Atomic Structure: theory | Physics Circle

CONCEPT

Atomic Models & Alpha-Particle Scattering

In one line
Thomson spread the positive charge over the whole atom; Rutherford’s α-scattering showed it sits in a tiny, heavy nucleus.

ATOMIC MODELS

25.1.1 Thomson and Rutherford

Plum pudding (left) and nuclear atom (right)

Thomson: a positive sphere with electrons embedded like seeds in a watermelon.
It could not explain large-angle α-scattering or the line spectrum of hydrogen.
Rutherford: all the positive charge and almost all the mass (99.95%) are in a tiny nucleus.
Nucleus ≈ 10⁻¹⁵ m (1 fermi); atom ≈ 10⁻¹⁰ m (1 Å). The atom is mostly empty space.
Electrons revolve round the nucleus like planets round the sun.

FAILURE OF RUTHERFORD’S MODEL

25.1.2 Why the atom should collapse

A revolving charge radiates and spirals in

A revolving electron is accelerated, so by classical theory it must radiate energy.
Its orbit would shrink and it would fall into the nucleus: atom not stable.
The emitted spectrum would be continuous, but hydrogen gives a line spectrum.

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α-scattering experiment (Geiger and Marsden)

Most α pass straight; a few turn back

Most α-particles pass undeflected; some turn through small angles.
About 1 in 8000 turns by more than 90°; very few return (180°).

Number scattered against angle

∝ 1 ÷ sin⁴(θ ÷ 2)

N falls very steeply as θ grows

For a foil of thickness t: Nt = constant.

Distance of closest approach

r₀ = 4kZe²mv² = 2kZe²K

r₀ ∝ Z, r₀ ∝ 1K, r₀ ∝ 1m, r₀ ∝ 1v².
r₀ is an upper limit of nuclear size (about 10⁻¹⁴ m for gold).

Impact parameter b

b = kZe²K cot(θ2)

Large b: small deflection. b = 0: head-on, θ = 180°.

Practise PYQs on this topicAtomic Models & Alpha-Particle Scattering · 16 questions with solutions

CONCEPT

Bohr Model: Postulates, Radius, Speed & de Broglie

In one line
Only orbits with angular momentum nh ÷ 2π are allowed. From this: r ∝ n² ÷ Z and v ∝ Z ÷ n.

BOHR’S POSTULATES

25.2.1 For H and H-like atoms

Electrons move only in certain stationary orbits and do not radiate there.
In these orbits the angular momentum is quantised:

L = mvr = nh2π · n = 1, 2, 3, …

Energy is emitted or absorbed only when the electron jumps between orbits:

hν = E₂ − E₁

Valid only for one-electron systems: H, ²H, He⁺, Li²⁺, Be³⁺.
Change in L between neighbouring orbits = h ÷ 2π ≈ 1.05 × 10⁻³⁴ J s.

RADIUS OF THE ORBIT

25.2.2 Coulomb force gives centripetal force

Electron in the nth orbit

mv²r = kZe²r² and mvr = nh2π

r_n = n²h²ε₀πmZe² = 0.53 n²Z Å

r ∝ n² ÷ Z: radii of H orbits are in the ratio 1 : 4 : 9 : 16.
Bohr radius a₀ = 0.53 Å (n = 1, Z = 1).
Area of orbit ∝ r² ∝ n⁴.

SPEED OF THE ELECTRON

25.2.3 Falls as n grows

v_n = Ze²2ε₀nh = c137 × Zn = 2.2 × 10⁶ Zn m/s

v ∝ Z ÷ n.
v₁c = e²2ε₀hc = 1137 (fine-structure constant).
Momentum p = mv ∝ Z ÷ n.

DE BROGLIE PICTURE

25.2.4 Why L is quantised

A stationary orbit holds a whole number of waves

2πr = nλ ⇒ λ = 2πrn = hmv

λ ∝ n ÷ Z (since r ∝ n² and v ∝ 1 ÷ n).
Going up to a higher orbit: λ increases. Coming down to n = 1: λ decreases.
Example: in n = 3 of H, λ = 2π(9a₀) ÷ 3 = 6πa₀.

Practise PYQs on this topicBohr Model: Orbits, Speed & Angular Momentum · 126 questions with solutions

CONCEPT

Orbital Frequency, Current, Magnetic Moment & Other Systems

In one line
Every orbit quantity follows from r ∝ n² ÷ Z and v ∝ Z ÷ n. Find the power of n first, then the ratio.

