answer: (d)

  1. E=hcλE = \frac{hc}{\lambda} with λ=0.21 m\lambda = 0.21\text{ m}.
  2. E=6.63×10−34×3×1080.21≈9.5×10−25 JE = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.21} \approx 9.5 \times 10^{-25}\text{ J}
  3. E≈10−24 JE \approx 10^{-24}\text{ J}
  4. So the answer is (d).
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Q509NEET 1991 · AIPMTEasyOld syllabus

Other / Mixed: Atoms & Bohr Links

An electron with (rest mass m0m_0) moves with a speed of 0.8 c0.8\,c. Its mass when it moves with this speed is

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Answer: (c)

  1. m=m01−v2/c2m = \frac{m_0}{\sqrt{1 - v^2/c^2}}
  2. 1−0.64=0.6\sqrt{1 - 0.64} = 0.6
  3. m=m00.6=5m03m = \frac{m_0}{0.6} = \frac{5m_0}{3}
  4. So the answer is (c) 5m03\frac{5m_0}{3}.
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