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Atomic Structure

Questions
360
MHT CET
139
JEE Main
146
NEET
75
Years
1988–2026
0 of 360 attemptedPick an answer to see the solution

Q1NEET 2026 · RE-NEETEasy

Atomic Models & Alpha-Particle Scattering · Alpha scattering: N against angle

In Geiger-Marsden experiment, the number of scattered α\alpha-particles N(θ)N(\theta) is plotted as a function of scattering angle θ\theta. Which of the following options represents the correct plot?

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Answer: (d)

  1. Rutherford's formula: N(θ)∝1sin⁡4(θ/2)N(\theta) \propto \dfrac{1}{\sin^4(\theta/2)}.
  2. NN is very large at small θ\theta and falls steeply as θ\theta increases; only a few particles reach θ≈180∘\theta \approx 180^\circ.
  3. The graph must decrease continuously from a huge value near 0∘0^\circ to a small value near 180∘180^\circ.
  4. So the answer is (d).
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Q2JEE Main 2026 · 4 APR · SHIFT 2Medium

Atomic Models & Alpha-Particle Scattering · Thomson & Rutherford models

In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because

A. The size of gold nucleus is very small as compared to the size of gold atom.

B. Alpha particle and gold nucleus have equal charge.

C. The impact parameter is minimum for a few alpha particles.

D. A few alpha particles have very high kinetic energy.

E. Only a few alpha particles undergo head-on collision with the nuclei.

Choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

  1. A: true, the nucleus is tiny compared with the atom, so very few particles come close to it.
  2. B: false, α\alpha has charge 2e2e while gold nucleus has 79e79e.
  3. C: true, only particles with nearly zero impact parameter turn back.
  4. D: false, all α\alpha-particles in the beam have about the same energy.
  5. E: true, rebound needs a nearly head-on collision, which is rare.
  6. So the answer is (d) A, C, E only.
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Q3JEE Main 2026 · 22 JAN · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

7.9 MeV α7.9\text{ MeV }\alpha-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ______ m.

[14πϵ0=9×109 Nm2/C2\Bigg[\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ Nm}^2/\text{C}^2 and electron charge =1.6×10−19 C]= 1.6 \times 10^{-19}\text{ C}\Bigg]

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Answer: (b)

  1. Closest approach: r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K} with K=7.9 MeV=7.9×106eK = 7.9\text{ MeV} = 7.9\times10^6 e J.
  2. r0=9×109×2×79×1.6×10−197.9×106=2.88×10−14r_0 = \dfrac{9\times10^9 \times 2 \times 79 \times 1.6\times10^{-19}}{7.9\times10^6} = 2.88\times10^{-14} m.
  3. This estimates the nuclear radius, so diameter =2r0=5.76×10−14= 2r_0 = 5.76\times10^{-14} m.
  4. So the answer is (b).
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Q4JEE Main 2026 · 21 JAN · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m. (Atomic number of gold =79= 79 and 14πϵ0=9×109\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} in SI units)

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Answer: (c)

  1. r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K}, with K=7.7×106eK = 7.7\times10^6 e J.
  2. r0=9×109×2×79×1.6×10−197.7×106r_0 = \dfrac{9\times10^9 \times 2 \times 79 \times 1.6\times10^{-19}}{7.7\times10^6}
  3. r0=2.275×10−77.7×106≈2.95×10−14r_0 = \dfrac{2.275\times10^{-7}}{7.7\times10^6} \approx 2.95\times10^{-14} m.
  4. So the answer is (c).
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Q5NEET 2024Easy

Atomic Models & Alpha-Particle Scattering · Thomson & Rutherford models

Given below are two statements:

Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.

Statement II: Atoms of each element are stable and emit their characteristic spectrum. In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (c)

  1. Statement I: an atom has equal positive (nuclear) and negative (electronic) charge, so it is neutral. True.
  2. Statement II: this is the experimental fact (NCERT) every atomic model must explain: atoms are stable and each element emits its own line spectrum. True.
  3. So both statements are correct, which is option (a).
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Q6JEE Main 2024 · 29 JAN · SHIFT 2Easy

Atomic Models & Alpha-Particle Scattering · Thomson & Rutherford models

Given below are two statements:

Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model.

Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model.

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

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Answer: (b)

  1. Statement I is exactly Rutherford's nuclear model: tiny massive positive nucleus with electrons revolving around it. True.
  2. Statement II describes Thomson's plum-pudding model, which is a different model, not a special case of Rutherford's. False.
  3. So the answer is (b).
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Q7JEE Main 2024 · 8 APR · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

In an alpha particle scattering experiment distance of closest approach for the α\alpha particle is 4.5×10−14 m4.5 \times 10^{-14}\text{ m}. If target nucleus has atomic number 80, then maximum velocity of α\alpha-particle is __________ ×105 m/s\times 10^{5}\text{ m/s} approximately.

( 14πϵ0=9×109 SI\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ SI} unit, mass of α\alpha particle =6.72×10−27 kg= 6.72 \times 10^{-27}\text{ kg})

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

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Answer: 156

  1. 12mv2=k(2e)(Ze)r0\frac{1}{2}mv^2 = \dfrac{k(2e)(Ze)}{r_0}
  2. 9×109×160×(1.6×10−19)24.5×10−14=8.19×10−13\dfrac{9\times10^9 \times 160 \times (1.6\times10^{-19})^2}{4.5\times10^{-14}} = 8.19\times10^{-13} J
  3. v2=2×8.19×10−136.72×10−27=2.44×1014v^2 = \dfrac{2 \times 8.19\times10^{-13}}{6.72\times10^{-27}} = 2.44\times10^{14}
  4. v≈1.56×107=156×105v \approx 1.56\times10^7 = 156\times10^5 m/s. Answer: 156.
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Q8JEE Main 2022 · 24 JUN · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering · Thomson & Rutherford models

Choose the correct option from the following options given below:

Show answer and solution

Answer: (c)

  1. (a) false: in Rutherford's model electrons revolve, so they are not in static stable equilibrium; in Thomson's model electrons can sit at rest in equilibrium.
  2. (b) false: it is the reverse, Thomson's model has nearly uniform mass, Rutherford's is highly concentrated.
  3. (c) true: an accelerating revolving electron radiates energy classically and spirals into the nucleus.
  4. (d) false: in both models the positive part carries most of the mass.
  5. So the answer is (c).
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Q9JEE Main 2022 · 30 JUL · SHIFT 2Medium

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy 3.2 Mev. The estimation of the radius of the nucleus of gold element (Z=79Z = 79) by him is _______ ×10−16 m\times 10^{-16}\text{ m}

(Take 14πϵ0=9×109 N m2C2\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \frac{\text{N m}^2}{\text{C}^2})

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

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Answer: 711

  1. Nuclear radius ≈\approx distance of closest approach: r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K}, K=3.2×106eK = 3.2\times10^6 e J.
  2. r0=9×109×158×1.6×10−193.2×106r_0 = \dfrac{9\times10^9 \times 158 \times 1.6\times10^{-19}}{3.2\times10^6}
  3. r0=7.11×10−14r_0 = 7.11\times10^{-14} m =711×10−16= 711\times10^{-16} m. Answer: 711.
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Q10JEE Main 2022 · 29 JUN · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

d1\sqrt{d_1} and d2\sqrt{d_2} are the impact parameters corresponding to scattering angles 60∘^\circ and 90∘^\circ respectively, when an α\alpha particle is approaching a gold nucleus. For d1 = x d2, the value of x will be ____________.

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

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Answer: 3

  1. Impact parameter b∝cot⁡(θ/2)b \propto \cot(\theta/2).
  2. d1d2=cot⁡30∘cot⁡45∘=3\dfrac{\sqrt{d_1}}{\sqrt{d_2}} = \dfrac{\cot 30^\circ}{\cot 45^\circ} = \sqrt{3}
  3. d1=3d2d_1 = 3d_2, so x=3x = 3. Answer: 3.
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Q11JEE Main 2020 · 8 JAN · SHIFT 1Easy

Atomic Models & Alpha-Particle Scattering · Alpha scattering: N against angle

The graph which depicts the results of Rutherford gold foil experiment with α\alpha-particales is: θ\theta: Scattering angle

Y: Number of scattered α\alpha-particles detected (Plots are schematic and not to scale)

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Answer: (c)

  1. N(θ)∝1sin⁡4(θ/2)N(\theta) \propto \dfrac{1}{\sin^4(\theta/2)}, so most α\alpha-particles go nearly undeviated and very few are scattered through large angles.
  2. The curve is very high near θ=0\theta = 0 and drops sharply, flattening out towards θ=π\theta = \pi.
  3. So the answer is (c).
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Q12NEET 2016 · PHASE 1Easy

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

When an α\alpha-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as

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Answer: (c)

  1. At closest approach all KE becomes PE: 12mv2=k(2e)(Ze)r0\frac{1}{2}mv^2 = \dfrac{k(2e)(Ze)}{r_0}.
  2. r0=4kZe2mv2r_0 = \dfrac{4kZe^2}{mv^2}, so for fixed vv, r0∝1mr_0 \propto \dfrac{1}{m}.
  3. So the answer is (c) 1m\frac{1}{m}.
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Q13NEET 2010 · AIPMT · PRELIMSEasy

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

An alpha nucleus of energy 12\frac{1}{2} mv2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

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Answer: (c)

  1. 12mv2=k(2e)(Ze)r0⇒r0=4kZe2mv2\frac{1}{2}mv^2 = \dfrac{k(2e)(Ze)}{r_0} \Rightarrow r_0 = \dfrac{4kZe^2}{mv^2}.
  2. So r0∝Zer_0 \propto Ze, r0∝1v2r_0 \propto \dfrac{1}{v^2} and r0∝1mr_0 \propto \dfrac{1}{m}.
  3. So the answer is (c) 1m\frac{1}{m}.
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Q14NEET 2009 · AIPMTEasy

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

In a Rutherford scattering experiment when a projectile of charge z1 and mass M1 approaches a target nucleus of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is

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Answer: (a)

  1. At closest approach, KE of projectile = PE: E=kz1z2e2r0E = \dfrac{k z_1 z_2 e^2}{r_0}.
  2. For a given r0r_0, E∝z1z2E \propto z_1 z_2; it does not depend on the masses.
  3. So the answer is (a).
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Q15JEE Main 2006 · AIEEEEasy

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

An alpha nucleus of energy 12mv2\frac{1}{2}mv^2 bombards a heavy nuclear target of charge ZeZe. Then the distance of closest approach for the alpha nucleus will be proportional to

Show answer and solution

Answer: (b)

  1. 12mv2=k(2e)(Ze)r0⇒r0=4kZe2mv2\frac{1}{2}mv^2 = \dfrac{k(2e)(Ze)}{r_0} \Rightarrow r_0 = \dfrac{4kZe^2}{mv^2}.
  2. Thus r0∝1mr_0 \propto \dfrac{1}{m}, r0∝1v2r_0 \propto \dfrac{1}{v^2} and r0∝Zer_0 \propto Ze.
  3. So the answer is (b) 1m\frac{1}{m}.
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Q16JEE Main 2004 · AIEEEMedium

Atomic Models & Alpha-Particle Scattering · Distance of closest approach & impact parameter

An α\alpha-particle of energy 5 MeV5\text{ MeV} is scattered through 180∘180^\circ by a fixed uranium nucleus. The distance of closest approach is of the order of

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Answer: (a)

  1. Head-on (180∘180^\circ): r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K}, uranium Z=92Z = 92, K=5×106eK = 5\times10^6 e J.
  2. r0=9×109×184×1.6×10−195×106≈5.3×10−14r_0 = \dfrac{9\times10^9 \times 184 \times 1.6\times10^{-19}}{5\times10^6} \approx 5.3\times10^{-14} m.
  3. 5.3×10−145.3\times10^{-14} m =5.3×10−12= 5.3\times10^{-12} cm.
  4. So the answer is (a) 10−1210^{-12} cm.
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Q17MHT CET 2026 · 13 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

When an electron in hydrogen atom revolves in a stationary orbit, it

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Answer: (a)

  1. In a stationary (Bohr) orbit the electron does not radiate energy, by Bohr's postulate.
  2. Its speed is constant but direction keeps changing, so velocity changes.
  3. So the answer is (a).
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Q18MHT CET 2026 · 20 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

Which of the following is an integral multiple of h2π\frac{h}{2\pi} in Bohr's model of an hydrogen atom?

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Answer: (d)

  1. Bohr's quantisation condition: L=mvr=nh2πL = mvr = n\dfrac{h}{2\pi}.
  2. So the answer is (d) angular momentum.
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Q19MHT CET 2026 · 18 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The angular momentum of the electron in the third Bohr orbit of a hydrogen atom is 'll' so its angular momentum in the fourth Bohr orbit is

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Answer: (c)

  1. L=nh2π⇒L∝nL = \dfrac{nh}{2\pi} \Rightarrow L \propto n.
  2. L4L3=43⇒L4=43l\dfrac{L_4}{L_3} = \dfrac{4}{3} \Rightarrow L_4 = \dfrac{4}{3}l.
  3. So the answer is (c).
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Q20MHT CET 2026 · 15 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

In the hydrogen atom, radii of the first four Bohr orbits are related as

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Answer: (c)

  1. rn=n2a0⇒r∝n2r_n = n^2 a_0 \Rightarrow r \propto n^2.
  2. r1:r2:r3:r4=1:4:9:16r_1 : r_2 : r_3 : r_4 = 1 : 4 : 9 : 16.
  3. So the answer is (c).
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Q21MHT CET 2026 · 11 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

In Bohr's theory of hydrogen atom, 'r' is the radius of the orbit, 'V' is the speed of electron and E is the total energy of the electron. Out of the following which physical quantity is inversely proportional to principal quantum number?

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Answer: (a)

  1. mVr=nh2πmVr = \frac{nh}{2\pi}, so Vr∝nVr \propto n.
  2. Hence 1Vr∝1n\frac{1}{Vr} \propto \frac{1}{n}.
  3. So the answer is (a) 1Vr\frac{1}{Vr}.
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Q22MHT CET 2026 · 19 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

The de Broglie wavelength of an electron moving in the nthn^{th} Bohr orbit of radius 'rr' is

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Answer: (c)

  1. Bohr condition: nn whole waves fit on the orbit, 2πr=nλ2\pi r = n\lambda.
  2. λ=2πrn\lambda = \frac{2\pi r}{n}
  3. So the answer is (c) 2πrn\frac{2\pi r}{n}.
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Q23MHT CET 2026 · 19 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

The de-Broglie wavelength of an electron moving in the nthn^{th} Bohr orbit of radius 'r' is

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Answer: (c)

  1. Bohr condition: 2πr=nλ2\pi r = n\lambda (whole number of de Broglie waves on the orbit).
  2. λ=2πrn\lambda = \frac{2\pi r}{n}
  3. So the answer is (c) 2πrn\frac{2\pi r}{n}.
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Q24MHT CET 2026 · 18 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

The de Broglie wavelength of the electron in the ground state is λ1\lambda_1 and that in the n=3n = 3 level is λ3\lambda_3 then λ3\lambda_3 is given by

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Answer: (d)

  1. λ=2πrn\lambda = \frac{2\pi r}{n} with r∝n2r \propto n^2 gives λ∝n\lambda \propto n (also λ=hmv\lambda = \frac{h}{mv}, v∝1nv \propto \frac{1}{n}).
  2. λ3=3λ1\lambda_3 = 3\lambda_1
  3. So the answer is (d) 3λ13\lambda_1.
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Q25MHT CET 2026 · 15 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in 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: (b)

  1. λ=hmv\lambda = \frac{h}{mv} and v∝1nv \propto \frac{1}{n}, so λ∝n\lambda \propto n.
  2. Ground state n=1n = 1 to third excited state n=4n = 4: λ\lambda becomes 44 times.
  3. So the answer is (b) will increase.
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Q26MHT CET 2026 · 19 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The magnetic moment of electron due to orbital motion is proportional to

(nn = principal quantum number)

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Answer: (a)

  1. Orbital magnetic moment μ=eL2m\mu = \frac{eL}{2m}, and L=nh2πL = \frac{nh}{2\pi}.
  2. μ=neh4πm∝n\mu = \frac{neh}{4\pi m} \propto n
  3. So the answer is (a) nn.
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Q27MHT CET 2026 · 17 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The orbital magnetic moment (morb)(m_{orb}) of a revolving electron around the nucleus varies with the principal quantum number (nn) as

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Answer: (b)

  1. morb=IA=ev2πrπr2=evr2=eL2mm_{orb} = IA = \frac{ev}{2\pi r}\pi r^2 = \frac{evr}{2} = \frac{eL}{2m}
  2. L=nh2πL = \frac{nh}{2\pi}, so morb∝nm_{orb} \propto n.
  3. So the answer is (b) morb∝nm_{orb} \propto n.
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Q28MHT CET 2026 · 16 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

The force acting on the electron in hydrogen atom (Bohr's theory) is related to the principal quantum number n as

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Answer: (b)

  1. F=ke2r2F = \frac{ke^2}{r^2} with r∝n2r \propto n^2.
  2. F∝1n4=n−4F \propto \frac{1}{n^4} = n^{-4}
  3. So the answer is (b) n−4n^{-4}.
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Q29MHT CET 2026 · 20 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The angular momentum of an electron in Bohr's hydrogen atom having energy (−0.544)(-0.544) eV is (hh = Planck's constant)

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Answer: (c)

  1. En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\text{ eV}: n2=13.60.544=25n^2 = \frac{13.6}{0.544} = 25, so n=5n = 5.
  2. L=nh2π=5h2πL = \frac{nh}{2\pi} = \frac{5h}{2\pi}
  3. So the answer is (c) 5h2π\frac{5h}{2\pi}.
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Q30JEE Main 2026 · 21 JAN · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The energy of an electron in an orbit of the Bohr's atom is −0.04E0-0.04E_0 eV where E0E_0 is the ground state energy. If LL is the angular momentum of the electron in this orbit and hh is the Planck's constant, then

2πLh\frac{2\pi L}{h} is ________:

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Answer: (c)

  1. En=E0n2E_n = \frac{E_0}{n^2}: 1n2=0.04⇒n=5\frac{1}{n^2} = 0.04 \Rightarrow n = 5.
  2. L=nh2π⇒2πLh=n=5L = \frac{nh}{2\pi} \Rightarrow \frac{2\pi L}{h} = n = 5
  3. So the answer is (c) 5.
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Q31MHT CET 2026 · 18 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

If the difference between (n+1)th(n+1)^{th} Bohr radius and nthn^{th} Bohr radius is given by (n−1)th(n-1)^{th} Bohr radius, then the value of n is

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Answer: (a)

  1. Bohr radius rn∝n2r_n \propto n^2.
  2. (n+1)2−n2=(n−1)2⇒2n+1=n2−2n+1(n+1)^2 - n^2 = (n-1)^2 \Rightarrow 2n + 1 = n^2 - 2n + 1
  3. n2=4n⇒n=4n^2 = 4n \Rightarrow n = 4
  4. So the answer is (a) 44.
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Q32MHT CET 2026 · 18 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The triply ionized beryllium (Be+++^{+++}) has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given Z=4Z = 4 for beryllium)

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Answer: (c)

  1. rn=n2Za0r_n = \frac{n^2}{Z} a_0
  2. Equal to a0a_0: n24=1⇒n=2\frac{n^2}{4} = 1 \Rightarrow n = 2
  3. So the answer is (c) n=2n = 2.
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Q33NEET 2026Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

In the first excited state of hydrogen atom, the energy of its electron is -3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately:

(Take 1 eV=1.6×10−19 J, e=1.6×10−19 C1\text{ eV} = 1.6 \times 10^{-19}\text{ J},\ e = 1.6 \times 10^{-19}\text{ C} and 14πε0=9×109 N m2/C2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2/\text{C}^2)

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Answer: (c)

  1. Total energy E=−ke22r⇒r=ke22∣E∣E = -\frac{ke^2}{2r} \Rightarrow r = \frac{ke^2}{2|E|}
  2. r=9×109×(1.6×10−19)22×3.4×1.6×10−19=14.4×10−106.8r = \frac{9\times10^9 \times (1.6\times10^{-19})^2}{2\times3.4\times1.6\times10^{-19}} = \frac{14.4\times10^{-10}}{6.8}
  3. r≈2.1×10−10 mr \approx 2.1\times10^{-10}\text{ m}
  4. So the answer is (c) 2.1×10−10 m2.1\times10^{-10}\text{ m}.
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Q34JEE Main 2026 · 24 JAN · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

Two electrons are moving in orbits of two hydrogen like atoms with speeds 3×105 m/s3 \times 10^{5}\text{ m/s} and 2.5×105 m/s2.5 \times 10^{5}\text{ m/s} respectively. If the radii of these orbits are nearly same then the possible order of energy states are ______ respectively.

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Answer: (d)

  1. r∝n2Zr \propto \frac{n^2}{Z} and v∝Znv \propto \frac{Z}{n}.
  2. From rr: Z∝n2rZ \propto n^2 r, so v∝n2rn=nrv \propto \frac{n^2 r}{n} = nr; with equal rr, v∝nv \propto n.
  3. n1n2=3×1052.5×105=65\frac{n_1}{n_2} = \frac{3\times10^5}{2.5\times10^5} = \frac{6}{5}
  4. So the answer is (d) 6 and 5.
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Q35MHT CET 2026 · 14 MAY · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

In hydrogenth atom, an electron jumps from higher orbit to lower orbit so that radius of the lower orbit is (14)\left(\frac{1}{4}\right) of radius of higher orbit. If speed of electron in higher orbit was V, then speed in lower orbit is

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Answer: (d)

  1. r∝n2r \propto n^2: rr becomes 14\frac{1}{4}, so nn becomes half.
  2. v∝1nv \propto \frac{1}{n}, so vv doubles.
  3. So the answer is (d) 2V2V.
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Q36NEET 2026 · RE-NEETMedium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity 25.6×105 m s−1\sqrt{25.6} \times 10^{5}\text{ m s}^{-1}. The radius of the orbit is x×10−9 mx \times 10^{-9}\text{ m}. The value of xx is:

[Take the mass of electron to the 9×10−31 kg9 \times 10^{-31}\text{ kg}, charge of electron =−1.6×10−19 C= -1.6 \times 10^{-19}\text{ C} and 14πε0=9×109 N m2C−2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2\text{C}^{-2} ]

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Answer: (a)

  1. mv2r=ke2r2⇒r=ke2mv2\frac{mv^2}{r} = \frac{ke^2}{r^2} \Rightarrow r = \frac{ke^2}{mv^2}
  2. r=9×109×2.56×10−389×10−31×25.6×1010=2.304×10−282.304×10−19r = \frac{9\times10^9\times2.56\times10^{-38}}{9\times10^{-31}\times25.6\times10^{10}} = \frac{2.304\times10^{-28}}{2.304\times10^{-19}}
  3. r=1×10−9 mr = 1\times10^{-9}\text{ m}
  4. So the answer is (a) 1.
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Q37MHT CET 2026 · 19 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

The moment of inertia of the electron moving in the second Bohr orbit of a hydrogen atom is [mm = mass of electron, hh = Planck's constant, ee = charge on electron, ε0\varepsilon_0 = permittivity of free space]

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Answer: (b)

  1. rn=ε0h2n2πme2r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}, so r2=4ε0h2πme2r_2 = \frac{4\varepsilon_0 h^2}{\pi m e^2}
  2. I=mr22=m⋅16ε02h4π2m2e4I = m r_2^2 = m\cdot\frac{16\varepsilon_0^2 h^4}{\pi^2 m^2 e^4}
  3. I=16ε02h4π2me4I = \frac{16\varepsilon_0^2 h^4}{\pi^2 m e^4}
  4. So the answer is (b).
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Q38MHT CET 2026 · 16 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

Using Bohr's atomic model, the orbital period of electron in hydrogen atom in the nnth orbit is (ϵ0\epsilon_0=permittivity of free space, h=Planck's constant, m = mass of electron, e = electronic charge)

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Answer: (a)

  1. rn=ε0n2h2πme2r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}, vn=e22ε0nhv_n = \frac{e^2}{2\varepsilon_0 nh}
  2. T=2πrnvn=2π⋅ε0n2h2πme2⋅2ε0nhe2T = \frac{2\pi r_n}{v_n} = 2\pi\cdot\frac{\varepsilon_0 n^2 h^2}{\pi m e^2}\cdot\frac{2\varepsilon_0 nh}{e^2}
  3. T=4ε02n3h3me4T = \frac{4\varepsilon_0^2 n^3 h^3}{m e^4}
  4. So the answer is (a).
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Q39MHT CET 2026 · 11 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is (hh = Planck's constant, mm is the mass of electron and rr is the radius of the orbit.)

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Answer: (a)

  1. mvr=h2π⇒v=h2πmrmvr = \frac{h}{2\pi} \Rightarrow v = \frac{h}{2\pi m r}
  2. f=v2πr=h4π2mr2f = \frac{v}{2\pi r} = \frac{h}{4\pi^2 m r^2}
  3. So the answer is (a) h/4π2mr2h/4\pi^2 m r^2.
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Q40MHT CET 2026 · 12 MAY · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is (μ0\mu_0 = permeability of free space, ϵ0\epsilon_0 = permittivity of free space, mm = mass of electron, ee = electronic charge)

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Answer: (b)

  1. B=μ0I2rB = \frac{\mu_0 I}{2r} with I=ev2πrI = \frac{ev}{2\pi r}, so B=μ0ev4πr2B = \frac{\mu_0 e v}{4\pi r^2}
  2. Ground state: v=e22ε0hv = \frac{e^2}{2\varepsilon_0 h}, r=ε0h2πme2r = \frac{\varepsilon_0 h^2}{\pi m e^2}
  3. B=μ0e4π⋅e22ε0h⋅π2m2e4ε02h4B = \frac{\mu_0 e}{4\pi}\cdot\frac{e^2}{2\varepsilon_0 h}\cdot\frac{\pi^2 m^2 e^4}{\varepsilon_0^2 h^4}
  4. B=μ0e7πm28ε03h5B = \frac{\mu_0 e^7 \pi m^2}{8\varepsilon_0^3 h^5}, so the answer is (b).
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Q41MHT CET 2026 · 13 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

An electron is revolving in a circular orbit of radius 'r' in hydrogen atom. Using Bohr's theory, the angular momentum of the electron is

(M = magnetic dipole moment, m = mass of electron)

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Answer: (c)

  1. M=IA=ev2πr⋅πr2=evr2M = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{evr}{2}
  2. L=mvr⇒M=e2mLL = mvr \Rightarrow M = \frac{e}{2m}L
  3. L=2mMeL = \frac{2mM}{e}, so the answer is (c).
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Q42JEE Main 2026 · 2 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

Using Bohr’s model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of hydrogen atom ________.