FREQUENCY AND TIME PERIOD

25.3.1 Revolutions per second

f = v2πr = mZ²e⁴4ε₀²n³h³ ∝ Z²n³

T = 1f = 4ε₀²n³h³mZ²e⁴ ∝ n³Z²

Using r directly (n = 1): v = h2πmr, so f = h4π²mr².
Ground state of H: T ≈ 1.5 × 10⁻¹⁶ s.

CURRENT AND MAGNETIC MOMENT

25.3.2 The orbit is a current loop

i = ef = ev2πr ∝ Z²n³

M = iA = evr2 = e2m L = neh4πm

M ∝ n and M ∝ L.
n = 1: Bohr magneton μ_B = eh4πm = 9.27 × 10⁻²⁴ A m².
Gyromagnetic ratio ML = e2m: the same in every orbit.
Electron is negative, so M points opposite to L (angle 180°).

FIELD AT THE NUCLEUS

25.3.3 From the orbital current

B = μ₀i2r ∝ Z³n⁵

B₂ : B₄ = 4⁵ : 2⁵ = 32 : 1 for H.

ACCELERATION AND FORCE

25.3.4 Centripetal values

F = kZe²r² ∝ Z³n⁴ · a = v²r ∝ Z³n⁴

With r given: a = h²4π²m²r³ (using v = h2πmr).
a₃ : a₅ = 5⁴ : 3⁴ = 625 : 81.

SUMMARY TABLE

25.3.5 Power of n and Z

Quantity	Depends on
r	n² ÷ Z
v, p	Z ÷ n
L	n
λ (de Broglie)	n ÷ Z
f, i	Z² ÷ n³
T	n³ ÷ Z²
M	n
B at nucleus	Z³ ÷ n⁵
F, a	Z³ ÷ n⁴
K, \|U\|, \|E\|	Z² ÷ n²

BOHR RULE FOR OTHER SYSTEMS

25.3.6 Same two equations

Write the force balance, and use mvr = nh ÷ 2π (or Iω = nh ÷ 2π).
Diatomic molecule: E = n²h²8π²I; second orbit: E = h²2π²I.
Particle in field B: E = nhqB4πm.
Muon in place of electron: r ÷ 207, energy × 207 (ionisation ≈ 2.8 keV).

Practise PYQs on this topicBohr Model: Orbits, Speed & Angular Momentum · 126 questions with solutions

CONCEPT

Energy of the Electron & Energy Levels

In one line
Total energy is negative: E = −13.6 Z² ÷ n² eV, with K = −E and U = 2E.

KINETIC, POTENTIAL AND TOTAL ENERGY

25.4.1 Three energies in one orbit

K = kZe²2r · U = −kZe²r · E = −kZe²2r

U = −2K · E = −K · K : U : E = 1 : −2 : −1

All three tend to zero as n grows

Going to a higher orbit: K decreases, U and E increase (less negative).
Coming down to a lower orbit: K increases, U and E decrease.
Example: E = −3.4 eV ⇒ K = 3.4 eV, U = −6.8 eV.

ENERGY OF THE nth LEVEL

25.4.2 Hydrogen and H-like ions

E_n = −mZ²e⁴8ε₀²n²h² = −13.6 Z²n² eV

E ∝ Z² ÷ n².
He⁺ (Z = 2) in n = 2 has −13.6 eV, the same as H in n = 1.
Given L = nh ÷ 2π, find n first, then E.

n	State	E for H (eV)
1	ground	−13.6
2	first excited	−3.4
3	second excited	−1.51
4	third excited	−0.85
∞	ionised	0

IONISATION AND EXCITATION

25.4.3 Energy needed

Ionisation energy from level n = 0 − E_n = 13.6 Z²n² eV

Excitation energy = E_final − E_initial · V = Ee

H from ground state: ionisation 13.6 eV; first excitation 10.2 eV.
He⁺: 54.4 eV; Li²⁺: 122.4 eV from the ground state.

ENERGY LEVEL DIAGRAM

25.4.4 Levels crowd together near zero

Energy levels of hydrogen

Larger gap ⇒ more energy ⇒ shorter wavelength.
Two successive jumps C → B → A: 1λ_CA = 1λ_CB + 1λ_BA.

