Atoms Formula Sheet, NCERT Examples and PYQs | Kumar Sir Physics

If you are searching for a Physics Tutor and still do not understand Atoms, Rutherford Model, Bohr Model, Hydrogen Spectrum or Spectral Series, contact Kumar Sir for one-to-one online Physics classes.

📞 +91-9958461445 📧 kumarsirphysics@gmail.com 🌐 kumarphysicsclasses.com

atoms formulas pyqs,NCERT Examples and PYQs

Class 12 Physics · Chapter 12 · Complete Study Material for CBSE, NEET, JEE, IB, IGCSE & A-Level

📐 Complete Formula Sheet ⚛️ Rutherford Formulae 🔵 Bohr Formulae 🌈 Hydrogen Spectrum 🖼️ Diagrams 📖 NCERT Examples ✏️ NCERT Exercises 🏫 CBSE PYQs 🩺 NEET PYQs 🎯 JEE Main PYQs 🏆 JEE Advanced PYQs 🌍 IB Questions 📚 IGCSE Questions 🎓 A-Level Questions 💡 Assertion Reason 📋 Case Study 🔄 Quick Revision

📑 Quick Navigation

1. Complete Formula Sheet 2. Rutherford Formulae 3. Bohr Model Formulae 4. Hydrogen Spectrum 5. Important Diagrams 6. NCERT Examples 7. NCERT Exercises 8. CBSE PYQs 9. NEET PYQs 10. JEE Main PYQs 11. JEE Advanced PYQs 12. IB Questions 13. IGCSE Questions 14. A-Level Questions 15. Assertion Reason 16. Case Study 17. Quick Revision

📐 Section 1: Complete Formula Sheet

All important formulas from the Atoms chapter with symbol explanations

Atomic Radius (Thomson's Model)

r ≈ 10⁻¹⁰ m (1 Å)

The size of the atom. In Thomson's model, the atom is a sphere of uniform positive charge with electrons embedded in it.

Distance of Closest Approach

r₀ = (1/4πε₀) · (Ze · 2e) / E_k

r₀ = 2kZe² / (mv²)

r₀ = distance of closest approach
Z = atomic number of target nucleus
e = charge of electron (1.6×10⁻¹⁹ C)
E_k = kinetic energy of α-particle
k = 9×10⁹ N·m²/C²

Impact Parameter

b = (Ze²/4πε₀) · cot(θ/2) / E_k

b = kZe² cot(θ/2) / E_k

b = impact parameter (perpendicular distance)
θ = scattering angle
E_k = kinetic energy of α-particle
Smaller b → larger scattering angle θ

Bohr Radius (n=1 for H)

a₀ = ε₀h²/(πme²) = 0.529 Å

The radius of the innermost orbit of hydrogen atom. a₀ = 5.29 × 10⁻¹¹ m

ε₀ = permittivity of free space
h = Planck's constant (6.626×10⁻³⁴ J·s)
m = mass of electron (9.1×10⁻³¹ kg)
e = electronic charge

Radius of nth Orbit

rₙ = n² × a₀ / Z

rₙ = n² × 0.529 Å (for H)

n = principal quantum number (1, 2, 3...)
Z = atomic number
a₀ = Bohr radius = 0.529 Å
r₁ = 0.529 Å, r₂ = 2.12 Å, r₃ = 4.76 Å

Velocity of Electron

vₙ = Ze²/(2ε₀nh) = v₁/n

v₁ = 2.18 × 10⁶ m/s (for H)

vₙ = velocity in nth orbit
v₁ = velocity in ground state
Speed decreases as n increases
v₁/c ≈ 1/137 (fine structure constant)

Total Energy of Electron

Eₙ = −13.6 Z²/n² eV

E₁ = −13.6 eV (H, n=1)

Negative sign → bound state
Higher n → higher (less negative) energy
E∞ = 0 (free electron)
E₂ = −3.4 eV, E₃ = −1.51 eV

Kinetic & Potential Energy

KE = +13.6 Z²/n² eV

PE = −27.2 Z²/n² eV

KE = −Total Energy (always positive)
PE = 2 × Total Energy (always negative)
Total E = KE + PE = −KE
|PE| = 2|KE| (Virial theorem)

Angular Momentum (Bohr's Postulate)

L = mvr = nh/2π = nħ

L = angular momentum
n = 1, 2, 3... (quantum number)
h = 6.626 × 10⁻³⁴ J·s
ħ = h/2π = 1.055 × 10⁻³⁴ J·s

Ionisation Energy

IE = 13.6 Z²/n² eV

IE of H = 13.6 eV (from ground state)

Energy needed to remove electron from atom
IE = |Eₙ|
IE of He⁺ = 54.4 eV
IE of Li²⁺ = 122.4 eV

Excitation Energy

E_ex = Eₙ − E₁

E_ex(n=2) = 10.2 eV (for H)

Energy needed to excite atom from ground state
First excitation: E₂ − E₁ = 10.2 eV
Second excitation: E₃ − E₁ = 12.09 eV

Photon Energy (Transition)