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

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Answer: 32

  1. B=μ0I2rB = \frac{\mu_0 I}{2r} with I=ev2πrI = \frac{ev}{2\pi r}, so B∝vr2B \propto \frac{v}{r^2}
  2. v∝1nv \propto \frac{1}{n}, r∝n2⇒B∝1n5r \propto n^2 \Rightarrow B \propto \frac{1}{n^5}
  3. B2B4=(42)5=32\frac{B_2}{B_4} = \left(\frac{4}{2}\right)^5 = 32
  4. So the answer is 32.
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Q43MHT CET 2026 · 13 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

The ratio of centripetal acceleration for an electron revolving in 3rd and 5th Bohr orbit of hydrogen atom is

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Answer: (c)

  1. a=v2ra = \frac{v^2}{r} with v∝1nv \propto \frac{1}{n}, r∝n2⇒a∝1n4r \propto n^2 \Rightarrow a \propto \frac{1}{n^4}
  2. a3a5=5434=62581\frac{a_3}{a_5} = \frac{5^4}{3^4} = \frac{625}{81}
  3. So the answer is (c) 625:81625 : 81.
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Q44MHT CET 2026 · 13 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Using Bohr's quantisation condition, what is the rotational kinetic energy in the second orbit for a diatomic molecule (I = moment of inertia of diatomic molecule, h = Planck's constant)

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Answer: (b)

  1. Quantisation: L=Iω=nh2πL = I\omega = \frac{nh}{2\pi}.
  2. Rotational KE =L22I=n2h28π2I= \frac{L^2}{2I} = \frac{n^2 h^2}{8\pi^2 I}.
  3. For n=2n = 2: KE=4h28π2I=h22π2IKE = \frac{4h^2}{8\pi^2 I} = \frac{h^2}{2\pi^2 I}
  4. So the answer is (b) h22Iπ2\frac{h^2}{2I\pi^2}.
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Q45MHT CET 2026 · 14 MAY · SHIFT 1Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Bohr model is applied to a particle of mass mm and charge qq is moving in a plane under the influence of a transverse magnetic field BB. The energy of the charged particle in nthn^{th} level will be (hh = Planck's constant)

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Answer: (c)

  1. Magnetic force gives circular motion: qvB=mv2r⇒r=mvqBqvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}.
  2. Bohr condition: mvr=m2v2qB=nh2π⇒v2=nhqB2πm2mvr = \frac{m^2 v^2}{qB} = \frac{nh}{2\pi} \Rightarrow v^2 = \frac{nhqB}{2\pi m^2}.
  3. Magnetic field does no work, so energy is only kinetic: E=12mv2=nhqB4πmE = \frac{1}{2}mv^2 = \frac{nhqB}{4\pi m}
  4. So the answer is (c) nhqB4πm\frac{nhqB}{4\pi m}.
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Q46MHT CET 2026 · 11 APR · SHIFT 2Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Bohr model is applied to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a transverse magnetic field B. The energy of the charged particle in the nthn^{th} level will be (h=Planck's constant)

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Answer: (d)

  1. qvB=mv2r⇒r=mvqBqvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}.
  2. Bohr condition: mvr=m2v2qB=nh2π⇒v2=nhqB2πm2mvr = \frac{m^2 v^2}{qB} = \frac{nh}{2\pi} \Rightarrow v^2 = \frac{nhqB}{2\pi m^2}.
  3. Energy is only kinetic: E=12mv2=nhqB4πmE = \frac{1}{2}mv^2 = \frac{nhqB}{4\pi m}
  4. So the answer is (d) nhqB4πm\frac{nhqB}{4\pi m}.
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Q47JEE Main 2025 · 4 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

Given below are two statements:

Statement (I): The dimensions of Planck's constant and angular momentum are same. Statement (II): In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.

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

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Answer: (b)

  1. Statement I: [h]=[E/f]=[h] = [E/f] = J s =[ML2T−1]=[L]= [ML^2T^{-1}] = [L]. True.
  2. Statement II: Bohr's condition is L=nh2πL = n\dfrac{h}{2\pi}, an integral multiple of h2π\dfrac{h}{2\pi}, not of hh. False.
  3. So the answer is (b).
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Q48JEE Main 2025 · 3 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

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

Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason R\textbf{R}: The formulation of Bohr model does not include repulsive force between electrons. In the light of the above statements, choose the correct answer from the options given below

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Answer: (b)

  1. A: true, Bohr's model works only for one-electron systems (H, He+^+, Li2+^{2+}...).
  2. R: true, it considers only electron-nucleus attraction and ignores electron-electron repulsion.
  3. That omission is exactly why it fails for many-electron atoms, so R explains A.
  4. So the answer is (b).
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Q49JEE Main 2025 · 4 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

Considering the Bohr model of hydrogen like atoms, the ratio of the radius of 55th orbit of the electron in Li2+\text{Li}^{2+} and He+\text{He}^{+}is

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Answer: (b)

  1. rn=n2a0Zr_n = \dfrac{n^2 a_0}{Z}; same n=5n = 5, so r∝1Zr \propto \dfrac{1}{Z}.
  2. rLi2+rHe+=ZHeZLi=23\dfrac{r_{\text{Li}^{2+}}}{r_{\text{He}^+}} = \dfrac{Z_{\text{He}}}{Z_{\text{Li}}} = \dfrac{2}{3}
  3. So the answer is (b).
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Q50JEE Main 2025 · 2 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of Li++\text{Li}^{++} ion in its ground state is given by 1Xa0\frac{1}{X}a_0, where X=X = __________ (Where a0a_0 is the first Bohr's radius.)

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Answer: (d)

  1. rn=n2a0Zr_n = \dfrac{n^2 a_0}{Z}; ground state n=1n = 1, Li++^{++} has Z=3Z = 3.
  2. r=a03⇒X=3r = \dfrac{a_0}{3} \Rightarrow X = 3.
  3. So the answer is (d) 3.
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Q51MHT CET 2025 · 19 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

The frequency of revolution of an electron in the nnth orbit of hydrogen atom is

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Answer: (d)

  1. f=v2πrf = \frac{v}{2\pi r} with v∝1nv \propto \frac{1}{n}, r∝n2r \propto n^2.
  2. f∝1/nn2=1n3f \propto \frac{1/n}{n^2} = \frac{1}{n^3}
  3. So the answer is (d) inversely proportional to n3n^3.
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Q52JEE Main 2025 · 28 JAN · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as:

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Answer: (c)

  1. f=v2πrf = \frac{v}{2\pi r}, v∝1nv \propto \frac{1}{n}, r∝n2r \propto n^2.
  2. f∝1n3f \propto \frac{1}{n^3}
  3. So the answer is (c) 1n3\frac{1}{n^3}.
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Q53MHT CET 2025 · 26 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The magnetic moment of electron due to orbital motion is proportional to ( n=n = principal quantum number)

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Answer: (a)

  1. μ=eL2m\mu = \frac{eL}{2m} with L=nh2πL = \frac{nh}{2\pi}.
  2. μ=neh4πm∝n\mu = n\frac{eh}{4\pi m} \propto n
  3. So the answer is (a) nn.
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Q54MHT CET 2025 · 25 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

In hydrogen atom in its ground state, the first Bohr orbit has radius ' r1r_1 '. When the atom is raised to one of its excited states, the electrons orbital velocity becomes one-third. The radius of that orbit is

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Answer: (d)

  1. vn∝1nv_n \propto \frac{1}{n}, so v=v13⇒n=3v = \frac{v_1}{3} \Rightarrow n = 3
  2. rn∝n2⇒r=9r1r_n \propto n^2 \Rightarrow r = 9r_1
  3. So the answer is (d) 9r19r_1.
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Q55MHT CET 2025 · 20 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

The ratio of angular momentum of an electron in nnth orbit of hydrogen atom to the velocity of electron in nnth orbit is proportional to

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Answer: (a)

  1. L=mvr⇒Lv=mrL = mvr \Rightarrow \frac{L}{v} = mr
  2. r∝n2r \propto n^2, so Lv∝n2\frac{L}{v} \propto n^2
  3. So the answer is (a) n2n^2.
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Q56MHT CET 2025 · 21 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( h=h = Planck's constant, I=I = moment of inertia of diatomic molecule)

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Answer: (a)

  1. Quantisation: L=nh2πL = \frac{nh}{2\pi}, so KE=L22I=n2h28π2IKE = \frac{L^2}{2I} = \frac{n^2 h^2}{8\pi^2 I}.
  2. For n=3n = 3: KE=9h28π2IKE = \frac{9h^2}{8\pi^2 I}
  3. So the answer is (a) 9h28π2I\frac{9h^2}{8\pi^2 I}.
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Q57NEET 2025Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

A particle of mass mm is moving around the origin with a constant force FF pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the nnth orbit and the particle's speed vv in the orbit depend on nn as

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Answer: (a)

  1. Constant force gives the centripetal force: mv2r=F⇒v2=Frm\frac{mv^2}{r} = F \Rightarrow v^2 = \frac{Fr}{m}.
  2. Bohr condition: v=nh2πmr⇒n2h24π2m2r2=Frmv = \frac{nh}{2\pi m r} \Rightarrow \frac{n^2h^2}{4\pi^2 m^2 r^2} = \frac{Fr}{m}.
  3. So r3∝n2⇒r∝n2/3r^3 \propto n^2 \Rightarrow r \propto n^{2/3}, and v∝r∝n1/3v \propto \sqrt{r} \propto n^{1/3}.
  4. So the answer is (a) r∝n2/3; v∝n1/3r \propto n^{2/3};\ v \propto n^{1/3}.
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Q58MHT CET 2024 · 11 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The angular momentum of the electron in the third Bohr orbit of hydrogen atom is ' ll '. Its angular momentum in the fourth Bohr orbit is

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Answer: (b)

  1. L=nh2π⇒L∝nL = \dfrac{nh}{2\pi} \Rightarrow L \propto n.
  2. L4L3=43⇒L4=43l\dfrac{L_4}{L_3} = \dfrac{4}{3} \Rightarrow L_4 = \dfrac{4}{3}l.
  3. So the answer is (b).
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Q59JEE Main 2024 · 4 APR · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

According to Bohr's theory, the moment of momentum of an electron revolving in 44th orbit of hydrogen atom is:

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Answer: (a)

  1. Moment of momentum =L=nh2π= L = \dfrac{nh}{2\pi}.
  2. For n=4n = 4: L=4h2π=2hπL = \dfrac{4h}{2\pi} = \dfrac{2h}{\pi}.
  3. So the answer is (a).
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Q60MHT CET 2024 · 3 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The ratio of the radius of the first Bohr orbit to that of the second Bohr orbit of the orbital electron is

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Answer: (c)

  1. r∝n2⇒r1r2=1222=14r \propto n^2 \Rightarrow \dfrac{r_1}{r_2} = \dfrac{1^2}{2^2} = \dfrac{1}{4}.
  2. So the answer is (c) 1:41 : 4.
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Q61MHT CET 2024 · 2 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The radius of innermost orbit of hydrogen atom is 5.3×10−11 m5.3 \times 10^{-11}\text{ m}. The radius of fourth allowed orbit of hydrogen atom is

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Answer: (a)

  1. rn=n2r1r_n = n^2 r_1
  2. r4=16×0.53 A˚=8.48 A˚r_4 = 16 \times 0.53\ \mathring{A} = 8.48\ \mathring{A}
  3. So the answer is (a).
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Q62JEE Main 2024 · 27 JAN · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The radius of third stationary orbit of electron for Bohr's atom is R. The radius of fourth stationary orbit will be:

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Answer: (b)

  1. Bohr radius rn∝n2r_n \propto n^2 (same atom).
  2. r4r3=4232=169\frac{r_4}{r_3} = \frac{4^2}{3^2} = \frac{16}{9}
  3. So the answer is (b) 169R\frac{16}{9}R.
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Q63MHT CET 2024 · 10 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

In the Bohr model of hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If ' r0r_0 ' is the radius of the ground state orbit, ' mm ' is the mass, ' e ' is the charge on the electron and ' ε0\varepsilon_0 ' is the permittivity of vacuum, the speed of the electron is

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Answer: (c)

  1. Centripetal force = Coulomb force: mv2r0=e24πε0r02\frac{mv^2}{r_0} = \frac{e^2}{4\pi\varepsilon_0 r_0^2}
  2. v2=e24πε0r0mv^2 = \frac{e^2}{4\pi\varepsilon_0 r_0 m}
  3. So the answer is (c) v=e4πε0r0mv = \frac{e}{\sqrt{4\pi\varepsilon_0 r_0 m}}.
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Q64MHT CET 2024 · 9 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

In hydrogen atom, if VnV_n and VpV_p are orbital velocities in nnth and ppth orbit respectively, then the ratio Vp:VnV_p : V_n is

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Answer: (b)

  1. Orbital speed vn∝1nv_n \propto \frac{1}{n}.
  2. VpVn=np\frac{V_p}{V_n} = \frac{n}{p}
  3. So the answer is (b) n:pn : p.
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Q65MHT CET 2024 · 16 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

When an electron orbiting in hydrogen atom in its ground state jumps to higher excited state, the de- Broglie wavelength associated with it

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Answer: (d)

  1. λ=hmv\lambda = \frac{h}{mv} and v∝1nv \propto \frac{1}{n}, so λ∝n\lambda \propto n.
  2. Moving to a higher nn lowers the speed, so λ\lambda grows.
  3. So the answer is (d) will increase.
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Q66MHT CET 2024 · 15 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The ratio of the areas of the electron orbits for the second excited state to the first excited state for the hydrogen atom is

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Answer: (d)

  1. Second excited state n=3n = 3, first excited state n=2n = 2.
  2. Area =πr2= \pi r^2 and r∝n2r \propto n^2, so area ∝n4\propto n^4.
  3. A3A2=3424=8116\frac{A_3}{A_2} = \frac{3^4}{2^4} = \frac{81}{16}
  4. So the answer is (d) 81:1681 : 16.
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Q67MHT CET 2024 · 11 MAY · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

In a hydrogen atom in its ground state, the first Bohr orbit has radius r1r_1. The electron's orbital speed becomes one-third when the atom is raised to one of its excited states. The radius of the orbit in that excited state is

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Answer: (c)

  1. vn∝1nv_n \propto \frac{1}{n}, so v=v13⇒n=3v = \frac{v_1}{3} \Rightarrow n = 3
  2. rn∝n2⇒r=9r1r_n \propto n^2 \Rightarrow r = 9r_1
  3. So the answer is (c) 9r19r_1.
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Q68JEE Main 2024 · 5 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

The angular momentum of an electron in a hydrogen atom is proportional to: (Where rr is the radius of orbit of electron)

Show answer and solution

Answer: (c)

  1. L=nh2π∝nL = \frac{nh}{2\pi} \propto n
  2. r∝n2⇒n∝rr \propto n^2 \Rightarrow n \propto \sqrt{r}
  3. So L∝rL \propto \sqrt{r}, answer (c).
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Q69MHT CET 2024 · 10 MAY · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

Radius of first orbit in H -atom is ' a0a_0 ' Then, de-Broglie wavelength of electron in the third orbit is

Show answer and solution

Answer: (b)

  1. Bohr condition: nλ=2πrn⇒λ=2πrnnn\lambda = 2\pi r_n \Rightarrow \lambda = \frac{2\pi r_n}{n}
  2. r3=9a0⇒λ=2π×9a03=6πa0r_3 = 9a_0 \Rightarrow \lambda = \frac{2\pi\times9a_0}{3} = 6\pi a_0
  3. So the answer is (b) 6πa06\pi a_0.
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Q70MHT CET 2024 · 10 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

Using Bohr's model, the orbital period of electron in hydrogen atom in nnth orbit is ( m=m = mass of electron, h=h = Planck's constant, e=e = electronic charge, ε0=\varepsilon_0 = permittivity of free space)

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Answer: (c)

  1. rn=ε0n2h2πme2r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}, vn=e22ε0nhv_n = \frac{e^2}{2\varepsilon_0 nh}
  2. T=2πrnvn=4ε02n3h3me4T = \frac{2\pi r_n}{v_n} = \frac{4\varepsilon_0^2 n^3 h^3}{m e^4}
  3. So the answer is (c).
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Q71JEE Main 2024 · 30 JAN · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

An electron revolving in nnth Bohr orbit has magnetic moment μn\mu_n. If μn∝nx\mu_n \propto n^x, the value of xx is

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Answer: (d)

  1. μn=e2mL=e2m⋅nh2π\mu_n = \frac{e}{2m}L = \frac{e}{2m}\cdot\frac{nh}{2\pi}
  2. So μn∝n1\mu_n \propto n^1
  3. So the answer is (d) x=1x = 1.
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Q72MHT CET 2024 · 15 MAY · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

Acceleration of an electron in the first Bohr's orbit is proportional to (m=m = mass of electron, r=r = radius of the orbit, h=h = Planck's constant)

Show answer and solution

Answer: (b)

  1. mvr=h2π⇒v=h2πmrmvr = \frac{h}{2\pi} \Rightarrow v = \frac{h}{2\pi m r}
  2. a=v2r=h24π2m2r3a = \frac{v^2}{r} = \frac{h^2}{4\pi^2 m^2 r^3}
  3. So a∝h2m2r3a \propto \frac{h^2}{m^2 r^3}, answer (b).
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Q73MHT CET 2024 · 3 MAY · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

A diatomic molecule has moment of inertia ' I ', By applying Bohr's quantization condition, its rotational energy in the nnth level is [n≥1][n \geq 1] [h= Planck's constant]

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Answer: (d)

  1. Quantisation: L=Iω=nh2πL = I\omega = \frac{nh}{2\pi}.
  2. E=L22I=n2h24π2⋅2I=n2(h28π2I)E = \frac{L^2}{2I} = \frac{n^2 h^2}{4\pi^2 \cdot 2I} = n^2\left(\frac{h^2}{8\pi^2 I}\right)
  3. So the answer is (d) n2(h28π2I)n^2\left(\frac{h^2}{8\pi^2 I}\right).
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Q74MHT CET 2023 · 13 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

When an electron is excited from its 4 th orbit to 5 th stationary orbit, the change in the angular momentum of electron is approximately.

(Planck's constant =h=6.63×10−34 J-s= h = 6.63 \times 10^{-34}\text{ J-s})

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Answer: (c)

  1. ΔL=(5−4)h2π=h2π\Delta L = (5-4)\dfrac{h}{2\pi} = \dfrac{h}{2\pi}
  2. ΔL=6.63×10−342×3.14≈1.06×10−34\Delta L = \dfrac{6.63\times10^{-34}}{2\times3.14} \approx 1.06\times10^{-34} J s
  3. So the answer is (c) 1×10−341\times10^{-34} J s.
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Q75JEE Main 2023 · 10 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be

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Answer: (a)

  1. Taking LL for the first orbit: L=h2πL = \dfrac{h}{2\pi}.
  2. Second orbit: L2=2h2π=2LL_2 = \dfrac{2h}{2\pi} = 2L.
  3. Change =2L−L=L= 2L - L = L. So the answer is (a).
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Q76NEET 2023Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The radius of inner most orbit of hydrogen atom is 5.3×10−11 m5.3 \times 10^{-11}\text{ m}. What is the radius of third allowed orbit of hydrogen atom?

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Answer: (c)

  1. rn=n2r1r_n = n^2 r_1
  2. r3=9×0.53 A˚=4.77 A˚r_3 = 9 \times 0.53\ \mathring{A} = 4.77\ \mathring{A}
  3. So the answer is (c).
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Q77JEE Main 2023 · 31 JAN · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be

Show answer and solution

Answer: (a)

  1. Bohr radius rn∝n2r_n \propto n^2.
  2. r3r2=94=2.25\frac{r_3}{r_2} = \frac{9}{4} = 2.25
  3. So the answer is (a) 2.25R2.25R.
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Q78JEE Main 2023 · 6 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The radius of fifth orbit of the Li++\text{Li}^{++} is __________ ×10−12 m\times 10^{-12}\text{ m}.

Take: radius of hydrogen atom =0.51 A˚= 0.51\ \mathring{A}

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

Show answer and solution

Answer: 425

  1. rn=r0n2Zr_n = r_0\frac{n^2}{Z}, with r0=0.51 A˚r_0 = 0.51\ \mathring{A}, Z=3Z = 3 for Li++\text{Li}^{++}, n=5n = 5.
  2. r5=0.51×253=4.25 A˚r_5 = 0.51 \times \frac{25}{3} = 4.25\ \mathring{A}
  3. 4.25 A˚=425×10−12 m4.25\ \mathring{A} = 425 \times 10^{-12}\text{ m}
  4. So the answer is 425425 (×10−12\times 10^{-12} m).
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Q79MHT CET 2023 · 10 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

The ratio of the velocity of the electron in the first Bohr orbit to that in the second Bohr orbit of hydrogen atom is

Show answer and solution

Answer: (b)

  1. vn∝1nv_n \propto \frac{1}{n}
  2. v1v2=21\frac{v_1}{v_2} = \frac{2}{1}
  3. So the answer is (b) 2:12 : 1.
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Q80JEE Main 2023 · 30 JAN · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

Speed of an electron in Bohr's 77th orbit for Hydrogen atom is 3.6×106 m/s3.6 \times 10^{6}\text{ m/s}. The corresponding speed of the electron in 33rd orbit, in m/s\text{m/s} is:

Show answer and solution

Answer: (c)

  1. vn∝1nv_n \propto \frac{1}{n}, so v3=v7×73v_3 = v_7 \times \frac{7}{3}.
  2. v3=3.6×106×73=8.4×106 m/sv_3 = 3.6 \times 10^6 \times \frac{7}{3} = 8.4 \times 10^6\text{ m/s}
  3. So the answer is (c) 8.4×1068.4 \times 10^6.
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Q81MHT CET 2023 · 9 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

For an electron moving in the nnth Bohr orbit the deBroglie wavelength of an electron is

Show answer and solution

Answer: (d)

  1. Standing-wave condition: 2πr=nλ2\pi r = n\lambda.
  2. λ=2πrn\lambda = \frac{2\pi r}{n}
  3. So the answer is (d) 2πrn\frac{2\pi r}{n}.
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Q82MHT CET 2023 · 14 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

Magnetic field at the centre of the hydrogen atom due to motion of electron in nnth orbit is proportional to

Show answer and solution

Answer: (d)

  1. B=μ0I2rB = \frac{\mu_0 I}{2r}, with I=ev2πrI = \frac{ev}{2\pi r}.
  2. v∝n−1v \propto n^{-1}, r∝n2r \propto n^2, so I∝n−3I \propto n^{-3}.
  3. B∝n−3n2=n−5B \propto \frac{n^{-3}}{n^2} = n^{-5}
  4. So the answer is (d) n−5n^{-5}.
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Q83MHT CET 2023 · 11 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The orbital magnetic moment associated with orbiting electron of charge 'ee' is

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Answer: (c)

  1. μ=IA=ev2πrπr2=e2m(mvr)=e2mL\mu = IA = \frac{ev}{2\pi r}\pi r^2 = \frac{e}{2m}(mvr) = \frac{e}{2m}L
  2. So μ∝L\mu \propto L; it is inversely (not directly) proportional to mm.
  3. So the answer is (c) directly proportional to angular momentum.
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Q84MHT CET 2023 · 9 MAY · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

The force acting on the electron in hydrogen atom (Bohr' theory) is related to the principle quantum number 'nn' as

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Answer: (b)

  1. Coulomb force F=ke2r2F = \frac{ke^2}{r^2}, r∝n2r \propto n^2.
  2. F∝n−4F \propto n^{-4}
  3. So the answer is (b) n−4n^{-4}.
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Q85JEE Main 2023 · 1 FEB · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

A light of energy 12.75 eV12.75\text{ eV} is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is xπ×10−17 eVs\frac{x}{\pi} \times 10^{-17}\text{ eVs}. The value of xx is ___________ (use h=4.14×10−15 eVs, c=3×108 ms−1h = 4.14 \times 10^{-15}\text{ eVs},\ c = 3 \times 10^{8}\text{ ms}^{-1} ).

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

Show answer and solution

Answer: 828

  1. En=−13.6+12.75=−0.85 eV=−13.6n2⇒n=4E_n = -13.6 + 12.75 = -0.85\text{ eV} = -\frac{13.6}{n^2} \Rightarrow n = 4
  2. L=nh2π=4h2π=2hπL = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}
  3. L=2×4.14×10−15π=828π×10−17 eVsL = \frac{2\times4.14\times10^{-15}}{\pi} = \frac{828}{\pi}\times10^{-17}\text{ eVs}
  4. So x=828x = 828.
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Q86JEE Main 2023 · 13 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

The radius of 22nd orbit of He+\text{He}^{+} of Bohr's model is r1r_1 and that of fourth orbit of Be3+\text{Be}^{3+} is represented as r2r_2. Now the ratio r2r1\frac{r_2}{r_1} is x:1x : 1. The value of xx is ___________.

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

Show answer and solution

Answer: 2

  1. r∝n2Zr \propto \frac{n^2}{Z}
  2. He+\text{He}^+, n=2n=2: r1∝42=2r_1 \propto \frac{4}{2} = 2; Be3+\text{Be}^{3+}, n=4n=4: r2∝164=4r_2 \propto \frac{16}{4} = 4
  3. r2r1=42=2\frac{r_2}{r_1} = \frac{4}{2} = 2
  4. So x=2x = 2.
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Q87JEE Main 2023 · 6 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

Experimentally it is found that 12.8 eV12.8\text{ eV} energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is 9x×10−10 m\frac{9}{x} \times 10^{-10}\text{ m}. The value of the xx is __________.

(1 eV=1.6×10−19 J, 14πϵ0=9×109 Nm2/C2\Bigg(1\text{ eV} = 1.6 \times 10^{-19}\text{ J},\ \frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ Nm}^2/\text{C}^2 and electronic charge =1.6×10−19 C)= 1.6 \times 10^{-19}\text{ C}\Bigg)

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

Show answer and solution

Answer: 16

  1. Ionisation energy =ke22r⇒r=ke22E= \frac{ke^2}{2r} \Rightarrow r = \frac{ke^2}{2E}
  2. r=9×109×(1.6×10−19)22×12.8×1.6×10−19=14.4×10−1025.6r = \frac{9\times10^9\times(1.6\times10^{-19})^2}{2\times12.8\times1.6\times10^{-19}} = \frac{14.4\times10^{-10}}{25.6}
  3. r=0.5625×10−10=916×10−10 mr = 0.5625\times10^{-10} = \frac{9}{16}\times10^{-10}\text{ m}
  4. So x=16x = 16.
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Q88MHT CET 2023 · 11 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

An electron in the hydrogen atom jumps from the first excited state to the ground state. What will be the percentage change in the speed of electron?

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Answer: (b)

  1. v∝1nv \propto \frac{1}{n}: from n=2n = 2 to n=1n = 1, v1=2v2v_1 = 2v_2.
  2. Change relative to the initial speed: v1−v2v2×100=100%\frac{v_1 - v_2}{v_2}\times100 = 100\%
  3. Change relative to the final speed: v1−v2v1×100=50%\frac{v_1 - v_2}{v_1}\times100 = 50\%
  4. The key takes the second form: (b) 50%.
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Q89MHT CET 2023 · 14 MAY · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

If the radius of the first Bohr orbit is 'rr' then the de-Broglie wavelength of the electron in the 44th orbit will be

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Answer: (c)

  1. λ=2πrnn\lambda = \frac{2\pi r_n}{n} with r4=16rr_4 = 16r
  2. λ=2π×16r4=8πr\lambda = \frac{2\pi\times16r}{4} = 8\pi r
  3. So the answer is (c) 8πr8\pi r.
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Q90MHT CET 2023 · 11 MAY · SHIFT 2Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Bohr model is applied to a particle of mass 'mm' and charge 'qq' moving in a plane under the influence of a transverse magnetic field 'BB'. The energy of the charged particle in the nnth leve will be [h=[h = Planck's constant ]]

Show answer and solution

Answer: (a)

  1. qvB=mv2r⇒r=mvqBqvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}.
  2. Bohr condition: mvr=m2v2qB=nh2π⇒v2=nhqB2πm2mvr = \frac{m^2 v^2}{qB} = \frac{nh}{2\pi} \Rightarrow v^2 = \frac{nhqB}{2\pi m^2}.
  3. Energy is only kinetic: E=12mv2=nhqB4πmE = \frac{1}{2}mv^2 = \frac{nhqB}{4\pi m}
  4. So the answer is (a) nhqB4πm\frac{nhqB}{4\pi m}.
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Q91JEE Main 2023 · 6 APR · SHIFT 2Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

A small particle of mass mm moves in such a way that its potential energy U=12mω2r2U = \frac{1}{2} m\omega^2 r^2 where ω\omega is constant and rr is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of nnth orbit will be proportional to,

Show answer and solution

Answer: (a)

  1. Force F=−dUdr=mω2rF = -\frac{dU}{dr} = m\omega^2 r gives mv2r=mω2r⇒v=ωr\frac{mv^2}{r} = m\omega^2 r \Rightarrow v = \omega r.
  2. Bohr condition: mvr=mωr2=nh2πmvr = m\omega r^2 = \frac{nh}{2\pi}.
  3. So r2∝n⇒r∝nr^2 \propto n \Rightarrow r \propto \sqrt{n}.
  4. So the answer is (a) n\sqrt{n}.
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Q92JEE Main 2022 · 27 JUN · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

Given below are two statements:

Statement I: In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E1) to higher energy orbit (E2), is given as hf = E1 −- E2

Statement II: The jumping of electron from higher energy orbit (E2) to lower energy orbit (E1) is associated with frequency of radiation given as f = (E2 −- E1)/h

This condition is Bohr's frequency condition.