Practise PYQs on this topicEnergy of the Electron & Energy Levels · 63 questions with solutions

CONCEPT

Transitions, Photon Emission & Excitation

In one line
A jump from n₂ down to n₁ emits one photon of energy 13.6 Z²(1 ÷ n₁² − 1 ÷ n₂²) eV.

ENERGY OF THE PHOTON

25.5.1 Electron falls to a lower level

The energy difference leaves as one photon

ΔE = hν = 13.6 Z² (1n₁² − 1n₂²) eV

1λ = RZ² (1n₁² − 1n₂²) · R = 1.097 × 10⁷ m⁻¹

λ (in nm) = 1240ΔE (in eV)

Highest frequency: largest gap, usually a jump ending at n = 1.
Only a downward jump emits; an upward jump absorbs.
Same jump in different ions: λ ∝ 1 ÷ Z².
Ratio questions: write (1 ÷ n₁² − 1 ÷ n₂²) for each jump and divide.

Jump	ΔE for H (eV)
2 → 1	10.2
3 → 1	12.09
4 → 1	12.75
3 → 2	1.89
4 → 3	0.66

A photon energy not equal to any level gap (e.g. 11.1 eV) cannot be emitted.

NUMBER OF SPECTRAL LINES

25.5.2 All possible downward jumps

Every pair of levels gives one line

N = n(n − 1)2 · between n₂ and n₁: N = (n₂ − n₁)(n₂ − n₁ + 1)2

n = 3: 3 lines. n = 4: 6 lines. n = 5: 10 lines.
A single atom gives at most n − 1 lines.

EXCITATION BY PHOTONS AND ELECTRONS

25.5.3 Find the level reached

A photon is absorbed only if its energy equals a level gap exactly.
An electron beam can give part of its energy: the atom reaches the highest level below the beam energy.
Photon 12.75 eV (λ ≈ 975 Å) takes H from n = 1 to n = 4.

photon from a jump on a metal: hν = φ + K_max = φ + eV₀

RECOIL OF THE ATOM

25.5.4 Photon carries momentum

Mv = hλ = hRZ² (1n₁² − 1n₂²)

Recoil energy = p²2M, very small compared with hν.

Practise PYQs on this topicTransitions, Photon Emission & Excitation · 82 questions with solutions

CONCEPT

Hydrogen Spectrum & Spectral Series

In one line
Every jump that ends on the same lower level belongs to one series. First line = longest λ; series limit = shortest λ.

LINE SPECTRUM

25.6.1 Light from excited hydrogen

Visible (Balmer) lines of hydrogen

Each element gives its own set of sharp bright lines.
A neon sign gives a line spectrum; a hot solid (sun, heater) gives a continuous one.

THE FIVE SERIES

25.6.2 Named by the lower level

Jumps ending on n = 1, 2, 3, 4

Series	n₁	n₂	Region
Lyman	1	2, 3, …	ultraviolet
Balmer	2	3, 4, …	visible
Paschen	3	4, 5, …	infrared
Brackett	4	5, 6, …	infrared
Pfund	5	6, 7, …	far infrared

n → n − 1 jump in a series is its first line; n + 2 → n is the second line.
Example: 4 → 2 is the second line of Balmer.

FIRST LINE AND SERIES LIMIT

25.6.3 Longest and shortest wavelength

λ_max = n²(n + 1)²(2n + 1)R · λ_min = n²R

λ_maxλ_min = (n + 1)²2n + 1

Series	λ_max	λ_min	Ratio
Lyman	43R	1R	43
Balmer	365R	4R	95
Paschen	1447R	9R	167
Brackett	4009R	16R	259
Pfund	90011R	25R	3611

1R = 912 Å (Lyman limit). Balmer limit = 4 × 912 = 3648 Å.
Series-limit frequency: ν = Rcn²; Balmer limit = Rc4.
Wave number of Balmer limit = R4 = 2.5 × 10⁶ m⁻¹.
Lyman limit − first Lyman line: 1λ values give the Balmer limit.

HYDROGEN-LIKE IONS

25.6.4 Same lines, scaled by Z²

1λ = RZ² (1n₁² − 1n₂²)

Equal wavelengths in two atoms: set Z²(1 ÷ n₁² − 1 ÷ n₂²) equal.
Example: Balmer limit of H (4 ÷ R) = Brackett limit of Z = 2 (16 ÷ 4R).