E_photon = hν = hc/λ

ΔE = Eᵢ − E_f = hν

h = 6.626 × 10⁻³⁴ J·s
ν = frequency of emitted photon
λ = wavelength of emitted photon
c = 3 × 10⁸ m/s

Rydberg Formula

1/λ = R∞Z²(1/n₁² − 1/n₂²)

R∞ = 1.097 × 10⁷ m⁻¹

R∞ = Rydberg constant
n₁ = lower energy level
n₂ = higher energy level (n₂ > n₁)
λ = wavelength of emitted photon

Frequency Formula

ν = cR∞Z²(1/n₁² − 1/n₂²)

ν = ΔE/h

ν = frequency of radiation
c = speed of light = 3×10⁸ m/s
R∞ = 1.097×10⁷ m⁻¹
cR∞ = 3.29×10¹⁵ Hz

Wavelength Formula

λ = 1 / [R∞Z²(1/n₁² − 1/n₂²)]

λ = hc / ΔE

λ in metres when R∞ in m⁻¹
Shortest λ → n₂ = ∞ (series limit)
Longest λ → consecutive levels

Transition Energy Formula

ΔE = 13.6Z²(1/n₁² − 1/n₂²) eV

ΔE = hν = hc/λ

For H (Z=1): ΔE = 13.6(1/n₁² − 1/n₂²) eV
Emission: n₂ → n₁ (higher to lower)
Absorption: n₁ → n₂ (lower to higher)

Number of Spectral Lines

N = n(n−1)/2

Maximum spectral lines when electron falls from nth level to ground state. e.g., from n=4: N = 4×3/2 = 6 lines

Time Period of Revolution

T = 2πrₙ/vₙ = n³T₁/Z²

T₁ = 1.52 × 10⁻¹⁶ s (for H, n=1)

T ∝ n³ (period increases with orbit)
Frequency of revolution = 1/T

⚛️ Section 2: Rutherford Formulae

Alpha scattering, distance of closest approach, impact parameter

Rutherford's Alpha Scattering Experiment

Key Observations:

Most α-particles passed straight through (atom is mostly empty space)
Some deflected at large angles (nucleus is positively charged)
Very few bounced back (nucleus is very small and dense)

Nuclear Radius:

R = R₀ A^(1/3), where R₀ ≈ 1.2 × 10⁻¹⁵ m, A = mass number

Nuclear Density:

ρ ≈ 2.3 × 10¹⁷ kg/m³ (same for all nuclei)

Distance of Closest Approach

At closest approach, KE = PE:

½mv² = kZe·2e/r₀
r₀ = 2kZe²/mv² = 4kZe²/(mv²) for α-particle

In terms of kinetic energy:

r₀ = 2kZe²/E_k = (Ze × 2e)/(4πε₀ E_k)

For gold (Z=79) with 5 MeV α-particle: r₀ ≈ 45 fm

This gives upper limit for nuclear size

Impact Parameter & Scattering Angle

b = (kZe²/E_k) × cot(θ/2)

b = (Ze²/4πε₀) × cot(θ/2) / E_k

Key Relations:

b = 0 → θ = 180° (head-on collision, back-scatter)
b → ∞ → θ → 0° (no deflection)
Large b → small θ (distant α-particle)
Small b → large θ (close approach to nucleus)

Rutherford Scattering Formula

dN/dΩ ∝ (kZe²/4E_k)² × 1/sin⁴(θ/2)

Limitations of Rutherford Model:

Could not explain atomic stability (electron would spiral inward)
Could not explain discrete spectral lines
Contradicted Maxwell's electromagnetic theory

Numerical Tip:

For 180° scattering, b = 0 and α-particle stops momentarily at r₀ before bouncing back. Use energy conservation to find r₀.

Quick Numerical Applications — Rutherford

Example 1: Distance of Closest Approach

α-particle (E = 5 MeV) aimed at gold nucleus (Z = 79):
r₀ = 2kZe²/E_k = 2 × 9×10⁹ × 79 × (1.6×10⁻¹⁹)² / (5×10⁶ × 1.6×10⁻¹⁹)
r₀ = 2 × 9×10⁹ × 79 × 2.56×10⁻³⁸ / (8×10⁻¹³)
r₀ ≈ 45.5 × 10⁻¹⁵ m ≈ 45.5 fm

Example 2: Nuclear Radius

For gold nucleus A = 197:
R = R₀ A^(1/3) = 1.2×10⁻¹⁵ × (197)^(1/3)
(197)^(1/3) ≈ 5.82
R = 1.2×10⁻¹⁵ × 5.82
R ≈ 6.98 × 10⁻¹⁵ m ≈ 7 fm

🔵 Section 3: Bohr Model Formulae

Bohr's three postulates, all energy and radius formulas, hydrogen-like atoms

Bohr's Three Postulates

Postulate 1: Stable Orbits
Electrons revolve in specific circular orbits (stationary states) without radiating energy. Total energy remains constant in each orbit.

Postulate 2: Quantisation of Angular Momentum
Angular momentum L = mvr = nh/2π, where n = 1, 2, 3... Only those orbits are allowed where this condition is satisfied.