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

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Answer: (d)

  1. Statement I: absorption from E1E_1 to E2E_2 needs hf=E2−E1hf = E_2 - E_1; E1−E2E_1 - E_2 is negative. False.
  2. Statement II: emission from E2E_2 to E1E_1 gives f=E2−E1hf = \dfrac{E_2 - E_1}{h}, Bohr's frequency condition. True.
  3. So the answer is (d).
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Q93NEET 2022 · PHASE 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

Let R1 be the radius of the second stationary orbit and R2 be the radius of the fourth stationary orbit of an electron in Bohr's model. The ratio R1R2\frac{R_1}{R_2} is:

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Answer: (b)

  1. r∝n2⇒R1R2=2242=416r \propto n^2 \Rightarrow \dfrac{R_1}{R_2} = \dfrac{2^2}{4^2} = \dfrac{4}{16}
  2. R1R2=0.25\dfrac{R_1}{R_2} = 0.25. So the answer is (b).
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Q94JEE Main 2022 · 30 JUL · SHIFT 2Easy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

Using Boh'r model of quantization of angular momentum the relation between the radius 'r' of the nnth allowed orbit of quantum number 'n' for an electron in hydrogen atom is:

Show answer and solution

Answer: (b)

  1. Quantization mvr=nh2πmvr = \frac{nh}{2\pi} with mv2r=ke2r2\frac{mv^2}{r} = \frac{ke^2}{r^2} gives r=n2h2ε0πme2r = \frac{n^2h^2\varepsilon_0}{\pi m e^2}.
  2. So the answer is (b) r∝n2r \propto n^2.
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Q95JEE Main 2022 · 25 JUN · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

The ratio for the speed of the electron in the 3rd orbit of He+ to the speed of the electron in the 3rd orbit of hydrogen atom will be:

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Answer: (d)

  1. vn=2.18×106Zn m/sv_n = 2.18 \times 10^6 \frac{Z}{n}\text{ m/s}, so v∝Zv \propto Z for the same nn.
  2. vHe+vH=21\frac{v_{He^+}}{v_H} = \frac{2}{1}
  3. So the answer is (d) 2:12 : 1.
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Q96JEE Main 2022 · 25 JUL · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

The momentum of an electron revolving in nnth orbit is given by:

(Symbols have their usual meanings)

Show answer and solution

Answer: (a)

  1. Angular momentum L=pr=nh2πL = pr = \frac{nh}{2\pi}.
  2. p=nh2πrp = \frac{nh}{2\pi r}
  3. So the answer is (a) nh2πr\frac{nh}{2\pi r}.
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Q97JEE Main 2022 · 25 JUL · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:

Show answer and solution

Answer: (b)

  1. μ=IA=e2mL\mu = IA = \frac{e}{2m}L in magnitude.
  2. Electron charge is −e-e, so μ⃗L\vec{\mu}_L is opposite to L⃗\vec{L}.
  3. So the answer is (b) μ⃗L=−eL⃗2m\vec{\mu}_L = -\frac{e\vec{L}}{2m}.
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Q98JEE Main 2022 · 30 JUN · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by:

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Answer: (b)

  1. Final energy =−13.6+12.09=−1.51 eV=−13.69= -13.6 + 12.09 = -1.51\text{ eV} = -\frac{13.6}{9}, so n=3n = 3.
  2. ΔL=(3−1)h2π=hπ\Delta L = (3 - 1)\frac{h}{2\pi} = \frac{h}{\pi}
  3. ΔL=6.63×10−343.14=2.11×10−34 Js\Delta L = \frac{6.63 \times 10^{-34}}{3.14} = 2.11 \times 10^{-34}\text{ Js}
  4. So the answer is (b) 2.11×10−342.11 \times 10^{-34} Js.
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Q99JEE Main 2022 · 27 JUN · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of:

(Given, Planck's constant = 6.6 ×\times 10−-34 Js).

Show answer and solution

Answer: (b)

  1. E1=−13.6 eVE_1 = -13.6\text{ eV}, E2=−3.4 eVE_2 = -3.4\text{ eV}, so 10.2 eV10.2\text{ eV} takes the electron from n=1n=1 to n=2n=2.
  2. Bohr: L=nh2πL = \frac{nh}{2\pi}, so ΔL=h2π\Delta L = \frac{h}{2\pi}
  3. ΔL=6.6×10−342×3.14≈1.05×10−34 Js\Delta L = \frac{6.6\times10^{-34}}{2\times3.14} \approx 1.05\times10^{-34}\text{ Js}
  4. So the answer is (b) 1.05×10−34 Js1.05\times10^{-34}\text{ Js}.
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Q100MHT CET 2022 · 11 AUG · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

Using Bohr's model, the orbital period of electron in hydrogen atom in the nnth orbit is (ε0=(\varepsilon_0 = permittivity of vacuum, h=h = Planck's constant, m=m = mass of electron, e=e = electronic charge)

Show answer and solution

Answer: (c)

  1. rn=ε0n2h2πme2r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}, vn=e22ε0nhv_n = \frac{e^2}{2\varepsilon_0 nh}
  2. T=2πrnvn=4ε02n3h3me4T = \frac{2\pi r_n}{v_n} = \frac{4\varepsilon_0^2 n^3 h^3}{m e^4}
  3. So the answer is (c).
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Q101JEE Main 2021 · 17 MAR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

If an electron is moving in the nth orbit of the hydrogen atom, then its velocity (vn) for the nth orbit is given as:

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Answer: (a)

  1. vn=e22ε0nhv_n = \frac{e^2}{2\varepsilon_0 nh}, so speed falls as nn rises.
  2. So the answer is (a) vn∝1nv_n \propto \frac{1}{n}.
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Q102MHT CET 2021 · 23 SEP · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The energy of an electron in the excited hydrogen atom is −3.4 eV-3.4\text{ eV}. Then according to Bohr's theory, the angular momentum of the electron in that excited state is (h=h = Plank's constant)

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Answer: (c)

  1. En=−13.6n2E_n = -\frac{13.6}{n^2}: n2=13.63.4=4n^2 = \frac{13.6}{3.4} = 4, so n=2n = 2.
  2. L=2h2π=hπL = \frac{2h}{2\pi} = \frac{h}{\pi}
  3. So the answer is (c) hπ\frac{h}{\pi}.
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Q103MHT CET 2021 · 21 SEP · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

When an electron in hydrogen atom jumps from third excited state to the ground state, the de- Broglie wavelength associated with the electron becomes

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Answer: (b)

  1. Third excited state is n=4n = 4; ground state n=1n = 1.
  2. λ=2πrnn∝n2n=n\lambda = \frac{2\pi r_n}{n} \propto \frac{n^2}{n} = n
  3. λ1λ4=14\frac{\lambda_1}{\lambda_4} = \frac{1}{4}
  4. So the answer is (b) 14\frac{1}{4}th.
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Q104JEE Main 2021 · 27 JUL · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 A˚\mathring{A}. If the speed of electron is 2.2 ×\times 106 m/s, then the current associated with the electron will be _____________ ×\times 10−-2 mA. [Take π\pi as 227\frac{22}{7} ]

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

Show answer and solution

Answer: 112

  1. I=eT=ev2πrI = \frac{e}{T} = \frac{ev}{2\pi r}
  2. I=1.6×10−19×2.2×1062×227×0.5×10−10=3.52×10−133.143×10−10I = \frac{1.6\times10^{-19}\times2.2\times10^6}{2\times\frac{22}{7}\times0.5\times10^{-10}} = \frac{3.52\times10^{-13}}{3.143\times10^{-10}}
  3. I=1.12×10−3 A=1.12 mA=112×10−2 mAI = 1.12\times10^{-3}\text{ A} = 1.12\text{ mA} = 112\times10^{-2}\text{ mA}
  4. So the answer is 112.
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Q105MHT CET 2021 · 23 SEP · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

In the Bohr model, an electron moves in a circular orbit around the nucleus. Considering an orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nth excited state, is

(e = electronic charge, me{}_{e} = mass of the electron, h = Planck's constant)

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Answer: (d)

  1. M=evr2=e2me(mevr)=e2meLM = \frac{evr}{2} = \frac{e}{2m_e}(m_e v r) = \frac{e}{2m_e}L
  2. Bohr: L=nh2πL = \frac{nh}{2\pi}
  3. M=(e2me)nh2πM = \left(\frac{e}{2m_e}\right)\frac{nh}{2\pi}, so the answer is (d).
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Q106MHT CET 2021 · 20 SEP · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The gyromagnetic ratio of an electron in an hydrogen atom, according to Bohr model is

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Answer: (b)

  1. Gyromagnetic ratio =ML=e2m= \frac{M}{L} = \frac{e}{2m}
  2. It contains no nn, so it is the same for every orbit.
  3. So the answer is (b) independent of which orbit it is in.
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Q107MHT CET 2021 · 21 SEP · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

Ratio centripetal acceleration for an electron revolving in 3rd and 5th Bohr orbit of hydrogen atom is

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Answer: (b)

  1. Centripetal acceleration a=v2ra = \frac{v^2}{r} with v∝1nv \propto \frac{1}{n} and r∝n2r \propto n^2.
  2. So a∝1/n2n2=1n4a \propto \frac{1/n^2}{n^2} = \frac{1}{n^4}.
  3. a3a5=5434=62581\frac{a_3}{a_5} = \frac{5^4}{3^4} = \frac{625}{81}
  4. So the answer is (b) 625:81625 : 81.
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Q108JEE Main 2021 · 18 MAR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Imagine that the electron in a hydrogen atom is replaced by a muon (μ\mu). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be:

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Answer: (b)

  1. Ionisation energy E=me48ϵ02h2∝mE = \frac{me^4}{8\epsilon_0^2 h^2} \propto m.
  2. Eμ=207×13.6=2815.2 eVE_\mu = 207 \times 13.6 = 2815.2\text{ eV}, so the ionisation potential is 2815.2 V2815.2\text{ V} (2815.2 eV2815.2\text{ eV}).
  3. So the answer is (b) 2815.2 eV.
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Q109JEE Main 2021 · 17 MAR · SHIFT 2Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

A particle of mass m moves in a circular orbit in a

central potential field U(r) = U0r4. If Bohr's quantization conditions

are applied, radii of possible

orbitals rn vary with nαn^{\alpha}, where α\alpha is ____________.

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

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Answer: 3

  1. F=dUdr=4U0r3=mv2r⇒v=2r2U0/mF = \frac{dU}{dr} = 4U_0 r^3 = \frac{mv^2}{r} \Rightarrow v = 2r^2\sqrt{U_0/m}, so v∝r2v \propto r^2.
  2. Bohr condition: mvr=nh2πmvr = \frac{nh}{2\pi} with mvr∝r3mvr \propto r^3, so r3∝nr^3 \propto n.
  3. rn∝n1/3r_n \propto n^{1/3}, which is the original question's form rn∝n1/αr_n \propto n^{1/\alpha}.
  4. So α=3\alpha = 3.
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Q110NEET 2020 · PHASE 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

For which one of the following, Bohr model is not valid?

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Answer: (c)

  1. Bohr model works only for one-electron (hydrogen-like) systems.
  2. He+^+, deuterium and hydrogen each have one electron; Ne+^+ has 9 electrons.
  3. So the answer is (c) Ne+^+.
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Q111MHT CET 2020 · 16 OCT · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

The force acting on the electrons in hydrogen atom (Bohr's theory) is related to the principle quantum number nn as

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Answer: (a)

  1. F=ke2r2F = \frac{ke^2}{r^2} and r∝n2r \propto n^2.
  2. F∝n−4F \propto n^{-4}
  3. So the answer is (a) n−4n^{-4}.
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Q112MHT CET 2020 · 16 OCT · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

The ratio of speed of an electron in the ground state in the Bohr's first orbit of hydrogen atom to velocity of light (c)(c) is ( h=h = Planck's constant, ε0=\varepsilon_0 = permittivity of free space, e=e = charge on electron)

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Answer: (c)

  1. Bohr speed vn=Ze22ε0nhv_n = \frac{Ze^2}{2\varepsilon_0 nh}; for H, n=1n = 1: v1=e22ε0hv_1 = \frac{e^2}{2\varepsilon_0 h}
  2. v1c=e22ε0hc\frac{v_1}{c} = \frac{e^2}{2\varepsilon_0 hc}
  3. So the answer is (c) e22ε0hc\frac{e^2}{2\varepsilon_0 hc}.
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Q113MHT CET 2020 · 19 OCT · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

Using Bohr's model, the orbital period of electron in hydrogen atom in nnth orbit is ( ε0=\varepsilon_0 = permittivity of free space, h=h = Planck's constant, m=m = mass of electron and e=e = electronic charge)

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Answer: (d)

  1. rn=ε0n2h2πme2r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}, vn=e22ε0nhv_n = \frac{e^2}{2\varepsilon_0 nh}
  2. T=2πrnvn=4ε02n3h3me4T = \frac{2\pi r_n}{v_n} = \frac{4\varepsilon_0^2 n^3 h^3}{m e^4}
  3. So the answer is (d).
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Q114JEE Main 2020 · 7 JAN · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 ×\times 10-16 s. The frequency of revolution of the electron in its first excited state (in s-1) is:

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Answer: (c)

  1. T∝n3T \propto n^3, so T2=8T1=8×1.6×10−16=1.28×10−15 sT_2 = 8T_1 = 8\times1.6\times10^{-16} = 1.28\times10^{-15}\text{ s}
  2. f2=1T2=11.28×10−15≈7.8×1014 s−1f_2 = \frac{1}{T_2} = \frac{1}{1.28\times10^{-15}} \approx 7.8\times10^{14}\text{ s}^{-1}
  3. So the answer is (c) 7.8×10147.8\times10^{14}.
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Q115MHT CET 2020 · 16 OCT · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Using Bohr's quantisation condition, what is the rotational energy in the second orbit for a diatomic molecule? (I=I = moment of inertia of diatomic molecule and, h=h = Planck's constant)

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Answer: (b)

  1. Quantisation: L=nh2πL = \frac{nh}{2\pi}, so E=L22I=n2h28π2IE = \frac{L^2}{2I} = \frac{n^2 h^2}{8\pi^2 I}.
  2. For n=2n = 2: E=4h28π2I=h22π2IE = \frac{4h^2}{8\pi^2 I} = \frac{h^2}{2\pi^2 I}
  3. So the answer is (b) h22Iπ2\frac{h^2}{2I\pi^2}.
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Q116MHT CET 2019 · 3 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

If the speed of an electron of hydrogen atom in the ground state is 2.2×106 m/s2.2 \times 10^{6}\text{ m/s}, then its speed in the third excited state will be

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Answer: (b)

  1. Third excited state means n=4n = 4; vn=v1nv_n = \frac{v_1}{n}.
  2. v4=2.2×1064=5.5×105 m/sv_4 = \frac{2.2 \times 10^6}{4} = 5.5 \times 10^5\text{ m/s}
  3. So the answer is (b) 5.5×105 m/s5.5 \times 10^5\text{ m/s}.
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Q117MHT CET 2019 · 2 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

The angle made by orbital angular momentum of electron with the direction of the orbital magnetic moment is

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Answer: (c)

  1. μ⃗=−e2mL⃗\vec{\mu} = -\frac{e}{2m}\vec{L}: the electron is negative, so its current is opposite to its motion.
  2. Hence μ⃗\vec{\mu} and L⃗\vec{L} are antiparallel.
  3. So the answer is (c) 180∘180^\circ.
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Q118JEE Main 2019 · 12 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 A˚\mathring{A}). The de-Broglie wavelength of this electron is:

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Answer: (c)

  1. Second excited state: n=3n = 3.
  2. λ=2πrn=2×3.14×4.653 A˚\lambda = \frac{2\pi r}{n} = \frac{2\times3.14\times4.65}{3}\ \mathring{A}
  3. λ≈9.7 A˚\lambda \approx 9.7\ \mathring{A}
  4. So the answer is (c) 9.7 A˚9.7\ \mathring{A}.
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Q119NEET 2019Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

The radius of the first permitted Bohr orbit, for the electron, in a hydrogen atom equals 0.51 Å and its ground state energy equals −13.6 eV-13.6\text{ eV}. If the electron in the hydrogen atom is replaced by muon (μ−\mu^-) [charge same as electron and mass 207 me207\, m_e], the first Bohr radius and ground state energy will be,

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Answer: (d)

  1. Bohr radius r∝1mr \propto \frac{1}{m} and energy E∝mE \propto m.
  2. rμ=0.53×10−10207≈2.56×10−13 mr_\mu = \frac{0.53 \times 10^{-10}}{207} \approx 2.56 \times 10^{-13}\text{ m}
  3. Eμ=−13.6×207≈−2815 eV≈−2.8 keVE_\mu = -13.6 \times 207 \approx -2815\text{ eV} \approx -2.8\text{ keV}
  4. So the answer is (d) 2.56×10−13 m,−2.8 keV2.56 \times 10^{-13}\text{ m}, -2.8\text{ keV}.
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Q120MHT CET 2019 · 2 MAY · SHIFT 1Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Bohr model is applied to a particle of mass ' mm ' and charge ' qq ' is moving in a plane under the influence of a transverse magnetic field ' BB '. The energy of the charged particle in the nnth level will be ( h=h = Planck's constant)

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Answer: (c)

  1. qvB=mv2r⇒r=mvqBqvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}.
  2. Bohr condition: mvr=m2v2qB=nh2π⇒v2=nhqB2πm2mvr = \frac{m^2 v^2}{qB} = \frac{nh}{2\pi} \Rightarrow v^2 = \frac{nhqB}{2\pi m^2}.
  3. Energy is only kinetic: E=12mv2=nhqB4πmE = \frac{1}{2}mv^2 = \frac{nhqB}{4\pi m}
  4. So the answer is (c) nhqB/4πmnhqB/4\pi m.
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Q121JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

A particle of mass m moves in a circular orbit in a central potential field U(r) = 12\frac{1}{2} kr2. If Bohr 's quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as:

Show answer and solution

Answer: (a)

  1. F=kr=mv2r⇒v=rk/mF = kr = \frac{mv^2}{r} \Rightarrow v = r\sqrt{k/m}.
  2. Bohr condition: mvr=mk r2=nh2π⇒rn∝nmvr = \sqrt{mk}\, r^2 = \frac{nh}{2\pi} \Rightarrow r_n \propto \sqrt{n}.
  3. E=12mv2+12kr2=kr2∝nE = \frac{1}{2}mv^2 + \frac{1}{2}kr^2 = kr^2 \propto n
  4. So the answer is (a) rn∝nr_n \propto \sqrt{n}, En∝nE_n \propto n.
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Q122JEE Main 2018 · 15 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Muon (μ−\mu^-) is a negatively charged (|q| = |e|) particle with a mass mμ\mu = 200 me, where me is the mass of the electron and e is the electronic charge. If μ−\mu^- is bound to a proton to form a hydrogen like atom, identify the correct statements.

(A) Radius of the muonic orbit is 200 times smaller than that of the electron.

(B) The speed of the μ−\mu^- in the nth orbit is 1200\frac{1}{200} times that of the electron in the nth orbit. (C) The ionization energy of muonic atom is 200 times more than that of a hydrogen atom. (D) The momentum of the muon in the nth orbit is 200 times more than that of the electron.

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Answer: (b)

  1. (A) r=ϵ0n2h2πme2∝1mr = \frac{\epsilon_0 n^2 h^2}{\pi m e^2} \propto \frac{1}{m}: 200 times smaller. True.
  2. (B) v=e22ϵ0nhv = \frac{e^2}{2\epsilon_0 n h} does not depend on mm: same speed. False.
  3. (C) E=me48ϵ02h2∝mE = \frac{me^4}{8\epsilon_0^2 h^2} \propto m: 200 times more. True.
  4. (D) p=mvp = mv with the same vv: 200 times more. True.
  5. So the answer is (b) (A), (C), (D).
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Q123JEE Main 2018Difficult

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn,λg\lambda_n, \lambda_g be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let Λn\Lambda_n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

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Answer: (a)

  1. 2πrn=nλn2\pi r_n = n\lambda_n with rn∝n2r_n \propto n^2, so λn∝n\lambda_n \propto n, i.e. λn=nλg\lambda_n = n\lambda_g.
  2. Photon: 1Λn=R(1−1n2)⇒Λn=1R(1−1n2)−1\frac{1}{\Lambda_n} = R\left(1 - \frac{1}{n^2}\right) \Rightarrow \Lambda_n = \frac{1}{R}\left(1 - \frac{1}{n^2}\right)^{-1}.
  3. For large nn: Λn≈1R(1+1n2)=1R+λg2R⋅1λn2\Lambda_n \approx \frac{1}{R}\left(1 + \frac{1}{n^2}\right) = \frac{1}{R} + \frac{\lambda_g^2}{R}\cdot\frac{1}{\lambda_n^2}
  4. So the answer is (a) Λn≈A+Bλn2\Lambda_n \approx A + \frac{B}{\lambda_n^2}.
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Q124JEE Main 2017 · 8 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to: (n = principal quantum number)

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Answer: (b)

  1. B=μ0I2rB = \frac{\mu_0 I}{2r}, I=ev2πr∝n−1n2=n−3I = \frac{ev}{2\pi r} \propto \frac{n^{-1}}{n^2} = n^{-3}.
  2. B∝n−3n2=n−5B \propto \frac{n^{-3}}{n^2} = n^{-5}
  3. So the answer is (b) n−5n^{-5}.
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Q125JEE Main 2017 · 9 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

The acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is:

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Answer: (c)

  1. Bohr condition for n=1n = 1: mvr=h2π⇒v=h2πmrmvr = \frac{h}{2\pi} \Rightarrow v = \frac{h}{2\pi m r}.
  2. a=v2r=h24π2m2r2⋅1ra = \frac{v^2}{r} = \frac{h^2}{4\pi^2 m^2 r^2} \cdot \frac{1}{r}
  3. So the answer is (c) h24π2m2r3\frac{h^2}{4\pi^2 m^2 r^3}.
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Q126JEE Main 2015 · 11 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

The de - Broglie wavelength associated with the electron in the n = 4 level is:

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Answer: (b)

  1. λ=hmv\lambda = \frac{h}{mv} with v∝1nv \propto \frac{1}{n}, so λ∝n\lambda \propto n.
  2. λ4=4λ1\lambda_4 = 4\lambda_1
  3. So the answer is (b) four times the ground-state wavelength.
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Q127NEET 2015 · AIPMT · CANCELLED PAPERMedium

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 ×\times 109 constant, Z = 2 and h (Planck's Constant) = 6.6 ×\times 10−-34 J s]

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Answer: (d)

  1. vn=2πkZe2nhv_n = \frac{2\pi kZe^2}{nh}
  2. v3=2π×9×109×2×(1.6×10−19)23×6.6×10−34v_3 = \frac{2\pi\times9\times10^9\times2\times(1.6\times10^{-19})^2}{3\times6.6\times10^{-34}}
  3. v3≈1.46×106 m/sv_3 \approx 1.46\times10^6\text{ m/s} (same as 2.19×106×232.19\times10^6\times\frac{2}{3})
  4. So the answer is (d) 1.46×106 m/s1.46\times10^6\text{ m/s}.
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Q128JEE Main 2015 · 10 APR · SHIFT 1Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

If one were to apply Bohr model to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the nth level will be:

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Answer: (b)

  1. qvB=mv2r⇒r=mvqBqvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}.
  2. Bohr condition: mvr=m2v2qB=nh2π⇒v2=nhqB2πm2mvr = \frac{m^2 v^2}{qB} = \frac{nh}{2\pi} \Rightarrow v^2 = \frac{nhqB}{2\pi m^2}.
  3. Energy is only kinetic: E=12mv2=n(hqB4πm)E = \frac{1}{2}mv^2 = n\left(\frac{hqB}{4\pi m}\right)
  4. So the answer is (b) n(hqB4πm)n\left(\frac{hqB}{4\pi m}\right).
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Q129NEET 2013Medium

Bohr Model: Orbits, Speed & Angular Momentum · Frequency, time period & current

An electron in hydrogen atom makes a transition n1 →\to n2 where n1 and n2 are principal quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are

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Answer: (d)

  1. T∝n3T \propto n^3
  2. T1T2=8⇒(n1n2)3=8⇒n1=2n2\frac{T_1}{T_2} = 8 \Rightarrow \left(\frac{n_1}{n_2}\right)^3 = 8 \Rightarrow n_1 = 2n_2
  3. Only n1=4, n2=2n_1 = 4,\ n_2 = 2 fits.
  4. So the answer is (d).
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Q130JEE Main 2013 · 9 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum · Magnetic moment & magnetic field at the nucleus

In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nthn^{th} excited state, is:

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Answer: (c)

  1. M=IA=ev2πr⋅πr2=e2m(mvr)M = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{e}{2m}(mvr)
  2. Bohr: mvr=nh2πmvr = \frac{nh}{2\pi}
  3. M=(e2m)nh2πM = \left(\frac{e}{2m}\right)\frac{nh}{2\pi}, so the answer is (c).
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Q131JEE Main 2013 · 23 APR · SHIFT 1Difficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as F=e24πϵ0(1r2+βr3)F = \frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{r^2} + \frac{\beta}{r^3}\right), where β\beta is a constant. For this atom, the radius of the nthn^{th} orbit in terms of the Bohr radius (a0=ϵ0h2mπe2)\left(a_0 = \frac{\epsilon_0 h^2}{m\pi e^2}\right) is:

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Answer: (c)

  1. mv2r=e24πϵ0⋅r+βr3\frac{mv^2}{r} = \frac{e^2}{4\pi\epsilon_0}\cdot\frac{r + \beta}{r^3} and Bohr condition v=nh2πmrv = \frac{nh}{2\pi m r}.
  2. Substitute: n2h24π2mr3=e24πϵ0⋅r+βr3\frac{n^2h^2}{4\pi^2 m r^3} = \frac{e^2}{4\pi\epsilon_0}\cdot\frac{r + \beta}{r^3}.
  3. r+β=ϵ0n2h2πme2=a0n2⇒rn=a0n2−βr + \beta = \frac{\epsilon_0 n^2 h^2}{\pi m e^2} = a_0 n^2 \Rightarrow r_n = a_0 n^2 - \beta
  4. So the answer is (c) rn=a0n2−βr_n = a_0 n^2 - \beta.
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Q132JEE Main 2012 · AIEEEDifficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

A diatomic molecule is made of two masses m1m_1 and m2m_2 which are separated by a distance r.r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: (nn is an integer)

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Answer: (d)

  1. Moment of inertia I=μr2I = \mu r^2 with reduced mass μ=m1m2m1+m2\mu = \frac{m_1 m_2}{m_1 + m_2}.
  2. Bohr rule L=nℏL = n\hbar, so E=L22I=n2ℏ2(m1+m2)2m1m2r2E = \frac{L^2}{2I} = \frac{n^2\hbar^2 (m_1 + m_2)}{2 m_1 m_2 r^2}.
  3. Writing hh for ℏ\hbar as the options do, E=(m1+m2)n2h22m1m2r2E = \frac{(m_1+m_2) n^2 h^2}{2 m_1 m_2 r^2}.
  4. So the answer is (d).
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Q133JEE Main 2008 · AIEEEDifficult

Bohr Model: Orbits, Speed & Angular Momentum · Bohr model for other systems

Suppose an electron is attracted towards the origin by a force kr\frac{k}{r} where ′k′'k' is a constant and ′r′'r' is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nthn^{th} orbital of the electron is found to be ′rn′'r_n' and the kinetic energy of the electron to be ′Tn′.'T_n'. Then which of the following is true?

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Answer: (b)

  1. mv2r=kr⇒mv2=k\frac{mv^2}{r} = \frac{k}{r} \Rightarrow mv^2 = k, so vv is constant.
  2. T=12mv2=k2T = \frac{1}{2}mv^2 = \frac{k}{2}, independent of nn.
  3. Bohr condition: mvr=nh2πmvr = \frac{nh}{2\pi} with constant vv, so rn∝nr_n \propto n.
  4. So the answer is (b) TnT_n independent of nn, rn∝nr_n \propto n.
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Q134NEET 2004 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

The Bohr model of atoms

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Answer: (a)

  1. Bohr postulated L=nh2πL = n\dfrac{h}{2\pi} (quantised angular momentum).
  2. It does not use the photoelectric equation, and it predicts discrete line spectra that differ from element to element.
  3. So the answer is (a).
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Q135NEET 2003 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

In which of the following systems will the radius of the first orbit (n = 1) be minimum?