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Quantum numbers

Number	Values	Tells us
n (principal)	1, 2, 3, …	shell, size, energy
l (orbital)	0 to n − 1	subshell s, p, d, f; shape
m_l (magnetic)	−l to +l	orientation; 2l + 1 values
m_s (spin)	+½ or −½	spin of the electron

Orbital angular momentum L = √l(l + 1) h2π

Electronic configuration

Pauli: no two electrons have all four quantum numbers the same; a shell holds at most 2n² electrons.
Aufbau: lower (n + l) fills first; equal (n + l): lower n first.
Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s …
Hund: electrons fill orbitals of a subshell singly first, with parallel spins.
Example: Si (Z = 14): 1s² 2s² 2p⁶ 3s² 3p².

Practise PYQs on this topicHydrogen Spectrum & Spectral Series · 60 questions with solutions

EXAM-DAY CARD

Just see this before the exam

mvr = nh2π

Bohr postulate

r = 0.53 n²Z Å

radius

v = 2.2 × 10⁶ Zn

speed (m/s)

E = −13.6 Z²n² eV

energy

1λ = RZ²(1n₁² − 1n₂²)

transition

N = n(n − 1)2

spectral lines

FORMULA SHEET

Whole chapter on one page

Rutherford

r₀ = 4kZe² ÷ mv²

N ∝ 1 ÷ sin⁴(θ ÷ 2)

b ∝ cot(θ ÷ 2)

nucleus ≈ 10⁻¹⁵ m

Orbit

r ∝ n² ÷ Z, v ∝ Z ÷ n

2πr = nλ

λ ∝ n ÷ Z

v₁ = c ÷ 137

Current & moment

f ∝ Z² ÷ n³, T ∝ n³ ÷ Z²

M = (e ÷ 2m)L ∝ n

B ∝ Z³ ÷ n⁵

F, a ∝ Z³ ÷ n⁴

Energy

E = −13.6Z² ÷ n² eV

K = −E, U = 2E

I.E. (H) = 13.6 eV

1st excitation = 10.2 eV

Transitions

hν = 13.6Z²(1 ÷ n₁² − 1 ÷ n₂²)

λ(nm) = 1240 ÷ ΔE(eV)

lines = n(n − 1) ÷ 2

recoil Mv = h ÷ λ

Series

Lyman UV, Balmer visible

λ_min = n² ÷ R

λ_max = n²(n + 1)² ÷ (2n + 1)R

1 ÷ R = 912 Å

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

"Second excited state" taken as n = 2

Going up: KE increases

He⁺ treated like H (Z forgotten)

First line = shortest λ

CORRECT ✓

Ground n = 1, first excited n = 2, second excited n = 3

Going up: K falls, U and E rise

E ∝ Z², r ∝ 1 ÷ Z, v ∝ Z

First line longest; series limit shortest

Atomic Structure

Atomic Models & Alpha-Particle Scattering

25.1.1 Thomson and Rutherford

25.1.2 Why the atom should collapse

Corner

Bohr Model: Postulates, Radius, Speed & de Broglie

25.2.1 For H and H-like atoms

25.2.2 Coulomb force gives centripetal force

25.2.3 Falls as n grows

25.2.4 Why L is quantised

Orbital Frequency, Current, Magnetic Moment & Other Systems

25.3.1 Revolutions per second

25.3.2 The orbit is a current loop

25.3.3 From the orbital current

25.3.4 Centripetal values

25.3.5 Power of n and Z

25.3.6 Same two equations

Energy of the Electron & Energy Levels

25.4.1 Three energies in one orbit

25.4.2 Hydrogen and H-like ions

25.4.3 Energy needed

25.4.4 Levels crowd together near zero

Transitions, Photon Emission & Excitation

25.5.1 Electron falls to a lower level

25.5.2 All possible downward jumps

25.5.3 Find the level reached

25.5.4 Photon carries momentum

Hydrogen Spectrum & Spectral Series

25.6.1 Light from excited hydrogen

25.6.2 Named by the lower level

25.6.3 Longest and shortest wavelength

25.6.4 Same lines, scaled by Z2

Corner

Just see this before the exam

Whole chapter on one page

Avoid these = easy marks

25.6.4 Same lines, scaled by Z²