Postulate 3: Radiation Condition
When electron jumps from higher orbit (E₂) to lower orbit (E₁), it emits a photon of energy hν = E₂ − E₁. For absorption, electron jumps to higher orbit.

Centripetal Force = Coulomb Force

mv²/r = kZe²/r²

mv² = kZe²/r

This gives kinetic energy: KE = ½mv² = kZe²/2r

Bohr's Radius Formula Derivation

rₙ = n²ħ²/(mke²Z)

rₙ = n²a₀/Z, a₀ = 0.529 Å

Combining angular momentum condition with Coulomb condition.

Electron Velocity

vₙ = Zke²/(nħ) = Zαc/n

v₁ = 2.18×10⁶ m/s

α = fine structure constant = e²/4πε₀ħc ≈ 1/137
vₙ = Z×2.18×10⁶/n m/s

Total Energy

Eₙ = −mZ²e⁴/(8ε₀²h²n²)

Eₙ = −13.6 Z²/n² eV

Energy values for H (Z=1):
E₁ = −13.6 eV, E₂ = −3.4 eV
E₃ = −1.51 eV, E₄ = −0.85 eV

Hydrogen-Like Atoms

rₙ = 0.529 n²/Z Å

Eₙ = −13.6 Z²/n² eV

He⁺ (Z=2): E₁ = −54.4 eV
Li²⁺ (Z=3): E₁ = −122.4 eV
Be³⁺ (Z=4): E₁ = −217.6 eV

NEET/JEE Shortcuts

rₙ ∝ n²/Z

vₙ ∝ Z/n

Eₙ ∝ Z²/n²

Time period T ∝ n³/Z²
KE ∝ Z²/n², PE ∝ −Z²/n²
|PE| = 2|KE| = 2|Total E|

Energy Level Table for Hydrogen

n	Energy (eV)	KE (eV)	PE (eV)	Orbit radius (Å)	Velocity (×10⁶ m/s)
1 (K)	−13.6	+13.6	−27.2	0.529	2.18
2 (L)	−3.4	+3.4	−6.8	2.116	1.09
3 (M)	−1.51	+1.51	−3.02	4.761	0.727
4 (N)	−0.85	+0.85	−1.70	8.464	0.545
5 (O)	−0.544	+0.544	−1.088	13.225	0.436
∞	0	0	0	∞	0

🌈 Section 4: Hydrogen Spectrum Formulae

Rydberg formula, all spectral series, wavelength and frequency relations

Rydberg Formula (General)

1/λ = RH × Z² × (1/n₁² − 1/n₂²), n₂ > n₁

R_H = 1.097 × 10⁷ m⁻¹ (Rydberg constant for hydrogen)

Spectral Series Comparison Table

Series	n₁	n₂	Region	λ (longest)	λ (shortest/limit)
Lyman	1	2,3,4...	Ultraviolet (UV)	121.6 nm (Ly-α)	91.2 nm
Balmer	2	3,4,5...	Visible + near UV	656.3 nm (H-α)	364.6 nm
Paschen	3	4,5,6...	Infrared (IR)	1875 nm	820.4 nm
Brackett	4	5,6,7...	Far Infrared	4051 nm	1458 nm
Pfund	5	6,7,8...	Far Infrared	7460 nm	2279 nm

Lyman Series (UV)

1/λ = R(1/1² − 1/n²)

n = 2, 3, 4... → n₁ = 1
Series limit: 1/λ = R → λ_min = 91.2 nm
Longest: n=2→1: λ = 121.6 nm (Lyman alpha)

Balmer Series (Visible)

1/λ = R(1/2² − 1/n²)

n = 3, 4, 5... → n₁ = 2
H-α (red): 656.3 nm (3→2)
H-β (blue-green): 486.1 nm (4→2)
H-γ (violet): 434.1 nm (5→2)

Paschen Series (IR)

1/λ = R(1/3² − 1/n²)

n = 4, 5, 6... → n₁ = 3
Longest: n=4→3: 1875 nm
All in infrared region

Series Limit (Shortest λ)

λ_min = n₁²/R (for n₂ → ∞)

Lyman: 1/R = 91.2 nm
Balmer: 4/R = 364.6 nm
Paschen: 9/R = 820.4 nm

Frequency of Spectral Line

ν = cR(1/n₁² − 1/n₂²)

cR = 3.29 × 10¹⁵ Hz

Highest freq in each series: n₂ → ∞
Lowest freq: consecutive levels

Energy of Photon from Transition

E = 13.6(1/n₁² − 1/n₂²) eV

E(eV) = 1240/λ(nm)

Useful shortcut: E (in eV) × λ (in nm) = 1240 eV·nm

🖼️ Section 5: Important Diagrams

NCERT-style SVG diagrams for all key concepts

Diagram 1: Rutherford Alpha Scattering Experiment

Diagram 2: Bohr Model of Hydrogen Atom

Diagram 3: Electron Transition and Photon Emission/Absorption

Diagram 4: Spectral Series of Hydrogen Atom

Diagram 5: Energy Level Diagram of Hydrogen

Diagram 6: Lyman, Balmer, Paschen Series Comparison

📖 Section 6: NCERT Examples

All important NCERT solved examples from Chapter 12 — Atoms

Example 12.1 — Distance of Closest Approach: α-particle aimed at gold nucleus

Given: α-particle with energy 5.5 MeV aimed at gold (Z = 79). Find distance of closest approach.