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Answer: (a)

  1. rn=n2a0Zr_n = \dfrac{n^2 a_0}{Z}, so for n=1n = 1, r∝1Zr \propto \dfrac{1}{Z}.
  2. Li2+^{2+} has Z=3Z = 3, the largest among the options (He+^+: 2, D and H: 1).
  3. So the answer is (a) doubly ionized lithium.
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Q136NEET 2003 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Acceleration & Coulomb force

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F⃗\vec{F}between the two is

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Answer: (d)

  1. Magnitude F=Ke2r2F = \frac{Ke^2}{r^2}; r^\hat{r} points from nucleus to electron.
  2. The force on the electron is attractive, towards the nucleus, i.e. along −r^-\hat{r}.
  3. So the answer is (d) −Ke2r2r^-K\frac{e^2}{r^2}\hat{r}.
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Q137NEET 1998 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Speed of the electron

In the Bohr's model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If a0a_0 is the radius of the ground state orbit, mm is the mass and ee is the charge on the electron, ε0\varepsilon_0 is the vacuum permittivity, the speed of the electron is

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Answer: (c)

  1. Centripetal force = Coulomb force: mv2a0=e24πε0a02\frac{mv^2}{a_0} = \frac{e^2}{4\pi\varepsilon_0 a_0^2}
  2. v2=e24πε0a0mv^2 = \frac{e^2}{4\pi\varepsilon_0 a_0 m}
  3. So the answer is (c) v=e4πε0a0mv = \frac{e}{\sqrt{4\pi\varepsilon_0 a_0 m}}.
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Q138NEET 1997 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

When hydrogen atom is in its first excited level, its radius is .......... of the Bohr radius.

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Answer: (b)

  1. First excited level means n=2n = 2.
  2. r2=22a0=4a0r_2 = 2^2 a_0 = 4a_0.
  3. So the answer is (b) 4 times.
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Q139NEET 1996 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

According to Bohr's principle, the relation between principal quantum number (nn) and radius of orbit (rr) is

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Answer: (d)

  1. rn=n2h2ϵ0πmZe2⇒r∝n2r_n = \dfrac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \Rightarrow r \propto n^2.
  2. So the answer is (d).
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Q140NEET 1992 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Radius of the orbit

In terms of Bohr radius a0a_0, the radius of the second Bohr orbit of a hydrogen atom is given by

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Answer: (a)

  1. rn=n2a0r_n = n^2 a_0
  2. r2=4a0r_2 = 4a_0. So the answer is (a).
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Q141NEET 1990 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · de Broglie waves in Bohr orbits

Consider an electron in the nnth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength λ\lambda of that electron as

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Answer: (d)

  1. mvr=nh2πmvr = \frac{nh}{2\pi} and λ=hmv\lambda = \frac{h}{mv} give 2πr=nhmv=nλ2\pi r = \frac{nh}{mv} = n\lambda.
  2. So the answer is (d) nλn\lambda.
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Q142NEET 1989 · AIPMTEasy

Bohr Model: Orbits, Speed & Angular Momentum · Bohr's postulates & angular momentum

To explain his theory, Bohr used

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Answer: (b)

  1. Bohr's key postulate: only orbits with mvr=nh2πmvr = n\dfrac{h}{2\pi} are allowed.
  2. So the answer is (b) quantisation of angular momentum.
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Q143MHT CET 2026 · 20 APR · SHIFT 2Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The kinetic energy of the electron in an orbit of radius rr in hydrogen atom is proportional to (ee = electronic charge)

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Answer: (c)

  1. Centripetal force: mv2r=ke2r2\frac{mv^2}{r} = \frac{ke^2}{r^2} with k=14πϵ0k = \frac{1}{4\pi\epsilon_0}.
  2. KE=12mv2=ke22r∝e22rKE = \frac{1}{2}mv^2 = \frac{ke^2}{2r} \propto \frac{e^2}{2r}
  3. So the answer is (c) e22r\frac{e^2}{2r}.
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Q144MHT CET 2026 · 16 APR · SHIFT 2Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

An electron makes a transition from an excited state to the ground state of a hydrogen like atom. Out of the following statements which one is correct?

(A) K.E, P. E. and T.E. decreases

(B) K.E. increases but P. E. and T.E. decreases

(C) K.E. and T.E. decreases but P.E. increases

(D) K.E. decreases, P.E. increases but T.E. remains the same

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Answer: (b)

  1. KE=ke22rKE = \frac{ke^2}{2r}, PE=−ke2rPE = -\frac{ke^2}{r}, TE=−ke22rTE = -\frac{ke^2}{2r}.
  2. Going to the ground state, rr decreases: KE increases, PE and TE become more negative (decrease).
  3. So the answer is (b) B.
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Q145MHT CET 2026 · 11 APR · SHIFT 1Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The kinetic energy of the electron in an orbit of radius rr in hydrogen atom is proportional to (ee = electronic charge)

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Answer: (c)

  1. Centripetal force: mv2r=ke2r2\frac{mv^2}{r} = \frac{ke^2}{r^2} with k=14πϵ0k = \frac{1}{4\pi\epsilon_0}.
  2. KE=12mv2=ke22r∝e22rKE = \frac{1}{2}mv^2 = \frac{ke^2}{2r} \propto \frac{e^2}{2r}
  3. So the answer is (c) e22r\frac{e^2}{2r}.
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Q146MHT CET 2026 · 19 APR · SHIFT 2Easy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

If the ionisation energy for the hydrogen atom is 13.613.6 eV, then the energy required to excite it from the ground state to the next higher state is nearly

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Answer: (c)

  1. Ionisation energy 13.6 eV means E1=−13.6E_1 = -13.6 eV, E2=−3.4E_2 = -3.4 eV.
  2. Excitation energy =E2−E1=−3.4−(−13.6)=10.2= E_2 - E_1 = -3.4 - (-13.6) = 10.2 eV (energy supplied is positive).
  3. So the answer is (c) 10.210.2 eV.
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Q147MHT CET 2026 · 15 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

If 'E' and 'L' denote the magnitude of total energy and angular momentum of revolving electron in nthn^{th} Bohr orbit then

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Answer: (c)

  1. Bohr model: E∝1n2E \propto \frac{1}{n^2} and L=nh2π∝nL = \frac{nh}{2\pi} \propto n.
  2. So E∝1L2=L−2E \propto \frac{1}{L^2} = L^{-2}
  3. So the answer is (c) E∝L−2E \propto L^{-2}.
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Q148MHT CET 2026 · 17 APR · SHIFT 2Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

In Bohr's atomic model, the energy of the electron is 'E' in the second orbit of hydrogen atom. So the energy of the electron in the third orbit of helium atom (Z=2)(Z = 2) will be

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Answer: (b)

  1. En=−13.6 Z2n2E_n = -13.6\,\frac{Z^2}{n^2} eV.
  2. H, n=2n = 2: E=−13.64E = -\frac{13.6}{4}.
  3. He, Z=2Z = 2, n=3n = 3: E′=−13.6×49E' = -13.6 \times \frac{4}{9}.
  4. E′E=4/91/4=169\frac{E'}{E} = \frac{4/9}{1/4} = \frac{16}{9}
  5. So the answer is (b) 16E9\frac{16E}{9}.
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Q149JEE Main 2026 · 2 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

Angular momentum of an electron in a hydrogen atom is 3hπ\frac{3h}{\pi}, then the energy of the electron is _____ eV.

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Answer: (c)

  1. L=nh2πL = \frac{nh}{2\pi}; given L=3hπ=6h2πL = \frac{3h}{\pi} = \frac{6h}{2\pi}, so n=6n = 6.
  2. E6=−13.636=−0.38E_6 = \frac{-13.6}{36} = -0.38 eV
  3. So the answer is (c) −0.38-0.38 eV.
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Q150MHT CET 2025 · 23 APR · SHIFT 1Easy

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

In hydrogen atom, the energy of electron in first and third orbit is ' E1E_1 ' and' E3E_3 ' respectively. If E3=xE1E_3 = xE_1 then the value of xx will be

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Answer: (a)

  1. In hydrogen, En=−13.6n2E_n = -\frac{13.6}{n^2} eV, so En∝1n2E_n \propto \frac{1}{n^2}.
  2. E3E1=1232=19\frac{E_3}{E_1} = \frac{1^2}{3^2} = \frac{1}{9}
  3. So the answer is (a) x=19x = \frac{1}{9}.
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Q151MHT CET 2025 · 5 MAY · SHIFT 2Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

The ratio of the total energy of the 22nd orbit electron for the hydrogen atom ( 1H{}^{1}\text{H}) to that of a helium ion (He+\text{He}^{+}) is:

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Answer: (d)

  1. En∝Z2n2E_n \propto \frac{Z^2}{n^2}; same n=2n = 2, so E∝Z2E \propto Z^2.
  2. EHEHe+=1222=14\frac{E_H}{E_{He^+}} = \frac{1^2}{2^2} = \frac{1}{4}
  3. So the answer is (d) 14\frac{1}{4}.
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Q152JEE Main 2025 · 7 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

In a hydrogen like ion, the energy difference between the 22nd excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

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Answer: (c)

  1. Second excited state is n=3n = 3.
  2. ΔE=13.6 Z2(1−19)=12.09 Z2\Delta E = 13.6\,Z^2\left(1 - \frac{1}{9}\right) = 12.09\,Z^2 eV
  3. 12.09 Z2=108.8⇒Z2=9⇒Z=312.09\,Z^2 = 108.8 \Rightarrow Z^2 = 9 \Rightarrow Z = 3
  4. So the answer is (c) 3.
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Q153JEE Main 2025 · 2 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

Considering Bohr's atomic model for hydrogen atom:

(A) the energy of H atom in ground state is same as energy of He+\text{He}^{+} ion in its first excited state. (B) the energy of H atom in ground state is same as that for Li++\text{Li}^{++} ion in its second excited state. (C) the energy of H atom in its ground state is same as that of He+\text{He}^{+} ion for its ground state. (D) the energy of He+\text{He}^{+} ion in its first excited state is same as that for Li++\text{Li}^{++} ion in its ground state. Choose the correct answer from the options given below:

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Answer: (a)

  1. E=−13.6 Z2n2E = -13.6\,\frac{Z^2}{n^2} eV; H ground state =−13.6= -13.6 eV.
  2. (A) He+^+, n=2n = 2: −13.6×44=−13.6-13.6 \times \frac{4}{4} = -13.6 eV. True.
  3. (B) Li++^{++}, second excited n=3n = 3: −13.6×99=−13.6-13.6 \times \frac{9}{9} = -13.6 eV. True.
  4. (C) He+^+ ground: −54.4-54.4 eV. False. (D) Li++^{++} ground: −122.4-122.4 eV ≠−13.6\ne -13.6 eV. False.
  5. So the answer is (a) (A), (B) only.
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Q154JEE Main 2025 · 24 JAN · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy level diagrams

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 A˚2000\ \text{\AA} and it becomes 6000 A˚6000\ \text{\AA} when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is

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Answer: (d)

  1. Energies add: EAC=EAB+EBCE_{AC} = E_{AB} + E_{BC}, so 1λAC=1λAB+1λBC\frac{1}{\lambda_{AC}} = \frac{1}{\lambda_{AB}} + \frac{1}{\lambda_{BC}}.
  2. 1λAB=12000−16000=26000\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} = \frac{2}{6000}
  3. λAB=3000 A˚\lambda_{AB} = 3000\ \text{\AA}
  4. So the answer is (d) 3000 A˚3000\ \text{\AA}.
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Q155MHT CET 2024 · 9 MAY · SHIFT 1Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

When a hydrogen atom is raised from the ground state to the excited state

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Answer: (a)

  1. KE=ke22rKE = \frac{ke^2}{2r} and PE=−ke2rPE = -\frac{ke^2}{r}.
  2. In an excited state rr is larger: KE falls, PE becomes less negative (increases).
  3. So the answer is (a) potential energy increases and K.E. decreases.
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Q156JEE Main 2024 · 5 APR · SHIFT 1Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

An electron rotates in a circle around a nucleus having positive charge ZeZe. Correct relation between total energy (E) of electron to its potential energy (U) is:

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Answer: (c)

  1. For a circular orbit U=−kZe2rU = -\frac{kZe^2}{r} and KE=kZe22r=−U2KE = \frac{kZe^2}{2r} = -\frac{U}{2}.
  2. E=KE+U=U2⇒2E=UE = KE + U = \frac{U}{2} \Rightarrow 2E = U
  3. So the answer is (c) 2E=U2E = U.
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Q157JEE Main 2024 · 30 JAN · SHIFT 1Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a hydrogen atom is:

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Answer: (c)

  1. In any Bohr orbit KE=ke22rKE = \frac{ke^2}{2r} and ∣U∣=ke2r|U| = \frac{ke^2}{r}.
  2. KE∣U∣=12\frac{KE}{|U|} = \frac{1}{2}, the same for every state (including n=6n = 6).
  3. So the answer is (c) 12\frac{1}{2}.
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Q158MHT CET 2024 · 10 MAY · SHIFT 1Easy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

If the ionisation energy for the hydrogen atom is 13.6 eV, then the energy required to excite it from the ground state to the next higher state is nearly

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Answer: (a)

  1. E1=−13.6E_1 = -13.6 eV, E2=−13.64=−3.4E_2 = \frac{-13.6}{4} = -3.4 eV.
  2. Energy required =E2−E1=10.2= E_2 - E_1 = 10.2 eV
  3. So the answer is (a) 10.210.2 eV.
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Q159MHT CET 2024 · 2 MAY · SHIFT 2Medium

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

Which of the following statements about the Bohr model of the hydrogen atom is false?

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Answer: (d)

  1. (a) a=v2ra = \frac{v^2}{r} with v∝1nv \propto \frac{1}{n}, r∝n2r \propto n^2, so a∝1n4a \propto \frac{1}{n^4}: smaller for n=2n = 2. True.
  2. (b) L=nh2πL = \frac{nh}{2\pi} is larger for n=2n = 2. True.
  3. (c) K=13.6n2K = \frac{13.6}{n^2} eV: 3.43.4 eV <13.6< 13.6 eV. True.
  4. (d) U=−27.2n2U = -\frac{27.2}{n^2} eV: U2=−6.8U_2 = -6.8 eV is greater than U1=−27.2U_1 = -27.2 eV, not less. False.
  5. So the answer is (d).
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Q160MHT CET 2024 · 16 MAY · SHIFT 2Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

In the second orbit of hydrogen atom, the energy of an electron is ' EE '. In the third orbit of helium atom, the energy of the electron will be (atomic number of helium =2= 2)

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Answer: (c)

  1. En∝Z2n2E_n \propto \frac{Z^2}{n^2}.
  2. EHe,3EH,2=22/3212/22=49×4=169\frac{E_{He,3}}{E_{H,2}} = \frac{2^2/3^2}{1^2/2^2} = \frac{4}{9} \times 4 = \frac{16}{9}
  3. So the answer is (c) 16E9\frac{16E}{9}.
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Q161MHT CET 2024 · 2 MAY · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

In the third orbit of hydrogen atom the energy of an electron ' EE '. In the fifth orbit of helium (Z=2)(Z = 2) the energy of an electron will be

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Answer: (b)

  1. En∝Z2n2E_n \propto \frac{Z^2}{n^2}.
  2. EHe,5EH,3=4/251/9=3625\frac{E_{He,5}}{E_{H,3}} = \frac{4/25}{1/9} = \frac{36}{25}
  3. So the answer is (b) 36E25\frac{36E}{25}.
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Q162JEE Main 2024 · 6 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

Radius of a certain orbit of hydrogen atom is 8.48 A˚\text{\AA}. If energy of electron in this orbit is E/xE/x. then x=x = ________ (Given a0=0.529 A˚, E=a_{0} = 0.529\ \text{\AA},\ E = energy of electron in ground state).

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

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Answer: 16

  1. rn=n2a0⇒n2=8.480.529≈16r_n = n^2 a_0 \Rightarrow n^2 = \frac{8.48}{0.529} \approx 16, so n=4n = 4.
  2. En=En2=E16E_n = \frac{E}{n^2} = \frac{E}{16}
  3. So x=16x = 16.
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Q163NEET 2024Medium

Energy of the Electron & Energy Levels · Energy level diagrams

Some energy levels of a molecule are shown in the figure with their wavelengths of transitions. Then:

Figure

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Answer: (d)

  1. λ=hcΔE\lambda = \frac{hc}{\Delta E}, so λ∝1ΔE\lambda \propto \frac{1}{\Delta E}.
  2. λ1\lambda_1: −5E2→−4E-\frac{5E}{2} \to -4E, ΔE=1.5E\Delta E = 1.5E. λ2\lambda_2: −2E→−3E-2E \to -3E, ΔE=E\Delta E = E. λ3\lambda_3: −2E→−4E-2E \to -4E, ΔE=2E\Delta E = 2E.
  3. So λ2>λ1>λ3\lambda_2 > \lambda_1 > \lambda_3 and λ2λ3=2EE=2\frac{\lambda_2}{\lambda_3} = \frac{2E}{E} = 2.
  4. So the answer is (d) λ2>λ1, λ2=2λ3\lambda_2 > \lambda_1,\ \lambda_2 = 2\lambda_3.
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Q164MHT CET 2023 · 11 MAY · SHIFT 2Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

According to Bohr's theory of hydrogen atom, the total energy of the electron in the nnth stationary orbit is

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Answer: (d)

  1. Bohr energy En=−me48ϵ02h2⋅1n2E_n = -\frac{me^4}{8\epsilon_0^2 h^2}\cdot\frac{1}{n^2}.
  2. So En∝1n2E_n \propto \frac{1}{n^2}.
  3. So the answer is (d) inversely proportional to n2n^2.
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Q165JEE Main 2023 · 11 APR · SHIFT 2Easy

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

The energy of He+\text{He}^+ ion in its first excited state is, (The ground state energy for the Hydrogen atom is −13.6 eV)-13.6\text{ eV}):

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Answer: (a)

  1. En=−13.6 Z2n2E_n = -13.6\,\frac{Z^2}{n^2} eV; for He+^+, Z=2Z = 2, first excited state n=2n = 2.
  2. E2=−13.6×44=−13.6E_2 = -13.6 \times \frac{4}{4} = -13.6 eV
  3. So the answer is (a) −13.6-13.6 eV.
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Q166NEET 2023Easy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

The ground state energy of hydrogen atom is −13.6 eV-13.6\text{ eV}. The energy needed to ionize hydrogen atom from its second excited state will be:

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Answer: (c)

  1. Second excited state is n=3n = 3: E3=−13.69=−1.51E_3 = \frac{-13.6}{9} = -1.51 eV.
  2. Ionisation energy from this state =0−E3=1.51= 0 - E_3 = 1.51 eV
  3. So the answer is (c) 1.511.51 eV.
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Q167JEE Main 2023 · 6 APR · SHIFT 1Easy

Energy of the Electron & Energy Levels · Energy level diagrams

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is:

Figure

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Answer: (d)

  1. Shortest wavelength means largest energy gap, since λ=hcΔE\lambda = \frac{hc}{\Delta E}.
  2. ΔE∝1nf2−1ni2\Delta E \propto \frac{1}{n_f^2} - \frac{1}{n_i^2}: A (4→34\to3) =0.049= 0.049, B (4→24\to2) =0.1875= 0.1875, C (3→23\to2) =0.139= 0.139, D (3→13\to1) =0.889= 0.889.
  3. D has the largest energy gap.
  4. So the answer is (d) D.
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Q168NEET 2023Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

The angular momentum of an electron moving in an orbit of hydrogen atom is 1.5(hπ)1.5\left(\frac{h}{\pi}\right). The energy in the same orbit is nearly.

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Answer: (a)

  1. L=nh2πL = \frac{nh}{2\pi}; given L=1.5 hπ=3h2πL = 1.5\,\frac{h}{\pi} = \frac{3h}{2\pi}, so n=3n = 3.
  2. E3=−13.69=−1.51E_3 = \frac{-13.6}{9} = -1.51 eV
  3. So the answer is (a) −1.5-1.5 eV.
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Q169JEE Main 2023 · 31 JAN · SHIFT 2Medium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

If the binding energy of ground state electron in a hydrogen atom is 13.6 eV13.6\text{ eV}, then, the energy required to remove the electron from the second excited state of Li2+\text{Li}^{2+} will be: x×10−1 eVx \times 10^{-1}\text{ eV}. The value of xx is ________.

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

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Answer: 136

  1. Li2+^{2+}: Z=3Z = 3; second excited state n=3n = 3.
  2. Energy required =13.6×99=13.6= 13.6 \times \frac{9}{9} = 13.6 eV =136×10−1= 136 \times 10^{-1} eV
  3. So x=136x = 136.
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Q170JEE Main 2023 · 25 JAN · SHIFT 2Medium

Energy of the Electron & Energy Levels · Energy level diagrams

The energy levels of an atom is shown in figure.

Figure

Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given (h = 6.62 ×\times 10 −34{}^{-34} Js)

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Answer: (d)

  1. Photon energy E=hcλ=1240 eV nm124.1 nm≈10 eVE = \frac{hc}{\lambda} = \frac{1240\ \text{eV nm}}{124.1\ \text{nm}} \approx 10\ \text{eV}.
  2. From the figure: A =2.2= 2.2 eV, B =5.2= 5.2 eV, C =3.0= 3.0 eV, D =0−(−10.0)=10.0= 0 - (-10.0) = 10.0 eV.
  3. So the answer is (d) D.
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Q171JEE Main 2022 · 24 JUN · SHIFT 2Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level:

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Answer: (b)

  1. K=ke22rK = \frac{ke^2}{2r}, P=−ke2rP = -\frac{ke^2}{r}, E=−ke22rE = -\frac{ke^2}{2r}.
  2. In a higher level rr is larger: KK decreases, while PP and EE become less negative (increase).
  3. So the answer is (b) K decreases, P and E increase.
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Q172NEET 2022 · PHASE 1Easy

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

Let T1 and T2 be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to the Bohr's model of an atom, the ratio T1: T2 is

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Answer: (d)

  1. First excited state is n=2n = 2, second excited state is n=3n = 3.
  2. En∝1n2E_n \propto \frac{1}{n^2}, so T1T2=1/41/9=94\frac{T_1}{T_2} = \frac{1/4}{1/9} = \frac{9}{4}
  3. So the answer is (d) 9:49 : 4.
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Q173MHT CET 2021 · 24 SEP · SHIFT 1Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

An electron makes a transition from an excited state to the ground state of a hydrogen like atom. Out of the following statements which one is correct?

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Answer: (c)

  1. KE=ke22rKE = \frac{ke^2}{2r}, PE=−ke2rPE = -\frac{ke^2}{r}, TE=−ke22rTE = -\frac{ke^2}{2r}.
  2. Moving to the ground state, rr decreases: KE increases, PE and TE decrease (more negative).
  3. So the answer is (c).
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Q174MHT CET 2021 · 22 SEP · SHIFT 2Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

In nnth Bohr orbit, the ratio of the kinetic energy of an electron to the total energy of it, is

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Answer: (b)

  1. In a Bohr orbit KE=ke22rKE = \frac{ke^2}{2r} and TE=−ke22rTE = -\frac{ke^2}{2r}, so KE=−TEKE = -TE.
  2. KETE=−1\frac{KE}{TE} = -1
  3. So the answer is (b) 1:−11 : -1.
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Q175MHT CET 2021 · 20 SEP · SHIFT 1Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

If the electron in a hydrogen atom moves from ground state orbit to 5th orbit, then the potential energy of the electron

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Answer: (a)

  1. PE=−ke2rPE = -\frac{ke^2}{r} with r∝n2r \propto n^2.
  2. From n=1n = 1 to n=5n = 5, rr grows 25 times, so PE becomes less negative (−27.2 eV→−1.09 eV-27.2\text{ eV} \to -1.09\text{ eV}).
  3. So the answer is (a) is increased.
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Q176MHT CET 2021 · 22 SEP · SHIFT 2Medium

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

If 'EE' and 'LL' denote the magnitude of total energy and angular momentum of revolving electron in nnth Bohr orbit, then

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Answer: (c)

  1. E=13.6n2E = \frac{13.6}{n^2} eV (magnitude), so E∝1n2E \propto \frac{1}{n^2}.
  2. L=nh2πL = \frac{nh}{2\pi}, so n∝Ln \propto L.
  3. Hence E∝L−2E \propto L^{-2}; the answer is (c).
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Q177MHT CET 2021 · 23 SEP · SHIFT 2Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

Energy of electron in the second orbit of hydrogen atom is EE. The energy of electron 'E3E_{3}' in the third orbit of helium (He)(\text{He}) atom will be

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Answer: (a)

  1. En∝Z2n2E_n \propto \frac{Z^2}{n^2}.
  2. H, n=2n = 2: E∝14E \propto \frac{1}{4}. He, Z=2Z = 2, n=3n = 3: E3∝49E_3 \propto \frac{4}{9}.
  3. E3E=4/91/4=169\frac{E_3}{E} = \frac{4/9}{1/4} = \frac{16}{9}
  4. So E3=16E9E_3 = \frac{16E}{9}, which is option (c).
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Q178JEE Main 2021 · 17 MAR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?

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Answer: (b)

  1. Treat the carbon ion as hydrogen-like with Z=6Z = 6: En=−13.6 36n2E_n = -13.6\,\frac{36}{n^2} eV.
  2. Set equal to −13.6-13.6 eV: 36n2=1⇒n=6\frac{36}{n^2} = 1 \Rightarrow n = 6
  3. So the answer is (b) 6.
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Q179JEE Main 2021 · 24 FEB · SHIFT 1Medium

Energy of the Electron & Energy Levels · Energy level diagrams

In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.

The transitions A, B and C respectively represent:

Figure

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Answer: (c)

  1. A: continuum (00 eV) to n=1n=1 is the series limit of the Lyman series.
  2. B: n=5→n=2n=5 \to n=2; Balmer members are 3→23\to2 (1st), 4→24\to2 (2nd), 5→25\to2 (3rd), so B is the third member.
  3. C: n=5→n=3n=5 \to n=3; Paschen members are 4→34\to3 (1st), 5→35\to3 (2nd), so C is the second member.
  4. So the answer is (c).
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Q180NEET 2020 · PHASE 2Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The total energy of an electron in the nnth stationary orbit of the hydrogen atom can be obtained by

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Answer: (c)

  1. Ground state energy of hydrogen is −13.6 eV-13.6\text{ eV} and En∝1n2E_n \propto \frac{1}{n^2}.
  2. En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\text{ eV}
  3. So the answer is (c).
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Q181MHT CET 2020 · 19 OCT · SHIFT 2Medium

Energy of the Electron & Energy Levels · Energy level diagrams

The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively

Figure

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Answer: (d)

  1. λ=hcΔE\lambda = \frac{hc}{\Delta E}: max wavelength for smallest ΔE\Delta E, min wavelength for largest ΔE\Delta E.
  2. A: 0→−20 \to -2: 2 eV. B: 0→−4.50 \to -4.5: 4.5 eV. C: −2→−4.5-2 \to -4.5: 2.5 eV. D: −2→−10-2 \to -10: 8 eV.
  3. Smallest gap is A, largest is D.
  4. So the answer is (d) A, D.
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Q182NEET 2019Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively.

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Answer: (a)

  1. In a Bohr orbit KE=−EKE = -E and PE=2EPE = 2E.
  2. KE=3.4 eVKE = 3.4\text{ eV}, PE=2(−3.4)=−6.8 eVPE = 2(-3.4) = -6.8\text{ eV}
  3. So the answer is (a) 3.4 eV, −6.8-6.8 eV.
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Q183JEE Main 2019 · 9 APR · SHIFT 2Medium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

A He+ ion is in its first excited state. Its ionization energy is :-

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Answer: (a)

  1. He+^+: Z=2Z = 2, first excited state n=2n = 2.
  2. E2=−13.6×44=−13.6E_2 = -13.6 \times \frac{4}{4} = -13.6 eV
  3. Ionisation energy =13.6= 13.6 eV; the answer is (a).
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Q184NEET 2018Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is

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Answer: (b)

  1. KE=ke22rKE = \frac{ke^2}{2r} and E=−ke22rE = -\frac{ke^2}{2r}, so KE=−EKE = -E.
  2. So the answer is (b) 1:−11 : -1.
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Q185JEE Main 2018 · 15 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

The energy required to remove the electron from a singly ionized Helium atom is 2.22.2 times the energies required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

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Answer: (c)

  1. Removing the electron from He+^+ (Z=2Z = 2, n=1n = 1): 13.6×4=54.413.6 \times 4 = 54.4 eV.
  2. First ionisation of He =54.42.2≈24.7= \frac{54.4}{2.2} \approx 24.7 eV.
  3. Total =54.4+24.7≈79= 54.4 + 24.7 \approx 79 eV
  4. So the answer is (c) 7979 eV.
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Q186JEE Main 2017Medium

Energy of the Electron & Energy Levels · Energy level diagrams

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ1/λ2\lambda_{1}/\lambda_{2}, is given by:

Figure

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Answer: (a)

  1. λ=hcΔE\lambda = \frac{hc}{\Delta E}, so λ∝1ΔE\lambda \propto \frac{1}{\Delta E}.
  2. λ1\lambda_1: −E→−2E-E \to -2E, ΔE1=E\Delta E_1 = E. λ2\lambda_2: −E→−43E-E \to -\frac{4}{3}E, ΔE2=E3\Delta E_2 = \frac{E}{3}.
  3. r=λ1λ2=ΔE2ΔE1=13r = \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{3}
  4. So the answer is (a) r=1/3r = 1/3.
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Q187JEE Main 2015Easy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion:

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Answer: (c)

  1. KE=kZe22rKE = \frac{kZe^2}{2r}, PE=−kZe2rPE = -\frac{kZe^2}{r}, TE=−kZe22rTE = -\frac{kZe^2}{2r}.
  2. Going to the ground state rr decreases: KE increases, PE and TE decrease (more negative).
  3. So the answer is (c).
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Q188JEE Main 2014 · 9 APR · SHIFT 1Medium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is:

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Answer: (b)

  1. Energy to remove electron from level nn: 13.6 Z2n213.6\,\frac{Z^2}{n^2} eV.
  2. Li++^{++}: Z=3Z = 3, first excited n=2n = 2: 13.6×94=30.613.6 \times \frac{9}{4} = 30.6 eV
  3. So the answer is (b) 30.630.6 eV.
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Q189JEE Main 2011 · AIEEEMedium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

Energy required for the electron excitation in Li++\text{Li}^{++} from the first to the third Bohr orbit is:

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Answer: (b)

  1. ΔE=13.6 Z2(112−132)\Delta E = 13.6\,Z^2\left(\frac{1}{1^2} - \frac{1}{3^2}\right) eV with Z=3Z = 3.
  2. ΔE=13.6×9×89=108.8\Delta E = 13.6 \times 9 \times \frac{8}{9} = 108.8 eV
  3. So the answer is (b) 108.8108.8 eV.
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Q190NEET 2010 · AIPMT · PRELIMSEasy

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

The energy of a hydrogen atom in the ground state is −- 13.6 eV. The energy of a He+ ion in the first excited state will be

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Answer: (a)

  1. For a hydrogen-like ion, En=−13.6 Z2n2E_n = -13.6\,\frac{Z^2}{n^2} eV.
  2. He+^+: Z=2Z = 2; first excited state n=2n = 2.
  3. E2=−13.6×44=−13.6E_2 = -13.6 \times \frac{4}{4} = -13.6 eV
  4. So the answer is (a) −13.6-13.6 eV.
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Q191NEET 2008 · AIPMTEasy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

The ground state energy of hydrogen atom is −- 13.6 eV. When its electron is in the first excited state, its excitation energy is

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Answer: (a)

  1. First excited state is n=2n = 2: E2=−3.4E_2 = -3.4 eV.
  2. Excitation energy =E2−E1=−3.4−(−13.6)=10.2= E_2 - E_1 = -3.4 - (-13.6) = 10.2 eV
  3. So the answer is (a) 10.210.2 eV.
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Q192NEET 2007 · AIPMTEasy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The total energy of electron in the ground state of hydrogen atom is −- 13.6 eV. The kinetic energy of an electron in the first excited state is

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Answer: (d)

  1. First excited state is n=2n = 2: E2=−13.64=−3.4 eVE_2 = -\frac{13.6}{4} = -3.4\text{ eV}.
  2. KE=−E2=3.4 eVKE = -E_2 = 3.4\text{ eV}
  3. So the answer is (d) 3.4 eV.
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Q193NEET 2005 · AIPMTEasy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

The total energy of an electron in the first excited state of hydrogen atom is about −- 3.4 eV. Its kinetic energy in this state is

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Answer: (a)

  1. In a Bohr orbit KE=−EKE = -E.
  2. KE=−(−3.4)=3.4 eVKE = -(-3.4) = 3.4\text{ eV}
  3. So the answer is (a) 3.4 eV.
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Q194JEE Main 2005 · AIEEEEasy

Energy of the Electron & Energy Levels · Energy level diagrams

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

Figure

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Answer: (b)

  1. Emission means a downward transition; (i) is upward (1→31\to3), so it is absorption.
  2. Downward transitions: (ii) 4→34\to3, (iii) 2→12\to1, (iv) 4→24\to2.
  3. In units of 13.6 eV: 4→34\to3: 0.0490.049, 2→12\to1: 0.750.75, 4→24\to2: 0.18750.1875.
  4. So the answer is (b) (iii).
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Q195NEET 2005 · AIPMTMedium

Energy of the Electron & Energy Levels · Energy level diagrams

Energy levels A, B and C of a certain atom corresponding to increasing values of energy i.e. EA < EB < EC. If λ\lambda1, λ\lambda2 and λ\lambda3 are wavelengths of radioations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?

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Answer: (b)

  1. Energy adds: EC−EA=(EC−EB)+(EB−EA)E_C - E_A = (E_C - E_B) + (E_B - E_A).
  2. hcλ3=hcλ1+hcλ2⇒1λ3=1λ1+1λ2\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} \Rightarrow \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}
  3. λ3=λ1λ2λ1+λ2\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}
  4. So the answer is (b).
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Q196JEE Main 2003 · AIEEEEasy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

Which of the following atoms has the lowest ionization potential?

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Answer: (b)

  1. Ionisation potential decreases down a group as the outer electron is farther from the nucleus and better shielded.
  2. Cs (Z=55Z = 55) is an alkali metal with a single loosely bound 6s6s electron; N, O, Ar have much higher values.
  3. So the answer is (b) 55133Cs{}^{133}_{55}\text{Cs}.
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Q197JEE Main 2003 · AIEEEMedium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

If the binding energy of the electron in a hydrogen atom is 13.6 eV,13.6\text{ eV}, the energy required to remove the electron from the first excited state of Li++\text{Li}^{++} is

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Answer: (a)

  1. Energy to remove electron from level nn: 13.6 Z2n213.6\,\frac{Z^2}{n^2} eV.
  2. Li++^{++}: Z=3Z = 3, n=2n = 2: 13.6×94=30.613.6 \times \frac{9}{4} = 30.6 eV
  3. So the answer is (a) 30.630.6 eV.
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Q198JEE Main 2002 · AIEEEMedium

Energy of the Electron & Energy Levels · Ionisation & excitation energy

If 13.6 eV13.6\text{ eV} energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n=2n = 2 is

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Answer: (c)

  1. Energy to remove electron from level nn =13.6n2= \frac{13.6}{n^2} eV.
  2. n=2n = 2: 13.64=3.4\frac{13.6}{4} = 3.4 eV
  3. So the answer is (c) 3.43.4 eV.
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Q199NEET 1997 · AIPMTEasy

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

The energy of the ground electronic state of hydrogen atom is −13.6-13.6 eV. The energy of the first excited state will be

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Answer: (c)

  1. En=E1n2E_n = \frac{E_1}{n^2}; first excited state is n=2n = 2.
  2. E2=−13.64=−3.4E_2 = \frac{-13.6}{4} = -3.4 eV
  3. So the answer is (c) −3.4-3.4 eV.
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Q200NEET 1996 · AIPMTEasy

Energy of the Electron & Energy Levels · Energy levels of H & H-like ions

The energy of a hydrogen atom in its ground state is −13.6-13.6 eV. The energy of the level corresponding to the quantum number n=2n = 2 in the hydrogen atom is

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Answer: (b)

  1. En=−13.6n2E_n = \frac{-13.6}{n^2} eV.
  2. E2=−13.64=−3.4E_2 = \frac{-13.6}{4} = -3.4 eV
  3. So the answer is (b) −3.4-3.4 eV.
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Q201NEET 1995 · AIPMTEasy

Energy of the Electron & Energy Levels · Kinetic, potential & total energy

When a hydrogen atom is raised from the ground state to an excited state,

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Answer: (d)

  1. KE=ke22rKE = \frac{ke^2}{2r} and PE=−ke2rPE = -\frac{ke^2}{r}.
  2. In an excited state rr is larger: KE decreases, PE becomes less negative (increases).
  3. So the answer is (d) the P.E. increases and K.E. decreases.
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Q202NEET 1995 · AIPMTMedium

Energy of the Electron & Energy Levels · Energy level diagrams

The figure indicates the energy level diagram of an atom and the origin of six spectral lines in emission (e.g. line no. 5 arises from the transition from level BB to AA). Which of the following spectral lines will occur in the absorption spectrum?

Figure

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Answer: (c)

  1. In absorption, atoms start in the ground state XX (almost all atoms are there).
  2. Only lines that end on XX in emission appear in absorption: lines 1 (C↔XC \leftrightarrow X), 2 (B↔XB \leftrightarrow X), 3 (A↔XA \leftrightarrow X).
  3. Lines 4, 5, 6 start from excited levels, which are nearly empty.
  4. So the answer is (c) 1, 2, 3.
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Q203NEET 1991 · AIPMTEasy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

The ground state energy of H-atom is 13.6 eV. The energy needed to ionize H-atom from its second excited state

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Answer: (a)

  1. Ground state energy is −13.6-13.6 eV (13.6 eV in magnitude).
  2. Second excited state is n=3n = 3: E3=−13.69=−1.51E_3 = \frac{-13.6}{9} = -1.51 eV.
  3. Energy needed to ionise =0−E3=1.51= 0 - E_3 = 1.51 eV
  4. So the answer is (a) 1.511.51 eV.
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Q204NEET 1990 · AIPMTMedium

Energy of the Electron & Energy Levels · Energy level diagrams

Energy levels A, BA,\ B and CC of a certain atom corresponding to increasing values of energy i.e. EA<EB<ECE_{A} < E_{B} < E_{C}. If λ1, λ2\lambda_{1},\ \lambda_{2} and λ3\lambda_{3} are wavelengths of radiations corresponding to transitions CC to B, BB,\ B to AA and CC to AA respectively, which of the following relations is correct?

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Answer: (b)

  1. Energies add: EC−EA=(EC−EB)+(EB−EA)E_C - E_A = (E_C - E_B) + (E_B - E_A).
  2. With E=hcλE = \frac{hc}{\lambda}: 1λ3=1λ1+1λ2\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}
  3. λ3=λ1λ2λ1+λ2\lambda_3 = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}
  4. So the answer is (b).
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Q205NEET 1988 · AIPMTEasy

Energy of the Electron & Energy Levels · Ionisation & excitation energy

The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of a singly ionised helium atom would be

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Answer: (d)

  1. Ionisation energy from ground state =13.6 Z2= 13.6\,Z^2 eV.
  2. He+^+ has Z=2Z = 2: 13.6×4=54.413.6 \times 4 = 54.4 eV
  3. So the answer is (d) 54.454.4 eV.
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Q206MHT CET 2026 · 12 MAY · SHIFT 1Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Which of the following transition in hydrogen atom emits photon of highest frequency?

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Answer: (a)

  1. Emitted frequency ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); emission needs a jump from higher nn to lower nn.
  2. (b) and (c) are absorptions. (a) 5→25\to2: 14−125=0.21\frac14-\frac1{25} = 0.21; (d) 3→23\to2: 14−19=0.139\frac14-\frac19 = 0.139.
  3. So the answer is (a) n=5n=5 to n=2n=2.
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Q207MHT CET 2026 · 15 APR · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Number of spectral lines

The electron in hydrogen atom is initially in the second excited state. When it finally moves to ground state, the maximum number of spectral lines emitted are

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Answer: (b)

  1. Second excited state means n=3n = 3.
  2. Maximum lines =n(n−1)2=3×22=3= \frac{n(n-1)}{2} = \frac{3\times2}{2} = 3
  3. So the answer is (b) 33.
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Q208MHT CET 2026 · 15 MAY · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In hydrogen like spectrum if wavelength of transition from 3rd3^{rd} to 2nd2^{nd} orbit is λ1\lambda_{1} then wavelength of transition from 4th4^{th} to 3rd3^{rd} orbit (Bohr's) is λ2\lambda_{2}. If λ2/λ1=20/x\lambda_{2}/\lambda_{1} = 20/x then value of xx is

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Answer: (c)

  1. 1λ∝(1nf2−1ni2)\frac{1}{\lambda} \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. 3→23\to2: 14−19=536\frac14-\frac19 = \frac{5}{36}. 4→34\to3: 19−116=7144\frac19-\frac1{16} = \frac{7}{144}.
  3. λ2λ1=5/367/144=207\frac{\lambda_2}{\lambda_1} = \frac{5/36}{7/144} = \frac{20}{7}
  4. So x=7x = 7, answer (c).
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Q209MHT CET 2026 · 13 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The wavelength of the radiation emitted is λ0\lambda_{0} when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be (20λ0/x)(20\lambda_{0}/x). The value of xx is

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Answer: (c)

  1. Second excited state is n=3n=3, first excited is n=2n=2, third excited is n=4n=4.
  2. 3→23\to2: 1λ0∝14−19=536\frac{1}{\lambda_0} \propto \frac14-\frac19 = \frac{5}{36}. 4→24\to2: 1λ′∝14−116=316\frac{1}{\lambda'} \propto \frac14-\frac1{16} = \frac{3}{16}.
  3. λ′λ0=5/363/16=80108=2027\frac{\lambda'}{\lambda_0} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}
  4. So x=27x = 27, answer (c) 2727.
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Q210MHT CET 2026 · 13 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

For hydrogen atom, λ1\lambda_{1} and λ2\lambda_{2} are the wavelengths corresponding to the transitions 11 and 22 respectively as shown in figure. The ratio of λ1\lambda_{1} and λ2\lambda_{2} is x/32x/32. The value of xx is

Figure

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Answer: (b)

  1. From the figure: transition 1 is n=3→1n=3\to1, transition 2 is n=2→1n=2\to1.
  2. 1λ1∝1−19=89\frac{1}{\lambda_1} \propto 1-\frac19 = \frac89; 1λ2∝1−14=34\frac{1}{\lambda_2} \propto 1-\frac14 = \frac34
  3. λ1λ2=3/48/9=2732\frac{\lambda_1}{\lambda_2} = \frac{3/4}{8/9} = \frac{27}{32}
  4. So x=27x = 27, answer (b) 2727.
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Q211JEE Main 2026 · 6 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ratio of momentum of the photons of the 11st and 22nd line of Balmer series of Hydrogen atoms is α/β\alpha/\beta. The possible values of α\alpha and β\beta are:-

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Answer: (d)

  1. Photon momentum p=hλ∝1λp = \frac{h}{\lambda} \propto \frac{1}{\lambda}.
  2. 1st Balmer line (3→23\to2): 1λ1=R(14−19)=5R36\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}
  3. 2nd Balmer line (4→24\to2): 1λ2=R(14−116)=3R16\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}
  4. p1p2=5/363/16=2027\frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{20}{27}
  5. So the answer is (d) 20 and 27.
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Q212JEE Main 2026 · 5 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f)(f). The ratio of the radius of the orbits in given by ri:rf=16:4r_{i} : r_{f} = 16 : 4. The wavelength of photon emitted due to this transition is ______ nm.

(Given Rydberg constant =1.0973×107 /m= 1.0973 \times 10^{7}\text{ /m})

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Answer: (c)

  1. r∝n2r \propto n^2, so ri:rf=16:4r_i : r_f = 16 : 4 gives ni=4n_i = 4, nf=2n_f = 2.
  2. 1λ=R(14−116)=3R16\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}
  3. λ=163×1.0973×107=4.86×10−7\lambda = \frac{16}{3 \times 1.0973 \times 10^{7}} = 4.86 \times 10^{-7} m
  4. So the answer is (c) 486 nm.
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Q213MHT CET 2026 · 20 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

An electron in the hydrogen atom jumps from nthn^{th} energy state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.652.65 eV. If the maximum kinetic energy of the emitted photoelectrons is 10.110.1 eV, then the value of 'nn' is

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Answer: (c)

  1. Photon energy =ϕ+Kmax=2.65+10.1=12.75= \phi + K_{max} = 2.65 + 10.1 = 12.75 eV.
  2. 13.6(1−1n2)=12.75⇒1n2=0.8513.6=11613.6\left(1 - \frac{1}{n^2}\right) = 12.75 \Rightarrow \frac{1}{n^2} = \frac{0.85}{13.6} = \frac{1}{16}
  3. So n=4n = 4, the answer is (c).
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Q214MHT CET 2026 · 15 MAY · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Recoil of the atom

An electron of stationary hydrogen atom jumps from 4th energy level to ground level. The recoil velocity of the atom as a result of electron transition will be

(hh = Planck's constant, RR = Rydberg's constant, m=m = mass of hydrogen atom)

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Answer: (d)

  1. Momentum is conserved: atom recoil momentum mvmv = photon momentum hλ\frac{h}{\lambda}.
  2. 1λ=R(1−116)=15R16\frac{1}{\lambda} = R\left(1 - \frac{1}{16}\right) = \frac{15R}{16}
  3. v=hmλ=15Rh16mv = \frac{h}{m\lambda} = \frac{15Rh}{16m}
  4. So the answer is (d).
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Q215MHT CET 2025 · 22 APR · SHIFT 1Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Out of the following transitions in hydrogen atom, identify the transition which emits photons of highest frequency.

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Answer: (b)

  1. Emitted frequency ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); emission needs a jump from higher nn to lower nn.
  2. (a) and (c) are absorptions. (b) 2→12\to1: 1−14=0.751-\frac14 = 0.75; (d) 6→26\to2: 14−136=0.22\frac14-\frac1{36} = 0.22.
  3. So the answer is (b) n=2n=2 to n=1n=1.
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Q216MHT CET 2025 · 21 APR · SHIFT 1Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Which of the following transitions in hydrogen atom emit photons of highest frequency? ( n=n = principle quantum number)

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Answer: (d)

  1. Emitted frequency ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); emission needs a jump from higher nn to lower nn.
  2. (a) and (b) are absorptions. (c) 5→35\to3: 19−125=0.071\frac19-\frac1{25} = 0.071; (d) 2→12\to1: 1−14=0.751-\frac14 = 0.75.
  3. So the answer is (d) n=2n=2 to n=1n=1.
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Q217JEE Main 2025 · 29 JAN · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Number of spectral lines

The number of spectral lines emitted by atomic hydrogen that is in the 4th energy level, is

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Answer: (b)

  1. Number of lines from level nn =n(n−1)2= \frac{n(n-1)}{2}
  2. n=4n = 4: 4×32=6\frac{4\times3}{2} = 6
  3. So the answer is (b) 66.
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Q218MHT CET 2025 · 26 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) third to 22nd energy level and (ii) highest energy level to 33rd level is

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Answer: (b)

  1. E∝(1nf2−1ni2)E \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. (i) 3→23\to2: 14−19=536\frac14-\frac19 = \frac{5}{36}. (ii) ∞→3\infty\to3: 19=436\frac19 = \frac{4}{36}.
  3. Ratio =5:4= 5:4
  4. So the answer is (b) 5:45:4.
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Q219MHT CET 2025 · 23 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In hydrogen atom, transition from the state n=6n = 6 to n=1n = 1 results in ultraviolet radiation. Infrared radiation will be obtained in the transition

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Answer: (d)

  1. Infrared lines in hydrogen belong to Paschen (nf=3n_f=3) and higher series.
  2. (a) is Lyman (UV); (b), (c) are Balmer (visible); (d) 5→35\to3 is Paschen (IR).
  3. So the answer is (d) n=5n=5 to n=3n=3.
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Q220MHT CET 2025 · 20 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In hydrogen atom spectrum, when an electron jumps from second excited state to the first excited state, the wavelength of radiation emitted is ' λ\lambda '. If the electron jumps from the third excited state to the second orbit, the wavelength of radiation emitted will be 20λx\frac{20\lambda}{x}. The value of xx is

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Answer: (b)

  1. Second excited state is n=3n=3, first excited is n=2n=2, third excited is n=4n=4.
  2. 3→23\to2: 1λ∝14−19=536\frac{1}{\lambda} \propto \frac14-\frac19 = \frac{5}{36}. 4→24\to2: 1λ′∝14−116=316\frac{1}{\lambda'} \propto \frac14-\frac1{16} = \frac{3}{16}.
  3. λ′λ=5/363/16=80108=2027\frac{\lambda'}{\lambda} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}
  4. So x=27x = 27, answer (b) 2727.
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Q221JEE Main 2025 · 3 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

An electron in the hydrogen atom initially in the fourth excited state makes a transition to nnth energy state by emitting a photon of energy 2.86 eV. The integer value of n will be__________.

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

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Answer: 2

  1. Fourth excited state is n=5n = 5: E5=−13.625=−0.544E_5 = -\frac{13.6}{25} = -0.544 eV.
  2. En=E5−2.86=−3.40E_n = E_5 - 2.86 = -3.40 eV
  3. −13.6n2=−3.40⇒n2=4-\frac{13.6}{n^2} = -3.40 \Rightarrow n^2 = 4
  4. So n=2n = 2.
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Q222MHT CET 2024 · 11 MAY · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (a) second to first energy level and (b) highest energy level to second level is

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Answer: (c)

  1. E∝(1nf2−1ni2)E \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. 2→12\to1: 1−14=341-\frac14 = \frac34. ∞→2\infty\to2: 14\frac14.
  3. Ratio =34:14=3:1= \frac34 : \frac14 = 3:1
  4. So the answer is (c) 3:13:1.
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Q223MHT CET 2024 · 4 MAY · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

For hydrogen atom, ' λ1\lambda_{1} ' and ' λ2\lambda_{2} ' are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of ' λ1\lambda_{1} ' and ' λ2\lambda_{2} ' is x32\frac{x}{32}. The value of ' xx ' is

Figure

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Answer: (c)

  1. From the figure: transition 1 is n=3→1n=3\to1, transition 2 is n=2→1n=2\to1.
  2. 1λ1∝1−19=89\frac{1}{\lambda_1} \propto 1-\frac19 = \frac89; 1λ2∝1−14=34\frac{1}{\lambda_2} \propto 1-\frac14 = \frac34
  3. λ1λ2=3/48/9=2732\frac{\lambda_1}{\lambda_2} = \frac{3/4}{8/9} = \frac{27}{32}
  4. So x=27x = 27, answer (c) 2727.
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Q224JEE Main 2024 · 1 FEB · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

A particular hydrogen-like ion emits the radiation of frequency 3×1015 Hz3 \times 10^{15}\text{ Hz} when it makes transition from n=2n = 2 to n=1n = 1. The frequency of radiation emitted in transition from n=3n = 3 to n=1n = 1 is x9×1015 Hz\frac{x}{9} \times 10^{15}\text{ Hz}, when x=x = ________.

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

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Answer: 32

  1. ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right): ν21∝34\nu_{21} \propto \frac{3}{4}, ν31∝89\nu_{31} \propto \frac{8}{9}
  2. ν31=3×1015×8/93/4=3×1015×3227=329×1015\nu_{31} = 3 \times 10^{15} \times \frac{8/9}{3/4} = 3 \times 10^{15} \times \frac{32}{27} = \frac{32}{9} \times 10^{15} Hz
  3. So x=32x = 32.
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Q225JEE Main 2024 · 9 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Number of spectral lines

A hydrogen atom in ground state is given an energy of 10.2 eV10.2\text{ eV}. How many spectral lines will be emitted due to transition of electrons?

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Answer: (d)

  1. E=−13.6+10.2=−3.4E = -13.6 + 10.2 = -3.4 eV =E2= E_2, so the atom goes to n=2n = 2.
  2. From n=2n = 2 only one transition (2→12\to1) is possible.
  3. So the answer is (d) 1.
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Q226JEE Main 2024 · 30 JAN · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Number of spectral lines

An electron of hydrogen atom in an excited state is having energy En=−0.85 eVE_n = -0.85\text{ eV}. The maximum number of allowed transitions to lower energy level is _________.

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

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Answer: 6

  1. En=−13.6n2=−0.85E_n = -\frac{13.6}{n^2} = -0.85 eV ⇒n2=16\Rightarrow n^2 = 16, n=4n = 4.
  2. Maximum number of transitions =n(n−1)2=4×32= \frac{n(n-1)}{2} = \frac{4 \times 3}{2}
  3. So the answer is 6.
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Q227JEE Main 2024 · 1 FEB · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:

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Answer: (c)

  1. Balmer lines end on n=2n = 2, so the atom must first reach at least n=3n = 3.
  2. Minimum energy =13.6(1−19)=12.09= 13.6\left(1 - \frac{1}{9}\right) = 12.09 eV
  3. So the answer is (c) ≈12.1\approx 12.1 eV.
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Q228MHT CET 2024 · 2 MAY · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Recoil of the atom

An electron of stationary Hydrogen atom passes from fifth energy level to ground level. The velocity that the atom acquired as a result of photo emission is

( m=m = mass of electron, R=R = Rydberg's constant)

( h=h = Planck's constant)

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Answer: (a)

  1. Momentum conservation: mv=hλmv = \frac{h}{\lambda}, where mm is the mass of the recoiling atom.
  2. 1λ=R(1−125)=24R25\frac{1}{\lambda} = R\left(1 - \frac{1}{25}\right) = \frac{24R}{25}
  3. v=24Rh25mv = \frac{24Rh}{25m}
  4. So the answer is (a).
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Q229JEE Main 2024 · 29 JAN · SHIFT 2Difficult

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

Hydrogen atom is bombarded with electrons accelerated through a potential difference of VV, which causes excitation of hydrogen atoms. If the experiment is being performed at T=0 KT = 0\text{ K}, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be α10 V\frac{\alpha}{10}\text{ V}, where α=\alpha = __________.