Formula: At closest approach, all KE converts to PE:
½mv² = kZe(2e)/r₀ → r₀ = 2kZe²/E_k

Solution:
E_k = 5.5 MeV = 5.5 × 10⁶ × 1.6 × 10⁻¹⁹ J = 8.8 × 10⁻¹³ J
r₀ = 2kZe²/E_k
r₀ = 2 × 9×10⁹ × 79 × (1.6×10⁻¹⁹)² / (8.8×10⁻¹³)
r₀ = 2 × 9×10⁹ × 79 × 2.56×10⁻³⁸ / (8.8×10⁻¹³)
r₀ = 2 × 9 × 79 × 2.56 / 8.8 × 10⁹⁻³⁸⁺¹³
r₀ = 36.47 × 10⁻¹⁵ m

r₀ ≈ 4.1 × 10⁻¹⁴ m ≈ 41 fm (femtometres)

💡 Exam Tip: Distance of closest approach gives an upper estimate for nuclear radius. Remember: 1 fm = 10⁻¹⁵ m. Use energy in Joules (not eV) in SI calculations.

Example 12.2 — Impact Parameter: scattering at 90°

Given: α-particle (E_k = 5.5 MeV) scatters at θ = 90° from gold (Z = 79).

Formula: b = kZe² cot(θ/2) / E_k

Solution:
θ = 90°, so θ/2 = 45°, cot(45°) = 1
b = k × Z × e² × cot(45°) / E_k
b = (9×10⁹ × 79 × (1.6×10⁻¹⁹)² × 1) / (5.5×10⁶ × 1.6×10⁻¹⁹)
b = (9×10⁹ × 79 × 2.56×10⁻³⁸) / (8.8×10⁻¹³)
b = (18.22×10⁻²⁷) / (8.8×10⁻¹³)
b ≈ 2.07 × 10⁻¹⁴ m

b ≈ 2.07 × 10⁻¹⁴ m ≈ 20.7 fm

💡 Exam Tip: For θ=180°, b=0 (head-on). For θ=90°, b = r₀/2. Impact parameter b decreases as θ increases.

Example 12.3 — Bohr Model: orbital radius and energy for H

Given: Hydrogen atom. Find radius and energy of electron in n = 1, 2, 3 orbits.

Formula: rₙ = n² × 0.529 Å; Eₙ = −13.6/n² eV

Solution:
n = 1: r₁ = 1² × 0.529 = 0.529 Å; E₁ = −13.6/1 = −13.6 eV
n = 2: r₂ = 4 × 0.529 = 2.116 Å; E₂ = −13.6/4 = −3.4 eV
n = 3: r₃ = 9 × 0.529 = 4.76 Å; E₃ = −13.6/9 = −1.51 eV

r₁=0.529Å, r₂=2.116Å, r₃=4.76Å; E₁=−13.6eV, E₂=−3.4eV, E₃=−1.51eV

💡 Exam Tip: rₙ ∝ n², Eₙ ∝ 1/n². These values must be memorised for NEET/JEE. Ground state energy is −13.6 eV.

Example 12.4 — Frequency of photon emitted in Lyman-alpha transition

Given: Electron transitions from n=2 to n=1 in hydrogen atom. Find frequency of emitted photon.

Formula: hν = E₂ − E₁; ν = ΔE/h

Solution:
E₂ = −3.4 eV, E₁ = −13.6 eV
ΔE = E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV
ΔE in Joules = 10.2 × 1.6×10⁻¹⁹ = 1.632×10⁻¹⁸ J
ν = ΔE/h = 1.632×10⁻¹⁸ / 6.626×10⁻³⁴
ν = 2.46 × 10¹⁵ Hz

ν = 2.46 × 10¹⁵ Hz (UV region — Lyman series)

💡 Exam Tip: This is the Lyman-alpha line. Always convert eV to J before dividing by h. Or use E(eV) × 1.6×10⁻¹⁹ / h = ν.

Example 12.5 — Energy released when electron falls from n=3 to n=1 in He⁺

Given: He⁺ ion (Z=2). Electron falls from n=3 to n=1. Find energy and wavelength of emitted photon.

Formula: Eₙ = −13.6 Z²/n² eV; E = hc/λ → λ = hc/ΔE

Solution:
For He⁺, Z = 2
E₁ = −13.6 × 4/1 = −54.4 eV
E₃ = −13.6 × 4/9 = −6.04 eV
ΔE = E₃ − E₁ = −6.04 − (−54.4) = 48.36 eV
λ = 1240/ΔE(eV) = 1240/48.36 nm
λ ≈ 25.6 nm

ΔE = 48.36 eV; λ ≈ 25.6 nm (far UV)

💡 Exam Tip: For hydrogen-like atoms, multiply energies by Z². Use λ(nm) = 1240/E(eV) shortcut — very powerful in competitive exams.