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

Show answer and solution

Answer: 121

  1. Balmer lines need the electron to reach at least n=3n = 3 (then 3→23 \to 2 is emitted).
  2. At T=0T = 0 K all atoms are in n=1n = 1, so the energy needed is E3−E1=13.6(1−19)=12.09 eVE_3 - E_1 = 13.6\left(1 - \frac{1}{9}\right) = 12.09\text{ eV}.
  3. Minimum accelerating voltage V≈12.1 V=12110 VV \approx 12.1\text{ V} = \frac{121}{10}\text{ V}.
  4. So α=121\alpha = 121.
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Q230JEE Main 2024 · 4 APR · SHIFT 1Difficult

Transitions, Photon Emission & Excitation · Recoil of the atom

A hydrogen atom changes its state from n=3n = 3 to n=2n = 2. Due to recoil, the percentage change in the wavelength of emitted light is approximately 1×10−n1 \times 10^{-n}. The value of nn is _______. [Given Rhc =13.6 eV, hc=1242 eV nm, h=6.6×10−34 J s= 13.6\text{ eV},\ hc = 1242\text{ eV nm},\ h = 6.6 \times 10^{-34}\text{ J s} mass of the hydrogen atom =1.6×10−27 kg= 1.6 \times 10^{-27}\text{ kg}]

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

Show answer and solution

Answer: 7

  1. Photon energy: E=13.6(14−19)=1.89 eVE = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 1.89\text{ eV}.
  2. Recoil energy =p22M=E22Mc2= \frac{p^2}{2M} = \frac{E^2}{2Mc^2}, so Δλλ=ΔEE=E2Mc2\frac{\Delta\lambda}{\lambda} = \frac{\Delta E}{E} = \frac{E}{2Mc^2}.
  3. Mc2=1.6×10−27×9×10161.6×10−19=9×108 eVMc^2 = \frac{1.6 \times 10^{-27} \times 9 \times 10^{16}}{1.6 \times 10^{-19}} = 9 \times 10^{8}\text{ eV}.
  4. Δλλ=1.891.8×109≈1×10−9\frac{\Delta\lambda}{\lambda} = \frac{1.89}{1.8 \times 10^{9}} \approx 1 \times 10^{-9}, i.e. ≈1×10−7 %\approx 1 \times 10^{-7}\ \%.
  5. So n=7n = 7.
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Q231JEE Main 2024 · 29 JAN · SHIFT 1Difficult

Transitions, Photon Emission & Excitation · Recoil of the atom

When a hydrogen atom going from n=2n = 2 to n=1n = 1 emits a photon, its recoil speed is x5 m/s\frac{x}{5}\text{ m/s}. Where x=x = ________. (Use, mass of hydrogen atom =1.6×10−27 kg= 1.6 \times 10^{-27}\text{ kg})

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

Show answer and solution

Answer: 17

  1. Photon energy E=10.2 eV=10.2×1.6×10−19=1.632×10−18E = 10.2\text{ eV} = 10.2 \times 1.6 \times 10^{-19} = 1.632 \times 10^{-18} J.
  2. Recoil momentum equals photon momentum: Mv=EcMv = \frac{E}{c}.
  3. v=1.632×10−183×108×1.6×10−27=3.4 m/s=175v = \frac{1.632 \times 10^{-18}}{3 \times 10^{8} \times 1.6 \times 10^{-27}} = 3.4\text{ m/s} = \frac{17}{5} m/s.
  4. So x=17x = 17.
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Q232MHT CET 2023 · 9 MAY · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The wavelength of light for the least energetic photons emitted in the Lyman series of the hydrogen spectrum is nearly [Take hc=1240 eV⋅nmhc = 1240\text{ eV}\cdot\text{nm}, change in energy of the levels =10.2 eV= 10.2\text{ eV} ]

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Answer: (b)

  1. Least energetic Lyman photon is n=2→n=1n=2 \to n=1 with ΔE=10.2\Delta E = 10.2 eV.
  2. λ=hcΔE=124010.2 nm≈121.6 nm\lambda = \frac{hc}{\Delta E} = \frac{1240}{10.2}\ \text{nm} \approx 121.6\ \text{nm}
  3. So the answer is (b) 122122 nm.
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Q233MHT CET 2023 · 9 MAY · SHIFT 1Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

If an electron in a hydrogen atom jumps from an orbit of level n=3n = 3 to orbit of level n=2n = 2, then the emitted radiation frequency is (where R = Rydberg's constant, C = Velocity of light)

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Answer: (d)

  1. ν=Rc(1nf2−1ni2)\nu = Rc\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. ν=Rc(14−19)=5Rc36\nu = Rc\left(\frac14-\frac19\right) = \frac{5Rc}{36}
  3. So the answer is (d) 5RC36\frac{5RC}{36}.
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Q234JEE Main 2023 · 24 JAN · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 ×\times 10 −15{}^{-15} eVs):

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Answer: (b)

  1. ΔE=13.6(1−116)=12.75\Delta E = 13.6\left(1-\frac{1}{16}\right) = 12.75 eV
  2. λ=hcΔE=4×10−15×3×10812.75\lambda = \frac{hc}{\Delta E} = \frac{4\times10^{-15}\times3\times10^{8}}{12.75} m
  3. λ=1.2×10−612.75 m≈94.1\lambda = \frac{1.2\times10^{-6}}{12.75}\ \text{m} \approx 94.1 nm
  4. So the answer is (b) 94.194.1 nm.
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Q235MHT CET 2023 · 13 MAY · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

An excited hydrogen atom emits a photon of wavelength λ\lambda in returning to ground state. The quantum number nn of the excited state is (R=R = Rydberg's constant)

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Answer: (b)

  1. Return to ground state: 1λ=R(1−1n2)\frac{1}{\lambda} = R\left(1-\frac{1}{n^2}\right)
  2. 1n2=1−1λR=λR−1λR\frac{1}{n^2} = 1-\frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}
  3. n=λRλR−1n = \sqrt{\frac{\lambda R}{\lambda R - 1}}
  4. So the answer is (b).
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Q236MHT CET 2023 · 12 MAY · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The wavelength of radiation emitted is 'λ0\lambda_{0}' when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be 20xλ0\frac{20}{x}\lambda_{0}. The value of xx is

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Answer: (d)

  1. Second excited state is n=3n=3, first excited is n=2n=2, third excited is n=4n=4.
  2. 3→23\to2: 1λ0∝14−19=536\frac{1}{\lambda_0} \propto \frac14-\frac19 = \frac{5}{36}. 4→24\to2: 1λ′∝14−116=316\frac{1}{\lambda'} \propto \frac14-\frac1{16} = \frac{3}{16}.
  3. λ′λ0=5/363/16=80108=2027\frac{\lambda'}{\lambda_0} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}
  4. So x=27x = 27, answer (d) 2727.
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Q237MHT CET 2023 · 9 MAY · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ratio of wavelengths for transition of electrons from 22nd orbit to 11st orbit of Helium (He+)\left(\text{He}^{+}\right) and Lithium (Li++)\left(\text{Li}^{++}\right) is (Atomic number of Helium =2= 2, Atomic number of Lithium =3= 3)

Show answer and solution

Answer: (a)

  1. For hydrogen-like ions 1λ=RZ2(1nf2−1ni2)\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); same transition gives λ∝1Z2\lambda \propto \frac{1}{Z^2}.
  2. λHe+λLi++=ZLi2ZHe2=94\frac{\lambda_{He^+}}{\lambda_{Li^{++}}} = \frac{Z_{Li}^2}{Z_{He}^2} = \frac{9}{4}
  3. So the answer is (a) 9:49:4.
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Q238JEE Main 2023 · 1 FEB · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

An electron of a hydrogen like atom, having Z=4Z = 4, jumps from 44th energy state to 22nd energy state. The energy released in this process, will be:

(Given Rch = 13.6 eV13.6\text{ eV})

Where R = Rydberg constant

c = Speed of light in vacuum

h = Planck's constant

Show answer and solution

Answer: (b)

  1. ΔE=Rhc Z2(1nf2−1ni2)\Delta E = Rhc\, Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)
  2. ΔE=13.6×16×(14−116)=13.6×16×316\Delta E = 13.6 \times 16 \times \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times 16 \times \frac{3}{16}
  3. ΔE=40.8\Delta E = 40.8 eV, so the answer is (b).
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Q239JEE Main 2023 · 15 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

As per given figure A,BA, B and CC are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths (λ1λ2)\left(\frac{\lambda_{1}}{\lambda_{2}}\right) i.e. is 74n\frac{7}{4n}, then the value of nn will be __________.

Figure

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

Show answer and solution

Answer: 5

  1. A, B, C are n=2,3,4n = 2, 3, 4. λ1\lambda_1: 3→23\to2, λ2\lambda_2: 4→34\to3.
  2. 1λ1=R(14−19)=5R36\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}, 1λ2=R(19−116)=7R144\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}
  3. λ1λ2=7/1445/36=720\frac{\lambda_1}{\lambda_2} = \frac{7/144}{5/36} = \frac{7}{20}
  4. 74n=720⇒n=5\frac{7}{4n} = \frac{7}{20} \Rightarrow n = 5
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Q240JEE Main 2023 · 31 JAN · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

For hydrogen atom, λ1\lambda_{1} and λ2\lambda_{2} are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of λ1\lambda_{1} and λ2\lambda_{2} is x32\frac{x}{32}. The value of xx is __________.

Figure

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

Show answer and solution

Answer: 27

  1. From the figure, transition 1 is 3→13\to1 and transition 2 is 2→12\to1.
  2. 1λ1=R(1−19)=8R9\frac{1}{\lambda_1} = R\left(1 - \frac{1}{9}\right) = \frac{8R}{9}, 1λ2=R(1−14)=3R4\frac{1}{\lambda_2} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}
  3. λ1λ2=3/48/9=2732\frac{\lambda_1}{\lambda_2} = \frac{3/4}{8/9} = \frac{27}{32}
  4. So x=27x = 27.
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Q241JEE Main 2023 · 25 JAN · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The wavelength of the radiation emitted is λ0\lambda_{0} when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be 20xλ0\frac{20}{x}\lambda_{0}. The value of xx is _____________.

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

Show answer and solution

Answer: 27

  1. λ0\lambda_0: second excited (n=3n=3) to first excited (n=2n=2): 1λ0=R(14−19)=5R36\frac{1}{\lambda_0} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}
  2. Third excited (n=4n=4) to n=2n=2: 1λ=R(14−116)=3R16\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}
  3. λ=λ0×5/363/16=2027λ0\lambda = \lambda_0 \times \frac{5/36}{3/16} = \frac{20}{27}\lambda_0
  4. So x=27x = 27.
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Q242JEE Main 2023 · 12 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Number of spectral lines

A 12.5 eV12.5\text{ eV} electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

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Answer: (c)

  1. An electron can give any part of its 12.5 eV. Excitation energies: n=3n=3 needs 12.09 eV, n=4n=4 needs 12.75 eV.
  2. So the highest level reached is n=3n = 3.
  3. Lines =3×22=3= \frac{3 \times 2}{2} = 3
  4. So the answer is (c) 3.
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Q243JEE Main 2023 · 11 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x×1015 Hzx \times 10^{15}\text{ Hz}. The value of xx is ____________.

(Given h =4.25×10−15 eV s= 4.25 \times 10^{-15}\text{ eV s})

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

Show answer and solution

Answer: 3

  1. Six wavelengths: n(n−1)2=6⇒n=4\frac{n(n-1)}{2} = 6 \Rightarrow n = 4.
  2. Absorbed energy =13.6(1−116)=12.75= 13.6\left(1 - \frac{1}{16}\right) = 12.75 eV
  3. ν=Eh=12.754.25×10−15=3×1015\nu = \frac{E}{h} = \frac{12.75}{4.25 \times 10^{-15}} = 3 \times 10^{15} Hz
  4. So x=3x = 3.
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Q244JEE Main 2022 · 29 JUL · SHIFT 1Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

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Answer: (a)

  1. E∝(1nf2−1ni2)E \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. (i) 2→12\to1: 1−14=341-\frac14 = \frac34. (ii) ∞→1\infty\to1: 1−0=11-0 = 1.
  3. Ratio =34:1=3:4= \frac34 : 1 = 3:4
  4. So the answer is (a) 3:43:4.
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Q245JEE Main 2022 · 25 JUL · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ\lambda. The value of principal quantum number 'nn' of the excited state will be: (R:R: Rydberg constant)

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Answer: (b)

  1. 1λ=R(1−1n2)\frac{1}{\lambda} = R\left(1 - \frac{1}{n^2}\right)
  2. 1n2=1−1λR=λR−1λR\frac{1}{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}
  3. n=λRλR−1n = \sqrt{\frac{\lambda R}{\lambda R - 1}}, so the answer is (b).
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Q246JEE Main 2022 · 25 JUL · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

xx+4\frac{x}{x+4} is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its

(i) third permitted energy level to the second level and

(ii) the highest permitted energy level to the second permitted level.

The value of x will be ____________.

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

Show answer and solution

Answer: 5

  1. E∝(1nf2−1ni2)E \propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)
  2. (i) 3→23\to2: 14−19=536\frac{1}{4} - \frac{1}{9} = \frac{5}{36}; (ii) ∞→2\infty\to2: 14=936\frac{1}{4} = \frac{9}{36}
  3. Ratio =59=xx+4= \frac{5}{9} = \frac{x}{x+4}
  4. So x=5x = 5.
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Q247JEE Main 2022 · 30 JUN · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

A hydrogen atom in its first excited state absorbs a photon of energy x ×\times 10−-2 eV and excited to a higher energy state where the potential energy of electron is−1{}^{-1}.08 eV. The value of x is ______________.

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

Show answer and solution

Answer: 286

  1. Total energy =12×= \frac{1}{2} \times PE =−1.082=−0.54= \frac{-1.08}{2} = -0.54 eV (this is n=5n = 5).
  2. First excited state energy =−3.4= -3.4 eV.
  3. Absorbed energy =−0.54−(−3.4)=2.86= -0.54 - (-3.4) = 2.86 eV =286×10−2= 286 \times 10^{-2} eV
  4. So x=286x = 286.
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Q248JEE Main 2022 · 27 JUN · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

A beam of monochromatic light is used to excite the electron in Li++ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be x ×\times 10−-10 m. The value of x is ___________.

[Given hc = 1242 eV nm]

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

Show answer and solution

Answer: 114

  1. For Li++^{++}, Z=3Z = 3: ΔE=13.6×9×(1−19)=108.8\Delta E = 13.6 \times 9 \times \left(1 - \frac{1}{9}\right) = 108.8 eV
  2. λ=hcΔE=1242108.8\lambda = \frac{hc}{\Delta E} = \frac{1242}{108.8} nm =11.4= 11.4 nm
  3. λ=114×10−10\lambda = 114 \times 10^{-10} m, so x=114x = 114.
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Q249JEE Main 2021 · 25 FEB · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is:

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Answer: (d)

  1. 1λ=R(1−14)=3R4\frac{1}{\lambda} = R\left(1-\frac14\right) = \frac{3R}{4}, with R=1.097×107 m−1R = 1.097\times10^{7}\ \text{m}^{-1}
  2. λ=43×1.097×107 m≈1.22×10−7\lambda = \frac{4}{3\times1.097\times10^{7}}\ \text{m} \approx 1.22\times10^{-7} m
  3. λ≈122\lambda \approx 122 nm
  4. So the answer is (d) 121.8121.8 nm.
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Q250JEE Main 2021 · 24 FEB · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

According to Bohr atomic model, in which of the following transitions will the frequency be maximum?

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Answer: (a)

  1. ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); the gap is largest for the lowest levels.
  2. 2→12\to1: 0.750.75; 3→23\to2: 0.1390.139; 4→34\to3: 0.0490.049; 5→45\to4: 0.0230.023
  3. So the answer is (a) n=2n=2 to n=1n=1.
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Q251MHT CET 2021 · 20 SEP · SHIFT 1Easy

Transitions, Photon Emission & Excitation · Number of spectral lines

The electron in hydrogen atom is initially in the third excited state. When it finally moves to ground state, the maximum number of spectral lines emitted are

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Answer: (d)

  1. Third excited state means n=4n = 4.
  2. Maximum lines =n(n−1)2=4×32=6= \frac{n(n-1)}{2} = \frac{4\times3}{2} = 6
  3. So the answer is (d) 66.
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Q252JEE Main 2021 · 27 AUG · SHIFT 2Easy

Transitions, Photon Emission & Excitation · Number of spectral lines

X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n = 6? The value of X is ______________.

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

Show answer and solution

Answer: 15

  1. Number of different wavelengths from level nn =n(n−1)2= \frac{n(n-1)}{2}
  2. n=6n = 6: 6×52=15\frac{6\times5}{2} = 15
  3. So X=15X = 15.
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Q253MHT CET 2021 · 24 SEP · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second energy level is

Show answer and solution

Answer: (b)

  1. E∝(1nf2−1ni2)E \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. 2→12\to1: 1−14=341-\frac14 = \frac34. ∞→2\infty\to2: 14\frac14.
  3. Ratio =34:14=3:1= \frac34 : \frac14 = 3:1
  4. So the answer is (b) 3:13:1.
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Q254JEE Main 2021 · 26 AUG · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

A particular hydrogen like ion emits radiation of frequency 2.92 ×\times 1015 Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be:

Show answer and solution

Answer: (d)

  1. ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) for the same ion.
  2. ν31∝1−19=89\nu_{31} \propto 1 - \frac{1}{9} = \frac{8}{9}, ν21∝1−14=34\nu_{21} \propto 1 - \frac{1}{4} = \frac{3}{4}
  3. ν21=2.92×1015×3/48/9=2.92×1015×2732\nu_{21} = 2.92 \times 10^{15} \times \frac{3/4}{8/9} = 2.92 \times 10^{15} \times \frac{27}{32}
  4. ν21=2.46×1015\nu_{21} = 2.46 \times 10^{15} Hz, so the answer is (d).
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Q255MHT CET 2020 · 16 OCT · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second level is respectively

Show answer and solution

Answer: (b)

  1. E∝(1nf2−1ni2)E \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. 2→12\to1: 1−14=341-\frac14 = \frac34. ∞→2\infty\to2: 14\frac14.
  3. Ratio =34:14=3:1= \frac34 : \frac14 = 3:1
  4. So the answer is (b) 3:13:1.
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Q256JEE Main 2020 · 9 JAN · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?

Show answer and solution

Answer: (b)

  1. Ionisation energy =Z2×1= Z^2 \times 1 Ry =9= 9 Ry ⇒Z=3\Rightarrow Z = 3.
  2. Second excited state is n=3n = 3: E=9×13.6×(1−19)=108.8E = 9 \times 13.6 \times \left(1 - \frac{1}{9}\right) = 108.8 eV
  3. λ=1240108.8\lambda = \frac{1240}{108.8} nm =11.4= 11.4 nm
  4. So the answer is (b) 11.4 nm.
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Q257JEE Main 2020 · 2 SEP · SHIFT 2Difficult

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In a hydrogen atom the electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to:

Show answer and solution

Answer: (c)

  1. ν∝1n2−1(n+1)2=2n+1n2(n+1)2\nu \propto \frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{2n+1}{n^2(n+1)^2}
  2. For n≫1n \gg 1: ν∝2nn4=2n3\nu \propto \frac{2n}{n^4} = \frac{2}{n^3}
  3. So the answer is (c) 1n3\frac{1}{n^3}.
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Q258JEE Main 2020 · 5 SEP · SHIFT 1Difficult

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

A particle of mass 200 MeV/c2 collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is

N4\frac{N}{4}. The value of N is:

(Given the mass of the hydrogen atom to be 1 GeV/c2) ______.

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

Show answer and solution

Answer: 51

  1. Momentum conservation: mu=Mv⇒v=mMu=u5mu = Mv \Rightarrow v = \frac{m}{M}u = \frac{u}{5} (since m=0.2m = 0.2 GeV/c2/c^2, M=1M = 1 GeV/c2/c^2).
  2. KE of recoiling atom =12Mv2=mMK=K5= \frac{1}{2}Mv^2 = \frac{m}{M}K = \frac{K}{5}, where KK is the initial KE.
  3. Energy used for excitation: K−K5=4K5=10.2 eVK - \frac{K}{5} = \frac{4K}{5} = 10.2\text{ eV} (1→21 \to 2 in H).
  4. K=12.75 eV=514K = 12.75\text{ eV} = \frac{51}{4} eV.
  5. So N=51N = 51.
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Q259JEE Main 2019 · 12 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, λ1λ2\frac{\lambda_{1}}{\lambda_{2}}, of the photons emitted in this process is:

Show answer and solution

Answer: (d)

  1. Third excited n=4n=4, second excited n=3n=3, first excited n=2n=2.
  2. 1λ1=R(19−116)=7R144\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}
  3. 1λ2=R(14−19)=20R144\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{20R}{144}
  4. λ1λ2=207\frac{\lambda_1}{\lambda_2} = \frac{20}{7}, so the answer is (d).
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Q260JEE Main 2019 · 11 JAN · SHIFT 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is λ\lambda. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:

Show answer and solution

Answer: (d)

  1. K, L, M, N shells are n=1,2,3,4n = 1, 2, 3, 4.
  2. M to L: 1λ∝14−19=536\frac{1}{\lambda} \propto \frac{1}{4} - \frac{1}{9} = \frac{5}{36}
  3. N to L: 1λ′∝14−116=316\frac{1}{\lambda'} \propto \frac{1}{4} - \frac{1}{16} = \frac{3}{16}
  4. λ′=λ×5/363/16=2027λ\lambda' = \lambda \times \frac{5/36}{3/16} = \frac{20}{27}\lambda, so the answer is (d).
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Q261JEE Main 2019 · 11 JAN · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

{} A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980A˚\text{Å}. The radius of the atom in the excited state, in terms of Bohr radius a0 will be: (hc = 12500 eVA˚\text{Å})

Show answer and solution

Answer: (d)

  1. Photon energy =12500980≈12.75= \frac{12500}{980} \approx 12.75 eV.
  2. En=−13.6+12.75=−0.85E_n = -13.6 + 12.75 = -0.85 eV ⇒n=4\Rightarrow n = 4.
  3. rn=n2a0=16a0r_n = n^2 a_0 = 16 a_0
  4. So the answer is (d) 16a016a_0.
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Q262JEE Main 2019 · 12 APR · SHIFT 1Difficult

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength λ\lambda, energy E=1240 eVλ(in nm)E = \frac{1240\text{ eV}}{\lambda(\text{in nm})}):

Show answer and solution

Answer: (c)

  1. E1=1240108.5=11.43E_1 = \frac{1240}{108.5} = 11.43 eV, E2=124030.4=40.79E_2 = \frac{1240}{30.4} = 40.79 eV; total =52.22= 52.22 eV.
  2. For He+^+, En=−54.4n2E_n = -\frac{54.4}{n^2} eV, ground state −54.4-54.4 eV.
  3. En=−54.4+52.22=−2.18E_n = -54.4 + 52.22 = -2.18 eV ⇒n2=54.42.18≈25\Rightarrow n^2 = \frac{54.4}{2.18} \approx 25
  4. So n=5n = 5, the answer is (c).
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Q263JEE Main 2019 · 10 APR · SHIFT 2Difficult

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

In Li+ +, electron in first Bohr orbit is excited to a level by a radiation of wavelength λ\lambda. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ\lambda?

(Given: H = 6.63 × 10–34 Js; c = 3 × 108 ms –1)

Show answer and solution

Answer: (a)

  1. Six lines: n(n−1)2=6⇒\frac{n(n-1)}{2} = 6 \Rightarrow excited to n=4n = 4.
  2. For Li++^{++} (Z=3Z = 3): E=13.6×9×(1−116)=114.75E = 13.6 \times 9 \times \left(1 - \frac{1}{16}\right) = 114.75 eV
  3. λ=hcE=6.63×10−34×3×108114.75×1.6×10−19=1.08×10−8\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{114.75 \times 1.6 \times 10^{-19}} = 1.08 \times 10^{-8} m
  4. So the answer is (a) 10.8 nm.
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Q264JEE Main 2019 · 8 APR · SHIFT 1Difficult

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is:

Show answer and solution

Answer: (c)

  1. Photon from H (2→12 \to 1): E=13.6(1−14)=10.2 eVE = 13.6\left(1 - \frac{1}{4}\right) = 10.2\text{ eV}.
  2. He+^+ levels: En=−54.4n2E_n = -\frac{54.4}{n^2} eV, so E1=−54.4E_1 = -54.4, E2=−13.6E_2 = -13.6, E3=−6.04E_3 = -6.04, E4=−3.4E_4 = -3.4, E5=−2.18E_5 = -2.18 eV.
  3. From n=2n = 2: E4−E2=−3.4+13.6=10.2 eVE_4 - E_2 = -3.4 + 13.6 = 10.2\text{ eV}, an exact match.
  4. No other listed jump needs exactly 10.210.2 eV.
  5. So the answer is (c) n=2→n=4n = 2 \rightarrow n = 4.
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Q265NEET 2016 · PHASE 2Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ\lambda. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

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Answer: (c)

  1. 1λ∝(1nf2−1ni2)\frac{1}{\lambda} \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
  2. 3→23\to2: 536\frac{5}{36}. 4→34\to3: 19−116=7144\frac19-\frac1{16} = \frac{7}{144}.
  3. λ′=λ×5/367/144=207λ\lambda' = \lambda \times \frac{5/36}{7/144} = \frac{20}{7}\lambda
  4. So the answer is (c) 207λ\frac{20}{7}\lambda.
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Q266JEE Main 2016 · 9 APR · SHIFT 1Difficult

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is:

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Answer: (c)

  1. Photon energy (H, 2→12 \to 1) =10.2 eV= 10.2\text{ eV}.
  2. Ionisation energy of Li2+^{2+} from level nn: 13.6×9n2=122.4n2\frac{13.6 \times 9}{n^2} = \frac{122.4}{n^2} eV.
  3. Need 122.4n2≤10.2⇒n2≥12⇒n≥3.46\frac{122.4}{n^2} \le 10.2 \Rightarrow n^2 \ge 12 \Rightarrow n \ge 3.46.
  4. Least integer n=4n = 4 (needs 7.657.65 eV).
  5. So the answer is (c) 4.
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Q267JEE Main 2014Medium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Hydrogen (1H1)\left({}_{1}H^{1}\right), Deuterium (1H2)\left({}_{1}H^{2}\right), singly ionised Helium (2He4)+\left({}_{2}He^{4}\right)^{+} and doubly ionised lithium (3Li6)++\left({}_{3}Li^{6}\right)^{++} all have one electron around the nucleus. Consider an electron transition from n=2n = 2 to n=1.n = 1. If the wavelengths of emitted radiation are λ1,λ2,λ3\lambda_{1}, \lambda_{2}, \lambda_{3} and λ4\lambda_{4} respectively then approximately which one of the following is correct?

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Answer: (c)

  1. 1λ=RZ2(1−14)\frac{1}{\lambda} = R Z^2 \left(1 - \frac{1}{4}\right), so λ∝1Z2\lambda \propto \frac{1}{Z^2} (nuclear mass effect neglected).
  2. H and D: Z=1Z = 1; He+^+: Z=2Z = 2; Li++^{++}: Z=3Z = 3.
  3. λ1=λ2=4λ3=9λ4\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4
  4. So the answer is (c).
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Q268NEET 2014 · AIPMTMedium

Transitions, Photon Emission & Excitation · Number of spectral lines

Hydrogen atom in ground state is excited by a monochromatic radiation of λ\lambda = 975 A˚\text{Å}. Number of spectral lines in the resulting spectrum emitted will be

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Answer: (c)

  1. Photon energy E=12400975≈12.75E = \frac{12400}{975} \approx 12.75 eV.
  2. En=−13.6+12.75=−0.85E_n = -13.6 + 12.75 = -0.85 eV ⇒n=4\Rightarrow n = 4.
  3. Number of lines =n(n−1)2=4×32=6= \frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6
  4. So the answer is (c) 6.
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Q269JEE Main 2014Difficult

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

The radiation corresponding to 3→23 \rightarrow 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field 3×10−4 T.3 \times 10^{-4}\text{ T}. If the radius of the larger circular path followed by these electrons is 10.0 mm10.0\text{ mm}, the work function of the metal is close to:

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Answer: (b)

  1. Photon energy (3→23 \to 2): E=13.6(14−19)=1.89 eVE = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 1.89\text{ eV}.
  2. Fastest electrons have the largest radius: p=eBr=1.6×10−19×3×10−4×0.01=4.8×10−25p = eBr = 1.6 \times 10^{-19} \times 3 \times 10^{-4} \times 0.01 = 4.8 \times 10^{-25} kg m/s.
  3. Kmax=p22m=(4.8×10−25)22×9.1×10−31=1.27×10−19 J≈0.79 eVK_{max} = \frac{p^2}{2m} = \frac{(4.8 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}} = 1.27 \times 10^{-19}\text{ J} \approx 0.79\text{ eV}.
  4. ϕ=E−Kmax=1.89−0.79≈1.1 eV\phi = E - K_{max} = 1.89 - 0.79 \approx 1.1\text{ eV}.
  5. So the answer is (b) 1.11.1 eV.
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Q270JEE Main 2013 · 25 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation · Number of spectral lines

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:

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Answer: (a)

n = 3n = 2n = 1LymanLymanBalmer
  1. 12.5 eV can excite hydrogen up to n=3n = 3 (needs 12.09 eV), not n=4n = 4 (needs 12.75 eV).
  2. Possible lines: 3→13\to1 and 2→12\to1 end on n=1n=1 (Lyman); 3→23\to2 ends on n=2n=2 (Balmer).
  3. So the answer is (a) 2 lines in Lyman and 1 line in Balmer series.
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Q271JEE Main 2013Difficult

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

In a hydrogen like atom electron make transition from an energy level with quantum number nn to another with quantum number (n−1)(n-1). If n≫1,n \gg 1, the frequency of radiation emitted is proportional to:

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Answer: (d)

  1. ν∝1(n−1)2−1n2=2n−1n2(n−1)2\nu \propto \frac{1}{(n-1)^2} - \frac{1}{n^2} = \frac{2n-1}{n^2(n-1)^2}
  2. For n≫1n \gg 1: ν∝2nn4=2n3\nu \propto \frac{2n}{n^4} = \frac{2}{n^3}
  3. So the answer is (d) 1n3\frac{1}{n^3}.
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Q272JEE Main 2012 · AIEEEEasy

Transitions, Photon Emission & Excitation · Number of spectral lines

Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4.4. Then the number of spectral lines in the emission spectra will be:

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Answer: (d)

  1. Number of lines from level nn =n(n−1)2= \frac{n(n-1)}{2}
  2. n=4n = 4: 4×32=6\frac{4\times3}{2} = 6
  3. So the answer is (d) 66.
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Q273NEET 2012 · AIPMT · MAINSMedium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

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Answer: (d)

  1. Photon energy ∝(1nf2−1ni2)\propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right). Infrared has much lower energy than UV, so pick the smallest bracket.
  2. 3→1:893\to1: \frac{8}{9}, 2→1:342\to1: \frac{3}{4}, 3→2:5363\to2: \frac{5}{36}, 4→2:3164\to2: \frac{3}{16}, 4→3:71444\to3: \frac{7}{144}
  3. 4→34\to3 gives the least energy (longest wavelength).
  4. So the answer is (d) 4→34 \rightarrow 3.
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Q274NEET 2012 · AIPMT · PRELIMSMedium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ\lambda1: λ\lambda2 emitted in the two cases is

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Answer: (d)

  1. Third excited state is n=4n=4, second excited is n=3n=3, first excited is n=2n=2.
  2. 1λ1=R(19−116)=7R144\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}
  3. 1λ2=R(14−19)=5R36=20R144\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36} = \frac{20R}{144}
  4. λ1λ2=207\frac{\lambda_1}{\lambda_2} = \frac{20}{7}, so the answer is (d).
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Q275NEET 2012 · AIPMT · PRELIMSMedium

Transitions, Photon Emission & Excitation · Recoil of the atom

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

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Answer: (a)

  1. Momentum conservation: mv=hλmv = \frac{h}{\lambda} (mm = mass of the atom).
  2. 1λ=R(1−125)=24R25\frac{1}{\lambda} = R\left(1 - \frac{1}{25}\right) = \frac{24R}{25}
  3. v=24hR25mv = \frac{24hR}{25m}
  4. So the answer is (a).
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Q276NEET 2011 · AIPMT · MAINSMedium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model?