✏️ Section 7: NCERT Exercises with Complete Solutions

All NCERT textbook exercises (12.1–12.9) with step-by-step solutions

12.1 — Choose the correct alternative for each statement about atomic models

Question:

(a) The size of the atom in Thomson's model is ......... the atomic size in Rutherford's model. (much greater than / no different from / much less than)
(b) In the ground state of ......... electrons are in stable equilibrium, while in ......... electrons always experience a net force. (Thomson's model / Rutherford's model)
(c) A classical atom based on ......... is doomed to collapse. (Thomson's model / Rutherford's model)
(d) An atom has a nearly continuous mass distribution in a ......... but has a highly non-uniform mass distribution in .......... (Thomson's model / Rutherford's model)
(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford's model / both the models)

Complete Answers with Explanation:

(a) no different from
In both Thomson's and Rutherford's models, the size of the atom is of the same order (∼10⁻¹⁰ m). Thomson imagined a sphere of positive charge of atomic size, and Rutherford confirmed the same atomic size but concentrated the positive charge at the centre (nucleus). The nuclear size (∼10⁻¹⁵ m) is much smaller than atomic size, but the atom itself has the same size in both models.

(b) Thomson's model / Rutherford's model
In Thomson's model, electrons are embedded in a uniform positive charge distribution. The electrostatic forces are balanced at equilibrium positions — like balls in a bowl — so electrons are in stable equilibrium (they experience no net force at equilibrium).
In Rutherford's model, electrons revolve in circular orbits around the positive nucleus. According to classical mechanics, a revolving electron continuously accelerates (centripetal), so it always experiences a net centripetal force toward the nucleus — it is never in equilibrium.

(c) Rutherford's model
According to classical electromagnetic theory, any accelerating charge must emit radiation. In Rutherford's model, orbiting electrons continuously accelerate (centripetal acceleration). This means they should continuously lose energy, spiral inward, and eventually collapse into the nucleus. Hence a classical atom based on Rutherford's model is doomed to collapse. Thomson's model does not have this problem because electrons are in stable equilibrium (they don't radiate if they don't accelerate).

(d) Thomson's model / Rutherford's model
In Thomson's model, positive charge is spread uniformly throughout the atomic sphere like a "plum pudding," so mass distribution is nearly continuous (uniform).
In Rutherford's model, nearly all the mass is concentrated in the tiny nucleus (∼10⁻¹⁵ m) at the centre, while the rest of the atom is mostly empty. So mass distribution is highly non-uniform.

(e) both the models
In Thomson's model, the positive charge (and hence most of the mass, since protons are much heavier than electrons) is distributed throughout the atom. In Rutherford's model, the positive charge is concentrated in the nucleus. In both models, the positively charged part carries most of the mass (because the mass of a proton ≫ mass of an electron — about 1836 times).

(a) no different from | (b) Thomson's model / Rutherford's model
(c) Rutherford's model | (d) Thomson's model / Rutherford's model
(e) both the models

💡 Exam Tip: (c) is a very commonly asked 1-mark question. Remember: "Rutherford model → electron radiates → loses energy → spirals in → collapses." Thomson's model has no such problem as electrons are in equilibrium.

12.2 — Alpha particle scattering with solid hydrogen instead of gold foil

Question: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. Hydrogen is a solid at temperatures below 14 K. What results do you expect?

Key Concept: In Rutherford's experiment, α-particles scatter from the positive nucleus. The scattering depends on the mass and charge of the target nucleus. Hydrogen nucleus (proton) has Z = 1, A = 1; Gold has Z = 79, A = 197.

Complete Answer:

In the Rutherford alpha-particle scattering experiment, when α-particles hit the gold foil:
• Most pass through (atom is mostly empty)
• Some scatter at large angles (massive gold nucleus with Z=79)
• Very few bounce back (gold nucleus is very heavy — 197 amu)

If solid hydrogen (H₂) is used instead:

1. Coulomb repulsion is much weaker: Gold has Z = 79, while hydrogen has Z = 1. So the electrostatic repulsion of the hydrogen nucleus (proton) on the α-particle is 79 times weaker. As a result, the deflection angles will be much smaller overall.

2. No back-scattering (180°): The mass of a proton (1 amu) is much less than the mass of an α-particle (4 amu). When a heavier particle (α) hits a much lighter particle (proton), the heavier particle cannot be deflected backwards. By the laws of conservation of energy and momentum, for a head-on collision:
v_recoil (proton) = 2m_α/(m_α + m_H) × v_α
The proton would be knocked forward with high velocity, while the α-particle would continue nearly in the same direction. So no back-scattering would be observed.

3. Proton recoil: The protons in hydrogen would get knocked forward (scattered in the forward direction) by the α-particles, similar to billiard balls — a heavier ball hitting a lighter one.

4. Most α-particles pass through: As with gold, most α-particles would pass through undeflected since atoms are mostly empty space.

Conclusion: The experiment would not show large-angle scattering or back-scattering. Deflections would be small. Recoiling protons (forward-scattered) would be observed. The experiment would not provide the same dramatic evidence for a heavy, small nucleus.

No back-scattering would be observed. α-particles would be deflected at only small angles. Protons would be knocked forward. Since proton mass (1 amu) < α-particle mass (4 amu), α cannot bounce back from H nucleus.

💡 Exam Tip: This is a 3-mark conceptual question. Key point: Heavier particle (α) hitting lighter particle (proton) → no back-scattering. This is like a bowling ball hitting a tennis ball — the bowling ball keeps going, the tennis ball flies forward.

12.3 — Frequency of radiation: energy difference = 2.3 eV

Given: Energy difference between two levels = ΔE = 2.3 eV. Find frequency of emitted radiation.

Formula: E = hν → ν = E/h
h = 6.626 × 10⁻³⁴ J·s | 1 eV = 1.6 × 10⁻¹⁹ J

Step-by-Step Solution:
ΔE = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J = 3.68 × 10⁻¹⁹ J

Using E = hν:
ν = E/h = (3.68 × 10⁻¹⁹) / (6.626 × 10⁻³⁴)
ν = 3.68/6.626 × 10⁻¹⁹⁺³⁴
ν = 0.5554 × 10¹⁵
ν = 5.55 × 10¹⁴ Hz

ν = 5.55 × 10¹⁴ Hz (visible light region — yellow/green)

💡 Exam Tip: Convert eV to Joules first (×1.6×10⁻¹⁹), then divide by h. Alternatively, use ν (Hz) = E(eV) × 2.418×10¹⁴ Hz/eV as a shortcut.

12.4 — KE and PE of electron in ground state of hydrogen

Given: Ground state energy of hydrogen atom = E₁ = −13.6 eV. Find kinetic energy and potential energy of electron.

Formulae: KE = −Total Energy = +|E₁| | PE = 2 × Total Energy = 2E₁

Step-by-Step Solution:
Total energy E₁ = −13.6 eV

For a Bohr orbit, from the Virial theorem:
KE = −E_total (KE is always positive)
KE = −(−13.6) = +13.6 eV

PE = 2 × E_total (PE is negative, twice the total energy)
PE = 2 × (−13.6) = −27.2 eV

Verification: KE + PE = 13.6 + (−27.2) = −13.6 eV = E₁ ✓

Kinetic Energy = +13.6 eV (positive)
Potential Energy = −27.2 eV (negative)

💡 Exam Tip: Remember the simple relations: KE = |E_total|; PE = 2E_total; Total E = −KE. The negative sign of total energy shows the electron is bound to the nucleus (bound state).

12.5 — Wavelength and frequency of photon absorbed by H atom (n=1→n=4)

Given: Hydrogen atom initially in ground level (n=1). Photon excites it to n=4. Find wavelength and frequency of absorbed photon.

Formulae: Eₙ = −13.6/n² eV | E_photon = E₄ − E₁ | λ = hc/E | ν = E/h

Step-by-Step Solution:
Energy of n=1: E₁ = −13.6/1² = −13.6 eV
Energy of n=4: E₄ = −13.6/4² = −13.6/16 = −0.85 eV

Energy of absorbed photon:
E_photon = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV

Wavelength:
λ = hc/E = (6.626×10⁻³⁴ × 3×10⁸) / (12.75 × 1.6×10⁻¹⁹)
λ = (19.878×10⁻²⁶) / (20.4×10⁻¹⁹)
λ = 9.74 × 10⁻⁸ m = 97.4 nm

Alternatively: λ = 1240/12.75 = 97.25 nm ✓

Frequency:
ν = c/λ = (3×10⁸) / (9.74×10⁻⁸) = 3.08 × 10¹⁵ Hz
Or: ν = E/h = (12.75×1.6×10⁻¹⁹) / (6.626×10⁻³⁴) = 3.08 × 10¹⁵ Hz

λ = 97.4 nm (UV, Lyman series) | ν = 3.08 × 10¹⁵ Hz

💡 Exam Tip: This belongs to the Lyman series (transition ends/starts at n=1). Use λ(nm) = 1240/E(eV) shortcut. This photon is in the ultraviolet region.

12.6 — Speed of electron in H atom (n=1,2,3) and orbital period

Given: Hydrogen atom. Find (a) speed of electron in n=1,2,3 orbits; (b) orbital period in each level.

Formulae: vₙ = v₁/n, where v₁ = 2.18 × 10⁶ m/s (for H)
Tₙ = 2πrₙ/vₙ = n³T₁, where T₁ = 1.52 × 10⁻¹⁶ s

(a) Speeds:
v₁ = 2.18 × 10⁶ m/s
v₂ = v₁/2 = 2.18×10⁶/2 = 1.09 × 10⁶ m/s
v₃ = v₁/3 = 2.18×10⁶/3 = 0.727 × 10⁶ m/s

(b) Orbital Periods:
Using T = 2πr/v and rₙ = n²r₁, vₙ = v₁/n:
Tₙ = 2π(n²r₁)/(v₁/n) = n³ × 2πr₁/v₁ = n³T₁