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Answer: (c)

  1. En=−13.6n2E_n = -\frac{13.6}{n^2} eV: E1=−13.6E_1 = -13.6, E2=−3.4E_2 = -3.4, E3=−1.51E_3 = -1.51, E4=−0.85E_4 = -0.85 eV.
  2. 4→34\to3: 0.660.66 eV; 3→23\to2: 1.891.89 eV; ∞→1\infty\to1: 13.613.6 eV.
  3. Lines ending on n=1n=1 have E≥10.2E \ge 10.2 eV (2→12\to1) and the next is 12.0912.09 eV (3→13\to1); nothing gives 11.111.1 eV.
  4. So the answer is (c) 11.111.1 eV.
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Q277NEET 2011 · AIPMT · MAINSMedium

Transitions, Photon Emission & Excitation · Excitation by photons & electrons

An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is

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Answer: (c)

  1. Kmax=eV0=10K_{max} = eV_0 = 10 eV, so photon energy =2.75+10=12.75= 2.75 + 10 = 12.75 eV.
  2. 13.6(1−1n2)=12.75⇒1n2=11613.6\left(1 - \frac{1}{n^2}\right) = 12.75 \Rightarrow \frac{1}{n^2} = \frac{1}{16}
  3. So n=4n = 4, the answer is (c).
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Q278JEE Main 2009 · AIEEEMedium

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The transition from the state n=4n = 4 to n=3n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

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Answer: (c)

  1. Infrared needs a photon of lower energy than the 4→34\to3 (UV) photon.
  2. 4→34\to3: 19−116=7144≈0.049\frac{1}{9} - \frac{1}{16} = \frac{7}{144} \approx 0.049
  3. 3→23\to2: 0.1390.139, 4→24\to2: 0.18750.1875, 2→12\to1: 0.750.75 are all larger; 5→45\to4: 116−125=9400≈0.0225\frac{1}{16} - \frac{1}{25} = \frac{9}{400} \approx 0.0225 is smaller.
  4. So the answer is (c) 5→45 \rightarrow 4.
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Q279NEET 2009 · AIPMTMedium

Transitions, Photon Emission & Excitation · Number of spectral lines

The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

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Answer: (c)

  1. Number of lines n(n−1)2=6⇒n=4\frac{n(n-1)}{2} = 6 \Rightarrow n = 4.
  2. Maximum wavelength means minimum energy gap, which is between the two highest levels.
  3. So the answer is (c) n=4n = 4 to n=3n = 3.
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Q280JEE Main 2007 · AIEEEEasy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Which of the following transitions in hydrogen atoms emit photons of highest frequency?

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Answer: (d)

  1. Emitted frequency ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); emission needs a jump from higher nn to lower nn.
  2. (a) and (b) are absorptions. (c) 6→26\to2: 14−136=0.22\frac14-\frac1{36} = 0.22; (d) 2→12\to1: 1−14=0.751-\frac14 = 0.75.
  3. So the answer is (d) n=2n=2 to n=1n=1.
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Q281NEET 2006 · AIPMTMedium

Transitions, Photon Emission & Excitation · Number of spectral lines

Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emited by hydrogen will be

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Answer: (c)

  1. After absorbing 12.1 eV: E=−13.6+12.1=−1.5E = -13.6 + 12.1 = -1.5 eV ≈E3\approx E_3, so n=3n = 3.
  2. Lines =n(n−1)2=3×22=3= \frac{n(n-1)}{2} = \frac{3 \times 2}{2} = 3 (3→13\to1, 3→23\to2, 2→12\to1).
  3. So the answer is (c) three.
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Q282NEET 2004 · AIPMTEasy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Energy EE of a hydrogen atom with principal quantum number nn is given by E=−13.6n2 eVE = \frac{-13.6}{n^{2}}\text{ eV}. The energy of a photon ejected when the electron jumps from n=3n = 3 state to n=2n = 2 state of hydrogen, is approximately

Show answer and solution

Answer: (d)

  1. E=13.6(122−132)E = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) eV
  2. E=13.6×536≈1.89E = 13.6 \times \frac{5}{36} \approx 1.89 eV
  3. So the answer is (d) 1.91.9 eV.
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Q283JEE Main 2003 · AIEEEEasy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The wavelengths involved in the spectrum of deuterium (12D)\left({}_{1}^{2}D\right) are slightly different from that of hydrogen spectrum, because

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Answer: (c)

  1. Rydberg constant depends on the reduced mass μ=meMme+M\mu = \frac{m_e M}{m_e + M} of electron and nucleus.
  2. Deuterium nucleus is about twice as heavy as the proton, so μ\mu and hence RR change slightly.
  3. Charge (attraction) is the same; nuclear size and nuclear forces do not affect the levels.
  4. So the answer is (c).
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Q284NEET 2000 · AIPMTEasy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

Maximum frequency of emission is obtained for the transition

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Answer: (a)

  1. Emitted frequency ν∝(1nf2−1ni2)\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right); emission needs a jump from higher nn to lower nn.
  2. (c) and (d) are absorptions. (a) 2→12\to1: 1−14=0.751-\frac14 = 0.75; (b) 6→26\to2: 14−136=0.22\frac14-\frac1{36} = 0.22.
  3. So the answer is (a) n=2n=2 to n=1n=1.
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Q285NEET 1995 · AIPMTEasy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

An electron makes a transition from orbit n=4n = 4 to the orbit n=2n = 2 of a hydrogen atom. What is the wavelength of the emitted radiations? (RR = Rydberg's constant)

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Answer: (d)

  1. 1λ=R(122−142)=R(4−116)=3R16\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) = R\left(\frac{4-1}{16}\right) = \frac{3R}{16}
  2. λ=163R\lambda = \frac{16}{3R}
  3. So the answer is (d) 163R\frac{16}{3R}.
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Q286NEET 1993 · AIPMTEasy

Transitions, Photon Emission & Excitation · Number of spectral lines

Hydrogen atoms are excited from ground state of the principle quantum number 4. Then the number of spectral lines observed will be

Show answer and solution

Answer: (b)

  1. Number of lines from level nn =n(n−1)2= \frac{n(n-1)}{2}
  2. n=4n = 4: 4×32=6\frac{4\times3}{2} = 6
  3. So the answer is (b) 66.
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Q287NEET 1992 · AIPMTEasy

Transitions, Photon Emission & Excitation · Photon energy & wavelength in a transition

The ionization energy of hydrogen atom is 13.6 eV. Following Bohr's theory, the energy corresponding to a transition between 33rd and 44th orbit is

Show answer and solution

Answer: (d)

  1. En=−13.6n2E_n = -\frac{13.6}{n^2} eV, so ΔE=13.6(19−116)\Delta E = 13.6\left(\frac19-\frac1{16}\right)
  2. ΔE=13.6×7144≈0.66\Delta E = 13.6 \times \frac{7}{144} \approx 0.66 eV
  3. So the answer is (d) 0.660.66 eV.
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Q288MHT CET 2026 · 20 MAY · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

Which one of the following spectral series of hydrogen atom lies in ultraviolet region?

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Answer: (a)

  1. Lyman series ends on n=1n = 1; its photons have energies 10.210.2 to 13.613.6 eV, i.e. wavelengths 9191 to 122122 nm.
  2. Paschen, Brackett and Pfund lie in the infrared.
  3. So the answer is (a) Lyman.
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Q289MHT CET 2026 · 13 MAY · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of minimum wavelengths of Lyman and Balmer series will be

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Answer: (b)

  1. Minimum wavelength is the series limit (n2→∞n_2 \to \infty): 1λmin=Rn12\frac{1}{\lambda_{min}} = \frac{R}{n_1^2}.
  2. Lyman: λmin=1R\lambda_{min} = \frac{1}{R}; Balmer: λmin=4R\lambda_{min} = \frac{4}{R}.
  3. Ratio =14=0.25= \frac{1}{4} = 0.25.
  4. So the answer is (b) 0.250.25.
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Q290MHT CET 2026 · 12 MAY · SHIFT 2Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In Bohr's theory of hydrogen atom, in any series of spectral lines, wavelength of first line and corresponding series limit is respectively

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Answer: (a)

  1. 1λ=R(1n12−1n22)\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right): the first line (n2=n1+1n_2 = n_1 + 1) has the smallest energy gap, so the longest λ\lambda.
  2. The series limit (n2→∞n_2 \to \infty) has the largest gap, so the shortest λ\lambda.
  3. So the answer is (a) longest and shortest.
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Q291MHT CET 2026 · 17 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

For wavelength of visible radiation of Hydrogen spectrum, Balmer gave an equation as λ=xm2m2−4\lambda = \frac{xm^2}{m^2 - 4}, where m is the integer value. The value of x in terms of Rydberg's constant R is

Show answer and solution

Answer: (b)

  1. Balmer: 1λ=R(14−1m2)=R(m2−4)4m2\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{m^2}\right) = \frac{R(m^2 - 4)}{4m^2}.
  2. λ=4R⋅m2m2−4\lambda = \frac{4}{R} \cdot \frac{m^2}{m^2 - 4}.
  3. Comparing with λ=xm2m2−4\lambda = \frac{xm^2}{m^2 - 4} gives x=4Rx = \frac{4}{R}.
  4. So the answer is (b) 4R\frac{4}{R}.
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Q292JEE Main 2026 · 23 JAN · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

In hydrogen atom spectrum, ( R→R \rightarrow Rydberg's constant)

A. the maximum wavelength of the radiation of Lyman series is 43R\frac{4}{3R}

B. the Balmer series lies in the visible region of the spectrum

C. the minimum wavelength of the radiation of Paschen series is 9R\frac{9}{R}

D. the minimum wavelength of Lyman series is 54R\frac{5}{4R}

Choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

  1. A: Lyman maximum λ\lambda (2→12 \to 1): 1λ=3R4⇒λ=43R\frac{1}{\lambda} = \frac{3R}{4} \Rightarrow \lambda = \frac{4}{3R}. Correct.
  2. B: Balmer lines (365365 to 656656 nm) are visible. Correct.
  3. C: Paschen minimum λ\lambda (limit): λ=9R\lambda = \frac{9}{R}. Correct.
  4. D: Lyman minimum λ\lambda (limit) is 1R\frac{1}{R}, not 54R\frac{5}{4R}. Wrong.
  5. So the answer is (d) A, B and C Only.
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Q293MHT CET 2026 · 14 MAY · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If λ1\lambda_1 and λ2\lambda_2 are the wavelengths of the first members of the Lyman and Paschen series respectively then λ1:λ2\lambda_1 : \lambda_2 is

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Answer: (d)

  1. Lyman first line: 1λ1=R(1−14)=3R4\frac{1}{\lambda_1} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}, so λ1=43R\lambda_1 = \frac{4}{3R}.
  2. Paschen first line: 1λ2=R(19−116)=7R144\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}, so λ2=1447R\lambda_2 = \frac{144}{7R}.
  3. λ1λ2=43×7144=7108\frac{\lambda_1}{\lambda_2} = \frac{4}{3} \times \frac{7}{144} = \frac{7}{108}.
  4. So the answer is (d) 7:1087 : 108.
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Q294MHT CET 2026 · 19 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number 'Z'. The value of 'Z' is

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Answer: (d)

  1. Shortest λ\lambda = series limit: 1λ=RZ2n12\frac{1}{\lambda} = \frac{RZ^2}{n_1^2}.
  2. H Balmer: λ=4R\lambda = \frac{4}{R}. Brackett (n1=4n_1 = 4) for atomic number ZZ: λ=16RZ2\lambda = \frac{16}{RZ^2}.
  3. 4R=16RZ2⇒Z2=4⇒Z=2\frac{4}{R} = \frac{16}{RZ^2} \Rightarrow Z^2 = 4 \Rightarrow Z = 2.
  4. So the answer is (d) 22.
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Q295JEE Main 2026 · 22 JAN · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm.

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Answer: (c)

  1. Lyman smallest: 1R=91\frac{1}{R} = 91 nm
  2. Paschen largest (4→34 \to 3): λP=1447R=1447×91=1872\lambda_P = \frac{144}{7R} = \frac{144}{7}\times 91 = 1872 nm
  3. Balmer largest (3→23 \to 2): λB=365R=365×91=655.2\lambda_B = \frac{36}{5R} = \frac{36}{5}\times 91 = 655.2 nm
  4. λP−λB=1216.8≈1217\lambda_P-\lambda_B = 1216.8 \approx 1217 nm
  5. So the answer is (c) 1217.
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Q296MHT CET 2025 · 20 APR · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In hydrogen spectrum, the ratio of wavelengths of the last line of Lyman series and that of the last line of Balmer series is

Show answer and solution

Answer: (c)

  1. Last line = series limit: λ=n12R\lambda = \frac{n_1^2}{R}.
  2. Lyman: 1R\frac{1}{R}; Balmer: 4R\frac{4}{R}.
  3. Ratio =14=0.25= \frac{1}{4} = 0.25.
  4. So the answer is (c) 0.25.
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Q297MHT CET 2025 · 19 APR · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of the wavelength of the last line of Paschen series to that of Balmer series is

Show answer and solution

Answer: (a)

  1. Last line = series limit: λ=n12R\lambda = \frac{n_1^2}{R}.
  2. Paschen: 9R\frac{9}{R}; Balmer: 4R\frac{4}{R}.
  3. Ratio =94= \frac{9}{4}.
  4. So the answer is (a) 94\frac{9}{4}.
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Q298JEE Main 2025 · 7 APR · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

Show answer and solution

Answer: (a)

  1. Largest wavelength = first line.
  2. Lyman (2→12 \to 1): 1λL=R(1−14)=3R4\frac{1}{\lambda_L} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}, so λL=43R\lambda_L = \frac{4}{3R}.
  3. Balmer (3→23 \to 2): 1λB=R(14−19)=5R36\frac{1}{\lambda_B} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}, so λB=365R\lambda_B = \frac{36}{5R}.
  4. λLλB=43×536=527\frac{\lambda_L}{\lambda_B} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}.
  5. So the answer is (a) 5:275 : 27.
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Q299MHT CET 2025 · 26 APR · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The frequencies for series limit of Balmer and Paschen series are ' ν1\nu_1 ' and ' ν3\nu_3 ' respectively. If frequency of first line of Balmer series ' ν2\nu_2 ' then the relation between ν1,ν2\nu_1, \nu_2 and ν3\nu_3 is

Show answer and solution

Answer: (c)

  1. ν1\nu_1 (Balmer limit) =Rc4= \frac{Rc}{4}; ν3\nu_3 (Paschen limit) =Rc9= \frac{Rc}{9}.
  2. ν2\nu_2 (Balmer first line, 3→23 \to 2) =Rc(14−19)= Rc\left(\frac{1}{4} - \frac{1}{9}\right).
  3. ν1−ν2=Rc9=ν3\nu_1 - \nu_2 = \frac{Rc}{9} = \nu_3 (energy of level 3 to infinity).
  4. So the answer is (c) ν1−ν2=ν3\nu_1 - \nu_2 = \nu_3.
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Q300MHT CET 2025 · 26 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In Paschen series, wavelength of first line is ' λ1\lambda_1 ' and for Brackett series, wavelength of first line is ' λ2\lambda_2 ' then ratio λ1λ2\frac{\lambda_1}{\lambda_2} is

Show answer and solution

Answer: (c)

  1. Paschen first line (4→34 \to 3): 1λ1=R(19−116)=7R144\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}.
  2. Brackett first line (5→45 \to 4): 1λ2=R(116−125)=9R400\frac{1}{\lambda_2} = R\left(\frac{1}{16} - \frac{1}{25}\right) = \frac{9R}{400}.
  3. λ1λ2=1447×9400=12962800=81175\frac{\lambda_1}{\lambda_2} = \frac{144}{7} \times \frac{9}{400} = \frac{1296}{2800} = \frac{81}{175}.
  4. So the answer is (c) 81175\frac{81}{175}.
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Q301MHT CET 2025 · 19 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If ' λ1\lambda_1 ' and ' λ2\lambda_2 ' are the wavelengths of the first member of the Balmer and Paschen series, in hydrogen atom respectively, then the ratio of respective frequencies, f1/f2f_1/f_2, is

Show answer and solution

Answer: (a)

  1. f∝1λ=R(1n12−1n22)f \propto \frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right).
  2. Balmer first line: f1∝14−19=536f_1 \propto \frac{1}{4} - \frac{1}{9} = \frac{5}{36}.
  3. Paschen first line: f2∝19−116=7144f_2 \propto \frac{1}{9} - \frac{1}{16} = \frac{7}{144}.
  4. f1f2=536×1447=207\frac{f_1}{f_2} = \frac{5}{36} \times \frac{144}{7} = \frac{20}{7}.
  5. So the answer is (a) 20:720 : 7.
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Q302MHT CET 2024 · 9 MAY · SHIFT 2Easy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

The spectral series observed for hydrogen atom found in visible region is

Show answer and solution

Answer: (b)

  1. Balmer series ends on n=2n = 2; wavelengths 365365 to 656656 nm, which is the visible region.
  2. Lyman is ultraviolet; Paschen and Brackett are infrared.
  3. So the answer is (b) Balmer.
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Q303NEET 2024Easy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

The spectral series which corresponds to the electronic transition from the levels n2=5,6,…n_2 = 5, 6, \ldots to the level n1=4n_1 = 4 is

Show answer and solution

Answer: (b)

  1. Series by lower level: n1=1n_1 = 1 Lyman, 22 Balmer, 33 Paschen, 44 Brackett, 55 Pfund.
  2. So the answer is (b) Brackett series.
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Q304MHT CET 2024 · 16 MAY · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

Frequency of the series limit of Balmer series of hydrogen atom of Rydberg's constant ' RR ' and velocity of light ' CC ' is

Show answer and solution

Answer: (a)

  1. Series limit of Balmer: 1λ=R(14−0)=R4\frac{1}{\lambda} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4}.
  2. ν=Cλ=RC4\nu = \frac{C}{\lambda} = \frac{RC}{4}.
  3. So the answer is (a) RC4\frac{RC}{4}.
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Q305MHT CET 2024 · 9 MAY · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is

Show answer and solution

Answer: (c)

  1. Shortest wavelength = series limit: λ=n12R\lambda = \frac{n_1^2}{R}.
  2. Balmer: 4R\frac{4}{R}; Paschen: 9R\frac{9}{R}.
  3. Ratio =4:9= 4 : 9.
  4. So the answer is (c) 4:94 : 9.
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Q306MHT CET 2024 · 4 MAY · SHIFT 2Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of minimum wavelengths of Lyman and Balmer series will be

Show answer and solution

Answer: (b)

  1. Minimum wavelength = series limit: λmin=n12R\lambda_{min} = \frac{n_1^2}{R}.
  2. Lyman: 1R\frac{1}{R}; Balmer: 4R\frac{4}{R}.
  3. Ratio =0.25= 0.25.
  4. So the answer is (b) 0.25.
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Q307JEE Main 2024 · 6 APR · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is:

Show answer and solution

Answer: (d)

  1. Shortest wavelength = series limit: λ=n12R\lambda = \frac{n_1^2}{R}.
  2. Balmer: 4R\frac{4}{R}; Lyman: 1R\frac{1}{R}.
  3. Ratio =4:1= 4 : 1.
  4. So the answer is (d) 4:14 : 1.
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Q308MHT CET 2024 · 9 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

According to Bohr's theory of hydrogen atom, the ratio of the maximum and minimum wavelength of Lyman series will be

Show answer and solution

Answer: (b)

  1. Lyman maximum λ\lambda (2→12 \to 1): λmax=43R\lambda_{max} = \frac{4}{3R}.
  2. Lyman minimum λ\lambda (limit): λmin=1R\lambda_{min} = \frac{1}{R}.
  3. λmaxλmin=43\frac{\lambda_{max}}{\lambda_{min}} = \frac{4}{3}.
  4. So the answer is (b) 4:34 : 3.
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Q309MHT CET 2024 · 3 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If ' λ1\lambda_1 ' and ' λ2\lambda_2 ' are the wavelengths of the first line of the Lyman and Paschen series respectively, then λ2:λ1\lambda_2 : \lambda_1 is

Show answer and solution

Answer: (d)

  1. Lyman first line: λ1=43R\lambda_1 = \frac{4}{3R}.
  2. Paschen first line: 1λ2=R(19−116)=7R144\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}, so λ2=1447R\lambda_2 = \frac{144}{7R}.
  3. λ2λ1=1447×34=1087\frac{\lambda_2}{\lambda_1} = \frac{144}{7} \times \frac{3}{4} = \frac{108}{7}.
  4. So the answer is (d) 108:7108 : 7.
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Q310JEE Main 2024 · 6 APR · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The longest wavelength associated with Paschen series is: (Given RH=1.097×107 SIR_H = 1.097 \times 10^{7}\text{ SI} unit)

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Answer: (b)

  1. Paschen longest (4→34 \to 3): 1λ=R(19−116)=7R144\frac{1}{\lambda} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  2. λ=1447×1.097×107\lambda = \frac{144}{7\times 1.097\times 10^{7}}
  3. λ≈1.876×10−6\lambda \approx 1.876\times 10^{-6} m
  4. So the answer is (b) 1.876×10−61.876 \times 10^{-6} m.
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Q311JEE Main 2024 · 31 JAN · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If the wavelength of the first member of Lyman series of hydrogen is λ\lambda. The wavelength of the second member will be

Show answer and solution

Answer: (c)

  1. First member (2→12 \to 1): 1λ=3R4\frac{1}{\lambda} = \frac{3R}{4}
  2. Second member (3→13 \to 1): 1λ′=R(1−19)=8R9\frac{1}{\lambda'} = R\left(1-\frac{1}{9}\right) = \frac{8R}{9}
  3. λ′λ=3/48/9=2732\frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{27}{32}
  4. So the answer is (c) 2732λ\frac{27}{32}\lambda.
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Q312JEE Main 2024 · 5 APR · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 A˚915\text{ Å}. The longest wavelength of spectral lines in the Balmer series will be _______ A˚\text{Å}.

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

Show answer and solution

Answer: 6588

  1. Lyman shortest: 1R=915\frac{1}{R} = 915 Å
  2. Balmer longest (3→23 \to 2): 1λ=5R36⇒λ=365R\frac{1}{\lambda} = \frac{5R}{36} \Rightarrow \lambda = \frac{36}{5R}
  3. λ=365×915=6588\lambda = \frac{36}{5}\times 915 = 6588 Å
  4. Answer: 6588 Å
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Q313JEE Main 2024 · 27 JAN · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If Rydberg's constant is RR, the longest wavelength of radiation in Paschen series will be α7R\frac{\alpha}{7R}, where α=\alpha = ________.