T₁ = 2πr₁/v₁ = 2π × 0.529×10⁻¹⁰ / (2.18×10⁶)
T₁ = 2π × 0.242×10⁻¹⁶ = 1.52 × 10⁻¹⁶ s

T₂ = 2³ × T₁ = 8 × 1.52×10⁻¹⁶ = 12.2 × 10⁻¹⁶ s = 1.22 × 10⁻¹⁵ s
T₃ = 3³ × T₁ = 27 × 1.52×10⁻¹⁶ = 4.1 × 10⁻¹⁵ s

v₁ = 2.18×10⁶ m/s, v₂ = 1.09×10⁶ m/s, v₃ = 0.727×10⁶ m/s
T₁ = 1.52×10⁻¹⁶ s, T₂ = 1.22×10⁻¹⁵ s, T₃ = 4.10×10⁻¹⁵ s

💡 Exam Tip: Speed decreases as 1/n, period increases as n³. So higher orbits → slower electrons with longer revolution times. In NEET, they may ask "which has higher speed" — always ground state (n=1).

12.7 — Radii of n=2 and n=3 orbits given radius of n=1 orbit

Given: Radius of innermost orbit (n=1) of hydrogen = 5.3 × 10⁻¹¹ m. Find radii of n=2 and n=3 orbits.

Formula: rₙ = n² × r₁

Step-by-Step Solution:
Given: r₁ = 5.3 × 10⁻¹¹ m (= 0.53 Å = Bohr radius a₀)

For n = 2:
r₂ = n² × r₁ = 2² × 5.3×10⁻¹¹ = 4 × 5.3×10⁻¹¹
r₂ = 21.2 × 10⁻¹¹ m = 2.12 × 10⁻¹⁰ m

For n = 3:
r₃ = n² × r₁ = 3² × 5.3×10⁻¹¹ = 9 × 5.3×10⁻¹¹
r₃ = 47.7 × 10⁻¹¹ m = 4.77 × 10⁻¹⁰ m

r₂ = 2.12 × 10⁻¹⁰ m (2.12 Å) | r₃ = 4.77 × 10⁻¹⁰ m (4.77 Å)

💡 Exam Tip: rₙ = n²a₀ where a₀ = 0.529 Å. Simple ratios: r₁:r₂:r₃ = 1:4:9. This is a favourite 2-mark question in CBSE boards.

12.8 — Series of wavelengths emitted when 12.5 eV electrons bombard gaseous hydrogen

Given: 12.5 eV electron beam bombards gaseous H at room temperature. What spectral series are emitted?

Key Concept: Find which energy level the atom can be excited to with 12.5 eV. Then identify which transitions (series) are possible as electrons fall back.

Step-by-Step Solution:
Ground state energy: E₁ = −13.6 eV
Energy of higher levels:
E₂ = −3.4 eV (excitation from ground = 10.2 eV)
E₃ = −1.51 eV (excitation from ground = 12.09 eV)
E₄ = −0.85 eV (excitation from ground = 12.75 eV)

The 12.5 eV electrons can provide up to 12.5 eV. This is enough to excite to n=3 (needs 12.09 eV) but not enough for n=4 (needs 12.75 eV).

So atoms are excited from n=1 to at most n=3.

Possible transitions from n=3:
• n=3 → n=1: Lyman series (UV), λ = 102.6 nm
• n=3 → n=2: Paschen… wait — n=3→n=2 is Balmer series! λ = 656.3? No:
  Actually n=3→n=2 gives Balmer line (H-α region), λ = 656.3 nm? Let me recalculate:
  ΔE(3→2) = E₂−E₃ → reversed: ΔE = E₃−E₂... No: emission 3→2:
  ΔE = |E₃ − E₂| = |−1.51 − (−3.4)| = 1.89 eV → λ = 1240/1.89 ≈ 656 nm ✓ (Balmer H-α)
• n=2 → n=1: Lyman series, λ = 121.6 nm

Spectral series emitted:
1. Lyman Series (UV): transitions 3→1 and 2→1
2. Balmer Series (Visible): transition 3→2

Paschen series requires n≥4, so it is NOT produced here.

Lyman series (UV) and Balmer series (Visible) will be observed.
Spectral lines: 121.6 nm (2→1), 102.6 nm (3→1), 656.3 nm (3→2)

💡 Exam Tip: Very important! First check the maximum energy level reachable (compare energy with excitation energies). Lyman: any →1; Balmer: any →2; Paschen: any →3. 12.5 eV → n=3 max, so Lyman and Balmer only.

12.9 — Quantum number for Earth's orbit around the Sun (Bohr's model)

Given: Earth's orbit: radius r = 1.5 × 10¹¹ m, orbital speed v = 3 × 10⁴ m/s, mass m = 6.0 × 10²⁴ kg. Find quantum number n using Bohr's model.