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

Show answer and solution

Answer: 144

  1. Paschen longest (4→34 \to 3): 1λ=R(19−116)=7R144\frac{1}{\lambda} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  2. λ=1447R\lambda = \frac{144}{7R}
  3. Answer: α=144\alpha = 144
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Q314NEET 2024Difficult

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

Match List I with List II:

Choose the correct answer from the options given below:

List I (Spectral Lines of Hydrogen for transitions from)List II (Wavelengths (nm))
A.n2=3n_2 = 3 to n1=2n_1 = 2I.410.2
B.n2=4n_2 = 4 to n1=2n_1 = 2II.434.1
C.n2=5n_2 = 5 to n1=2n_1 = 2III.656.3
D.n2=6n_2 = 6 to n1=2n_1 = 2IV.486.1
Show answer and solution

Answer: (b)

  1. Larger energy gap means shorter wavelength, so λ\lambda decreases as n2n_2 increases for the Balmer lines.
  2. 3→23 \to 2 (HαH_\alpha): 656.3 nm (III); 4→24 \to 2 (HβH_\beta): 486.1 nm (IV)
  3. 5→25 \to 2 (HγH_\gamma): 434.1 nm (II); 6→26 \to 2 (HδH_\delta): 410.2 nm (I)
  4. So the answer is (b) A-III, B-IV, C-II, D-I.
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Q315NEET 2023Easy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

The wavelength of Lyman series of hydrogen atom appears in:

Show answer and solution

Answer: (c)

  1. Lyman lines (n→1n \to 1) have λ\lambda from 9191 nm to 122122 nm.
  2. These wavelengths are in the ultraviolet.
  3. So the answer is (c) ultraviolet region.
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Q316MHT CET 2023 · 14 MAY · SHIFT 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

Frequency of the series limit of Balmer series of hydrogen atom in terms of Rydberg's constant (R) and velocity of light (c) is

Show answer and solution

Answer: (d)

  1. Balmer series limit: 1λ=R4\frac{1}{\lambda} = \frac{R}{4}.
  2. ν=cλ=Rc4\nu = \frac{c}{\lambda} = \frac{Rc}{4}.
  3. So the answer is (d) Rc4\frac{Rc}{4}.
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Q317MHT CET 2023 · 14 MAY · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum is

Show answer and solution

Answer: (c)

  1. Longest wavelength = first line of each series.
  2. Lyman: λL=43R\lambda_L = \frac{4}{3R}; Balmer: 1λB=R(14−19)=5R36\frac{1}{\lambda_B} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}, so λB=365R\lambda_B = \frac{36}{5R}.
  3. λLλB=43×536=527\frac{\lambda_L}{\lambda_B} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}.
  4. So the answer is (c) 527\frac{5}{27}.
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Q318MHT CET 2023 · 12 MAY · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In Balmer series, wavelength of the 22nd line is 'λ1\lambda_1' and for Paschen series, wavelength of the 11st line is 'λ2\lambda_2', then the ratio 'λ1\lambda_1 ' to 'λ2\lambda_2' is

Show answer and solution

Answer: (c)

  1. Balmer 2nd line (4→24 \to 2): 1λ1=R(14−116)=3R16\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}, so λ1=163R\lambda_1 = \frac{16}{3R}.
  2. Paschen 1st line (4→34 \to 3): 1λ2=7R144\frac{1}{\lambda_2} = \frac{7R}{144}, so λ2=1447R\lambda_2 = \frac{144}{7R}.
  3. λ1λ2=163×7144=727\frac{\lambda_1}{\lambda_2} = \frac{16}{3} \times \frac{7}{144} = \frac{7}{27}.
  4. So the answer is (c) 7:277 : 27.
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Q319MHT CET 2023 · 12 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In Lyman series, series limit of wavelength is λ1\lambda_1. The wavelength of first line of Lyman series is λ2\lambda_2 and in Balmer series, the series limit of wavelength is λ3\lambda_3. Then the relation between λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 is

Show answer and solution

Answer: (d)

  1. 1λ=R(1n12−1n22)\frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
  2. Lyman limit (∞→1\infty \to 1): 1λ1=R\frac{1}{\lambda_1} = R; Lyman first line (2→12 \to 1): 1λ2=3R4\frac{1}{\lambda_2} = \frac{3R}{4}
  3. Balmer limit (∞→2\infty \to 2): 1λ3=R4\frac{1}{\lambda_3} = \frac{R}{4}
  4. 1λ1−1λ2=R−3R4=R4=1λ3\frac{1}{\lambda_1}-\frac{1}{\lambda_2} = R-\frac{3R}{4} = \frac{R}{4} = \frac{1}{\lambda_3}
  5. So the answer is (d) 1λ1−1λ2=1λ3\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}.
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Q320MHT CET 2023 · 10 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number zz. The value of zz is

Show answer and solution

Answer: (a)

  1. For a hydrogen-like atom 1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right); shortest wavelength means n2=∞n_2 = \infty.
  2. Balmer (H, Z=1Z=1): 1λ=R4\frac{1}{\lambda} = \frac{R}{4}
  3. Brackett (n1=4n_1=4): 1λ=RZ216\frac{1}{\lambda} = \frac{RZ^2}{16}
  4. Z216=14⇒Z2=4⇒Z=2\frac{Z^2}{16} = \frac{1}{4} \Rightarrow Z^2 = 4 \Rightarrow Z = 2
  5. So the answer is (a) 2.
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Q321MHT CET 2023 · 10 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of longest to shortest wavelength emitted in Paschen series of hydrogen atom is

Show answer and solution

Answer: (a)

  1. Paschen series: n1=3n_1 = 3.
  2. Longest (4→34 \to 3): 1λmax=R(19−116)=7R144\frac{1}{\lambda_{max}} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  3. Shortest (∞→3\infty \to 3): 1λmin=R9\frac{1}{\lambda_{min}} = \frac{R}{9}
  4. λmaxλmin=144/79=14463\frac{\lambda_{max}}{\lambda_{min}} = \frac{144/7}{9} = \frac{144}{63}
  5. So the answer is (a) 14463\frac{144}{63}.
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Q322NEET 2023Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ\lambda. The shortest wavelength in the Brackett series is:

Show answer and solution

Answer: (a)

  1. Shortest wavelength of a series: λ=n12R\lambda = \frac{n_1^2}{R}
  2. Balmer (n1=2n_1=2): λ=4R\lambda = \frac{4}{R}; Brackett (n1=4n_1=4): λ′=16R\lambda' = \frac{16}{R}
  3. λ′=4λ\lambda' = 4\lambda
  4. So the answer is (a) 4λ4\lambda.
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Q323JEE Main 2023 · 10 APR · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If 917 A˚\text{Å} be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ___________ A˚\text{Å}.

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

Show answer and solution

Answer: 3668

  1. Lowest wavelength = series limit: λ=n12R\lambda = \frac{n_1^2}{R}
  2. Lyman: 1R=917\frac{1}{R} = 917 Å
  3. Balmer: λ=4R=4×917=3668\lambda = \frac{4}{R} = 4\times 917 = 3668 Å
  4. Answer: 3668 Å
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Q324JEE Main 2023 · 8 APR · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of wavelength of spectral lines Hα\text{H}_\alpha and Hβ\text{H}_\beta in the Balmer series is x20\frac{x}{20}. The value of xx is _________.

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

Show answer and solution

Answer: 27

  1. HαH_\alpha (3→23 \to 2): 1λα=5R36\frac{1}{\lambda_\alpha} = \frac{5R}{36}; HβH_\beta (4→24 \to 2): 1λβ=3R16\frac{1}{\lambda_\beta} = \frac{3R}{16}
  2. λαλβ=3/165/36=10880=2720\frac{\lambda_\alpha}{\lambda_\beta} = \frac{3/16}{5/36} = \frac{108}{80} = \frac{27}{20}
  3. Answer: x=27x = 27
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Q325MHT CET 2022 · 11 AUG · SHIFT 2Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The wave number of the last line of the Balmer series in hydrogen spectrum will be (Rydberg's constant =107 m−1= 10^{7}\text{ m}^{-1})

Show answer and solution

Answer: (b)

  1. Last line of Balmer is the series limit: νˉ=R(14−0)=R4\bar{\nu} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4}.
  2. νˉ=1074=2.5×106 m−1\bar{\nu} = \frac{10^{7}}{4} = 2.5 \times 10^{6}\text{ m}^{-1}.
  3. So the answer is (b) 2.5×106 m−12.5 \times 10^{6}\text{ m}^{-1}.
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Q326JEE Main 2022 · 26 JUL · SHIFT 1Difficult

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In the hydrogen spectrum, λ\lambda be the wavelength of first transition line of Lyman series. The wavelength difference will be "aλ\lambda'' between the wavelength of 33rd transition line of Paschen series and that of 22nd transition line of Balmer series where a=a = ___________.

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

Show answer and solution

Answer: 5

  1. Lyman first line (2→12 \to 1): 1λ=3R4⇒λ=43R\frac{1}{\lambda} = \frac{3R}{4} \Rightarrow \lambda = \frac{4}{3R}
  2. Paschen 3rd line (6→36 \to 3): 1λP=R(19−136)=R12⇒λP=12R\frac{1}{\lambda_P} = R\left(\frac{1}{9}-\frac{1}{36}\right) = \frac{R}{12} \Rightarrow \lambda_P = \frac{12}{R}
  3. Balmer 2nd line (4→24 \to 2): λB=163R\lambda_B = \frac{16}{3R}
  4. λP−λB=203R=5×43R=5λ\lambda_P-\lambda_B = \frac{20}{3R} = 5\times\frac{4}{3R} = 5\lambda
  5. Answer: a=5a = 5
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Q327JEE Main 2021 · 17 MAR · SHIFT 2Easy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?

Show answer and solution

Answer: (b)

  1. Balmer series (n→2n \to 2) has wavelengths 365365 to 656656 nm, the visible region.
  2. So the answer is (b) Balmer series.
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Q328MHT CET 2021 · 24 SEP · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of maximum to minimum wavelength in Balmer series of hydrogen atom is

Show answer and solution

Answer: (c)

  1. Balmer series: n1=2n_1 = 2.
  2. Maximum (3→23 \to 2): 1λmax=R(14−19)=5R36\frac{1}{\lambda_{max}} = R\left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5R}{36}
  3. Minimum (∞→2\infty \to 2): 1λmin=R4\frac{1}{\lambda_{min}} = \frac{R}{4}
  4. λmaxλmin=36/54=95\frac{\lambda_{max}}{\lambda_{min}} = \frac{36/5}{4} = \frac{9}{5}
  5. So the answer is (c) 9 : 5.
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Q329MHT CET 2021 · 23 SEP · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The shortest wavelength for Lyman series is 912 A˚\text{Å}. The longest wavelength in Paschen series is

Show answer and solution

Answer: (c)

  1. Lyman shortest: λ=1R=912 A˚\lambda = \frac{1}{R} = 912\ \text{Å}
  2. Paschen longest (4→34 \to 3): 1λ=R(19−116)=7R144\frac{1}{\lambda} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  3. λ=1447×912≈18760 A˚\lambda = \frac{144}{7}\times 912 \approx 18760\ \text{Å}
  4. So the answer is (c) 18760 Å.
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Q330MHT CET 2021 · 21 SEP · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

'λ1\lambda_1' is the wavelength of series limit of Lyman series, 'λ2\lambda_2 ' is the wavelength of the first line line of Lyman series and 'λ3\lambda_3' is the series limit of the Balmer series. Then the relation between λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 is

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Answer: (a)

  1. Lyman limit: 1λ1=R\frac{1}{\lambda_1} = R; Lyman first line (2→12 \to 1): 1λ2=3R4\frac{1}{\lambda_2} = \frac{3R}{4}
  2. Balmer limit: 1λ3=R4\frac{1}{\lambda_3} = \frac{R}{4}
  3. 1λ1−1λ2=R4=1λ3\frac{1}{\lambda_1}-\frac{1}{\lambda_2} = \frac{R}{4} = \frac{1}{\lambda_3}
  4. So the answer is (a) 1λ1−1λ2=1λ3\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}.
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Q331JEE Main 2021 · 26 FEB · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If λ\lambda1 and λ\lambda2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of λ\lambda1: λ\lambda2 is:

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Answer: (a)

  1. Lyman third member (4→14 \to 1): 1λ1=R(1−116)=15R16\frac{1}{\lambda_1} = R\left(1-\frac{1}{16}\right) = \frac{15R}{16}
  2. Paschen first member (4→34 \to 3): 1λ2=R(19−116)=7R144\frac{1}{\lambda_2} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  3. λ1λ2=7/14415/16=1122160=7135\frac{\lambda_1}{\lambda_2} = \frac{7/144}{15/16} = \frac{112}{2160} = \frac{7}{135}
  4. So the answer is (a) 7 : 135.
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Q332JEE Main 2021 · 16 MAR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The first three spectral lines of H-atom in the Balmer series are

given λ\lambda1, λ\lambda2, λ\lambda3 considering the Bohr atomic model, the wave lengths of first and third spectral lines (λ1λ3)\left(\frac{\lambda_1}{\lambda_3}\right) are related by a factor of approximately 'x' ×\times 10−-1.

The value of x, to the nearest integer, is _________.

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

Show answer and solution

Answer: 15

  1. First Balmer line (3→23 \to 2): 1λ1=R(14−19)=5R36\frac{1}{\lambda_1} = R\left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5R}{36}
  2. Third Balmer line (5→25 \to 2): 1λ3=R(14−125)=21R100\frac{1}{\lambda_3} = R\left(\frac{1}{4}-\frac{1}{25}\right) = \frac{21R}{100}
  3. λ1λ3=21/1005/36=756500=1.512≈15×10−1\frac{\lambda_1}{\lambda_3} = \frac{21/100}{5/36} = \frac{756}{500} = 1.512 \approx 15\times 10^{-1}
  4. Answer: x=15x = 15
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Q333JEE Main 2020 · 4 SEP · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths {} of the Lyman series is 304 A˚\text{Å}. The corresponding difference for the Paschan series in A˚\text{Å} is: ___________.

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

Show answer and solution

Answer: 10553

  1. Lyman: λmax=43R\lambda_{max} = \frac{4}{3R}, λmin=1R\lambda_{min} = \frac{1}{R}, difference =13R=304= \frac{1}{3R} = 304 Å, so 1R=912\frac{1}{R} = 912 Å
  2. Paschen: λmax=1447R\lambda_{max} = \frac{144}{7R}, λmin=9R\lambda_{min} = \frac{9}{R}
  3. Difference =(1447−9)1R=817×912= \left(\frac{144}{7}-9\right)\frac{1}{R} = \frac{81}{7}\times 912
  4. Answer: ≈10553\approx 10553 Å
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Q334JEE Main 2020 · 8 JAN · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is:

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

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Answer: 486

  1. 1λ1=5R36\frac{1}{\lambda_1} = \frac{5R}{36} (3→23 \to 2), 1λ2=3R16\frac{1}{\lambda_2} = \frac{3R}{16} (4→24 \to 2)
  2. λ2=2027λ1=2027×6561=4860\lambda_2 = \frac{20}{27}\lambda_1 = \frac{20}{27}\times 6561 = 4860 Å
  3. Answer: 486 nm
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Q335MHT CET 2019 · 2 MAY · SHIFT 2Easy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the

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Answer: (a)

  1. Lower level n=2n = 2 means the Balmer series.
  2. 3→23 \to 2 is the first line, 4→24 \to 2 is the second line.
  3. So the answer is (a) second line of Balmer series.
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Q336MHT CET 2019 · 3 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In hydrogen emission spectrum, for any series, the principal quantum number is nn. Corresponding maximum wavelength λ\lambda is ( R=R = Rydberg's constant)

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Answer: (b)

  1. Maximum wavelength of a series ending on level nn comes from (n+1)→n(n+1) \to n.
  2. 1λ=R(1n2−1(n+1)2)=R(2n+1)n2(n+1)2\frac{1}{\lambda} = R\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) = \frac{R(2n+1)}{n^2(n+1)^2}
  3. λ=n2(n+1)2R(2n+1)\lambda = \frac{n^2(n+1)^2}{R(2n+1)}
  4. So the answer is (b).
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Q337MHT CET 2019 · 2 MAY · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In Balmer series, wavelength of first line is ' λ1\lambda_1 ' and in Brackett series wavelength of first line is ' λ2\lambda_2 ' then λ1λ2\frac{\lambda_1}{\lambda_2} is

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Answer: (a)

  1. Balmer first line (3→23 \to 2): 1λ1=R(14−19)=5R36\frac{1}{\lambda_1} = R\left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5R}{36}
  2. Brackett first line (5→45 \to 4): 1λ2=R(116−125)=9R400\frac{1}{\lambda_2} = R\left(\frac{1}{16}-\frac{1}{25}\right) = \frac{9R}{400}
  3. λ1λ2=36/5400/9=3242000=0.162\frac{\lambda_1}{\lambda_2} = \frac{36/5}{400/9} = \frac{324}{2000} = 0.162
  4. So the answer is (a) 0.162.
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Q338MHT CET 2019 · 2 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The wavelength of the first line in Balmer series in the hydrogen spectrum is ' λ\lambda '. What is the wavelength of the second line in the same series?

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Answer: (a)

  1. First line (3→23 \to 2): 1λ=5R36\frac{1}{\lambda} = \frac{5R}{36}; second line (4→24 \to 2): 1λ′=R(14−116)=3R16\frac{1}{\lambda'} = R\left(\frac{1}{4}-\frac{1}{16}\right) = \frac{3R}{16}
  2. λ′λ=5/363/16=80108=2027\frac{\lambda'}{\lambda} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}
  3. So the answer is (a) 2027λ\frac{20}{27}\lambda.
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Q339JEE Main 2019 · 9 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:

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Answer: (b)

  1. 1λ3→2=5R36\frac{1}{\lambda_{3\to2}} = \frac{5R}{36}, 1λ4→2=3R16\frac{1}{\lambda_{4\to2}} = \frac{3R}{16}
  2. λ4→2=5/363/16 λ3→2=2027×660\lambda_{4\to2} = \frac{5/36}{3/16}\,\lambda_{3\to2} = \frac{20}{27}\times 660
  3. λ4→2≈488.9\lambda_{4\to2} \approx 488.9 nm
  4. So the answer is (b) 488.9 nm.
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Q340JEE Main 2018Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

If the series limit frequency of the Lyman series is νL\nu_L, then the series limit frequency of the Pfund series is:

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Answer: (a)

  1. Series limit frequency ν=Rcn12\nu = \frac{Rc}{n_1^2}.
  2. Lyman (n1=1n_1 = 1): νL=Rc\nu_L = Rc; Pfund (n1=5n_1 = 5): ν=Rc25\nu = \frac{Rc}{25}.
  3. So the answer is (a) νL/25\nu_L/25.
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Q341NEET 2017Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

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Answer: (b)

  1. Last line = series limit: λ=n12R\lambda = \frac{n_1^2}{R}.
  2. Balmer: 4R\frac{4}{R}; Lyman: 1R\frac{1}{R}.
  3. Ratio =4= 4.
  4. So the answer is (b) 4.
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Q342NEET 2016 · PHASE 1Easy

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

Given the value of Rydberg constant is 107 m−1{}^{-1}, the wave number of the last line of the Balmer series in hydrogen spectrum will be

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Answer: (a)

  1. Rydberg constant R=107 m−1R = 10^{7}\text{ m}^{-1}.
  2. Last line of Balmer: νˉ=R4=0.25×107 m−1\bar{\nu} = \frac{R}{4} = 0.25 \times 10^{7}\text{ m}^{-1}.
  3. So the answer is (a) 0.25×107 m−10.25 \times 10^{7}\text{ m}^{-1}.
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Q343NEET 2015 · AIPMTMedium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Show answer and solution

Answer: (b)

  1. Lyman longest (2→12 \to 1): 1λL=3R4\frac{1}{\lambda_L} = \frac{3R}{4}
  2. Balmer longest (3→23 \to 2): 1λB=5R36\frac{1}{\lambda_B} = \frac{5R}{36}
  3. λLλB=5/363/4=527\frac{\lambda_L}{\lambda_B} = \frac{5/36}{3/4} = \frac{5}{27}
  4. So the answer is (b) 527\frac{5}{27}.
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Q344NEET 2013Medium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

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Answer: (c)

  1. Lyman longest (2→12 \to 1): 1λL=3R4\frac{1}{\lambda_L} = \frac{3R}{4}
  2. Balmer longest (3→23 \to 2): 1λB=5R36\frac{1}{\lambda_B} = \frac{5R}{36}
  3. λLλB=5/363/4=527\frac{\lambda_L}{\lambda_B} = \frac{5/36}{3/4} = \frac{5}{27}
  4. So the answer is (c) 527\frac{5}{27}.
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Q345NEET 2011 · AIPMT · PRELIMSMedium

Hydrogen Spectrum & Spectral Series · Series limit, first line & wavelength ratios

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is

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Answer: (d)

  1. 1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
  2. H, Lyman first line (2→12 \to 1): 1λ=3R4\frac{1}{\lambda} = \frac{3R}{4}
  3. Ion, Balmer second line (4→24 \to 2): 1λ=RZ2⋅316\frac{1}{\lambda} = RZ^2\cdot\frac{3}{16}
  4. 3Z216=34⇒Z2=4⇒Z=2\frac{3Z^2}{16} = \frac{3}{4} \Rightarrow Z^2 = 4 \Rightarrow Z = 2
  5. So the answer is (d) 2.
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Q346NEET 2000 · AIPMTEasy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

When an electron does transition from n = 4 to n = 2, then emitted line spectrum will be

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Answer: (b)

  1. Ending on n=2n = 2 gives the Balmer series.
  2. 3→23 \to 2 is the first line, so 4→24 \to 2 is the second line.
  3. So the answer is (b) second line of Balmer series.
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Q347NEET 1993 · AIPMTEasy

Hydrogen Spectrum & Spectral Series · Spectral series & their regions

Which source is associated with a line emission spectrum?

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Answer: (b)

  1. A line spectrum comes from excited free atoms in a low-pressure gas.
  2. A neon sign is a gas discharge tube, so it gives bright lines.
  3. Electric fire, filament lamps and the Sun are hot dense sources giving continuous spectra.
  4. So the answer is (b) Neon street sign.
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Q348JEE Main 2025 · 24 JAN · SHIFT 2Easy

Quantum Numbers, X-rays & Other Questions · Other questions

The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation

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Answer: (c)

  1. Relativistic energy-momentum relation: E2=p2c2+m2c4E^2 = p^2c^2 + m^2c^4 (with mm the rest mass).
  2. Check: for p=0p = 0, E=mc2E = mc^2; each term has units of energy squared, which rules out the other options.
  3. So the answer is (c).
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Q349JEE Main 2023 · 8 APR · SHIFT 2Easy

Quantum Numbers, X-rays & Other Questions · X-rays

The waves emitted when a metal target is bombarded with high energy electrons are

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Answer: (d)

  1. High-energy electrons hitting a metal target decelerate and knock out inner-shell electrons.
  2. Both processes emit high-energy photons of very short wavelength, i.e. X-rays (the working of a Coolidge tube).
  3. So the answer is (d) X-rays.
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Q350NEET 2022 · PHASE 1EasyOld syllabus

Quantum Numbers, X-rays & Other Questions · Other questions

The graph which shows the variation of the de Broglie wavelength (λ\lambda) of a particle and its associated momentum (p) is

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Answer: (d)

  1. de Broglie relation: λ=hp\lambda = \frac{h}{p}, so λp=h\lambda p = h = constant.
  2. λ∝1p\lambda \propto \frac{1}{p} gives a rectangular hyperbola, falling steeply and approaching the axes.
  3. So the answer is (d).
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Q351NEET 2022 · PHASE 2Medium

Quantum Numbers, X-rays & Other Questions · Other questions

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is 2.4 ×\times 1039. The ratio of the proportionality constant, K=14πε0K = \frac{1}{4\pi\varepsilon_0} to the gravitational constant G is nearly (Given that the charge of the proton and electron each = 1.6 ×\times 10−-19 C, the mass of the electron = 9.11 ×\times 10−-31 kg, the mass of the proton = 1.67 ×\times 10−-27 kg):

Show answer and solution

Answer: (b)

  1. FeFg=Ke2Gmemp⇒KG=FeFg⋅mempe2\frac{F_e}{F_g} = \frac{Ke^2}{Gm_em_p} \Rightarrow \frac{K}{G} = \frac{F_e}{F_g}\cdot\frac{m_em_p}{e^2}
  2. memp=9.11×10−31×1.67×10−27≈1.52×10−57m_em_p = 9.11\times10^{-31}\times1.67\times10^{-27} \approx 1.52\times10^{-57}; e2=2.56×10−38e^2 = 2.56\times10^{-38}
  3. KG=2.4×1039×1.52×10−572.56×10−38≈1.4×1020\frac{K}{G} = 2.4\times10^{39}\times\frac{1.52\times10^{-57}}{2.56\times10^{-38}} \approx 1.4\times10^{20}
  4. So the answer is (b) 102010^{20}.
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Q352JEE Main 2021 · 27 JUL · SHIFT 2Medium

Quantum Numbers, X-rays & Other Questions · X-rays

The Kα\alpha X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be __________ keV. (Round off to the nearest integer)

[h = 4.14 ×\times 10−-15 eVs, c = 3 ×\times 108 ms−1{}^{-1}]

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

Show answer and solution

Answer: 10

  1. KαK_\alpha photon energy E=hcλ=4.14×10−15×3×1080.071×10−9E = \frac{hc}{\lambda} = \frac{4.14\times10^{-15}\times 3\times10^{8}}{0.071\times10^{-9}}
  2. E≈1.75×104E \approx 1.75\times 10^{4} eV =17.5= 17.5 keV
  3. KαK_\alpha is emitted when an L electron fills the K hole, so EL=EK−EKα=27.5−17.5E_L = E_K - E_{K_\alpha} = 27.5 - 17.5
  4. Answer: 10 keV
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Q353MHT CET 2021 · 20 SEP · SHIFT 2Medium

Quantum Numbers, X-rays & Other Questions · Other questions

The P.E. 'U' of a moving particle of mass 'm' varies with 'x'-axis as shown in figure. The deBroglie wavelength or the particle in the regions 0≤x≤10 \le x \le 1 and x>1x > 1 are λ1\lambda_1 and λ2\lambda_2 respectively. II the total energy of the particle is 'nEnE', then the ratio λ1/λ2\lambda_1/\lambda_2 is

Figure

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Answer: (c)

  1. Total energy nEnE; kinetic energy K=nE−UK = nE - U.
  2. 0≤x≤10 \le x \le 1: U=E⇒K1=(n−1)EU = E \Rightarrow K_1 = (n-1)E; x>1x > 1: U=0⇒K2=nEU = 0 \Rightarrow K_2 = nE
  3. λ=h2mK⇒λ1λ2=K2K1=nn−1\lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{n}{n-1}}
  4. So the answer is (c) nn−1\sqrt{\frac{n}{n-1}}.
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Q354JEE Main 2020 · 3 SEP · SHIFT 2Medium

Quantum Numbers, X-rays & Other Questions · Other questions

Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to:

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Answer: (a)

  1. qV=12mv2⇒v=2qVmqV = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2qV}{m}}
  2. Both ions have charge ee; mHe≈4mHm_{He} \approx 4m_H
  3. vHvHe=mHemH=4=2\frac{v_H}{v_{He}} = \sqrt{\frac{m_{He}}{m_H}} = \sqrt{4} = 2
  4. So the answer is (a) 2 : 1.
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Q355NEET 2007 · AIPMTMedium

Quantum Numbers, X-rays & Other Questions · Other questions

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential VV and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio (charge on the ionmass of the ion)\left(\frac{\text{charge on the ion}}{\text{mass of the ion}}\right) will be proportional to

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Answer: (a)

  1. Acceleration: qV=12mv2⇒v=2qVmqV = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2qV}{m}}
  2. Circular path: R=mvqB=1B2mVqR = \frac{mv}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}
  3. qm=2VB2R2∝1R2\frac{q}{m} = \frac{2V}{B^2R^2} \propto \frac{1}{R^2}
  4. So the answer is (a) 1/R21/R^2.
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Q356NEET 2001 · AIPMTMedium

Quantum Numbers, X-rays & Other Questions · X-rays

The interplanar distance in a crystal is 2.8 ×\times 10−-8 m. The value of maximum wavelength which can be diffracted

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Answer: (b)

  1. Bragg's law: 2dsin⁡θ=nλ2d\sin\theta = n\lambda
  2. Maximum λ\lambda when sin⁡θ=1\sin\theta = 1, n=1n = 1: λmax=2d\lambda_{max} = 2d
  3. λmax=2×2.8×10−8=5.6×10−8\lambda_{max} = 2\times 2.8\times 10^{-8} = 5.6\times 10^{-8} m
  4. So the answer is (b) 5.6×10−85.6 \times 10^{-8} m.
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Q357NEET 1995 · AIPMTEasyOld syllabus

Quantum Numbers, X-rays & Other Questions · X-rays

The figure represents the observed intensity of X-rays emitted by an X-ray tube, as a function of wavelength. The sharp peaks AA and BB denote

Figure

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Answer: (b)

  1. An X-ray tube gives a smooth continuous (bremsstrahlung) background.
  2. Sharp peaks on top of it come from electron transitions between inner shells of the target atoms (KαK_\alpha, KβK_\beta), which depend on the target element.
  3. So the answer is (b) characteristic radiations.
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Q358NEET 1992 · AIPMTEasy

Quantum Numbers, X-rays & Other Questions · Quantum numbers & electronic configuration

Of the following pairs of species which one will have the same electronic configuration for both members?

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Answer: (d)

  1. C (Z=6Z=6): 1s2 2s2 2p21s^2\,2s^2\,2p^2
  2. N+\text{N}^+ (7−1=67-1 = 6 electrons): 1s2 2s2 2p21s^2\,2s^2\,2p^2
  3. Other pairs have different electron counts (Li+\text{Li}^+: 2 vs Na+\text{Na}^+: 10; He: 2 vs Ne+\text{Ne}^+: 9; H: 1 vs Li: 3).
  4. So the answer is (d) C and N+\text{N}^+.
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Q359NEET 1990 · AIPMTEasy

Quantum Numbers, X-rays & Other Questions · Quantum numbers & electronic configuration

The valence electron in alkali metal is a

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Answer: (c)

  1. Alkali metals (group 1) have configuration [noble gas] ns1ns^1.
  2. The single valence electron is in an ss orbital.
  3. So the answer is (c) ss-electron.
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Q360NEET 1989 · AIPMTEasy

Quantum Numbers, X-rays & Other Questions · Quantum numbers & electronic configuration

The atomic number of silicon is 14. Its ground state electron configuration is

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Answer: (d)

  1. Fill 14 electrons in order 1s,2s,2p,3s,3p1s, 2s, 2p, 3s, 3p; 2p2p holds at most 6.
  2. 1s2 2s2 2p6 3s2 3p21s^2\,2s^2\,2p^6\,3s^2\,3p^2 (total 2+2+6+2+2=142+2+6+2+2 = 14)
  3. So the answer is (d).
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