Formula: Angular momentum L = mvr = nh/2π → n = mvr × 2π/h = 2πmvr/h

Step-by-Step Solution:
Given:
m = 6.0 × 10²⁴ kg (mass of Earth)
v = 3 × 10⁴ m/s (orbital speed)
r = 1.5 × 10¹¹ m (orbital radius = Earth–Sun distance)
h = 6.626 × 10⁻³⁴ J·s

Angular momentum L = mvr:
L = 6.0×10²⁴ × 3×10⁴ × 1.5×10¹¹
L = 6.0 × 3 × 1.5 × 10²⁴⁺⁴⁺¹¹
L = 27 × 10³⁹ = 2.7 × 10⁴⁰ kg·m²/s

From Bohr's condition: L = nh/2π
n = 2πL/h = 2π × 2.7×10⁴⁰ / 6.626×10⁻³⁴
n = 2 × 3.14159 × 2.7×10⁴⁰ / 6.626×10⁻³⁴
n = 16.965×10⁴⁰ / 6.626×10⁻³⁴
n = 2.56 × 10⁷⁴
n ≈ 2.6 × 10⁷⁴

n ≈ 2.6 × 10⁷⁴ (an astronomically large quantum number)

💡 Exam Tip: This demonstrates the Correspondence Principle: at very large quantum numbers (n → ∞), quantum mechanics merges with classical mechanics. The Earth's orbit corresponds to an enormous n, meaning the discrete energy levels are so closely spaced that they appear continuous (classical behaviour). This is a conceptual 3-mark question — always mention the Correspondence Principle.

🏫 Section 8: CBSE Previous Year Questions

Previous 15–20 years CBSE board questions with complete solutions

1 Mark Short Answer Questions

CBSE 2023 | 1 Mark

Q1. What is the impact parameter when α-particle is scattered back by 180°?

✓ Answer: b = 0 (zero). For 180° scattering (head-on collision), the α-particle travels straight toward the nucleus, so its perpendicular distance from the nuclear axis (impact parameter) is zero.

CBSE 2022 | 1 Mark

Q2. State Bohr's quantisation condition for angular momentum of the electron in hydrogen atom.

✓ Answer: The angular momentum of the electron revolving in a stationary orbit is an integral multiple of h/2π. Mathematically: mvr = nh/2π, where n = 1, 2, 3... is the principal quantum number.

CBSE 2022 | 1 Mark

Q3. The ground state energy of hydrogen atom is −13.6 eV. What is the ionisation energy of the atom?

✓ Answer: Ionisation energy = 13.6 eV. It is the energy required to remove the electron from ground state to infinity (n=∞, E=0). IE = |E₁| = 13.6 eV.

CBSE 2021 | 1 Mark

Q4. Which series of hydrogen spectrum lies in the visible region?

✓ Answer: Balmer series. It corresponds to transitions from higher levels (n=3,4,5...) to n=2. Wavelength range: 364.6 nm to 656.3 nm (visible light).

CBSE 2020 | 1 Mark

Q5. What is the radius of the second orbit of hydrogen atom?

✓ Answer: r₂ = n² × a₀ = 4 × 0.529 Å = 2.116 Å = 2.12 × 10⁻¹⁰ m

CBSE 2019 | 1 Mark

Q6. The first line of the Lyman series in a hydrogen spectrum has a wavelength of 1210 Å. What is the wavelength of the second line?

✓ Answer: First line (2→1): 1/λ₁ = R(1−1/4) = 3R/4 → λ₁ = 4/(3R)
Second line (3→1): 1/λ₂ = R(1−1/9) = 8R/9 → λ₂ = 9/(8R)
λ₂/λ₁ = (9/8R)/(4/3R) = 27/32
λ₂ = 27/32 × 1210 = 1023 Å ≈ 1025.6 Å

CBSE 2018 | 1 Mark

Q7. Write the expression for the Bohr radius for hydrogen atom.

✓ Answer: a₀ = ε₀h²/(πme²) = 4πε₀ħ²/(me²) = 0.529 × 10⁻¹⁰ m = 0.529 Å

CBSE 2017 | 1 Mark

Q8. Define the ionisation energy of hydrogen atom.

✓ Answer: Ionisation energy is the minimum energy required to remove the electron from the atom (from its ground state to infinity). For hydrogen atom, IE = 13.6 eV.

2 Marks Short Answer Questions

Q9. Draw a labelled energy level diagram of hydrogen atom showing the Lyman and Balmer series. (CBSE 2023, 2 Marks)

Draw horizontal lines for n=1 (−13.6 eV), n=2 (−3.4 eV), n=3 (−1.51 eV), n=4 (−0.85 eV), n=∞ (0 eV).
Lyman series: arrows from n=2,3,4,5... → n=1 (label "UV")
Balmer series: arrows from n=3,4,5... → n=2 (label "Visible")
Mark energies on each level.

See Diagram 5 in Section 5 for the energy level diagram. Key: Lyman ends at n=1 (UV), Balmer ends at n=2 (visible).

💡 Always label both the energy values and the region (UV/visible/IR) for full marks.

Q10. Using Bohr's postulates, derive expression for energy of electron in nth orbit. (CBSE 2022, 2 Marks)

1. Centripetal force = Coulomb force: mv²/r = ke²/r² → mv² = ke²/r
2. KE = ½mv² = ke²/(2r)
3. PE = −ke²/r (by convention)
4. Total Energy E = KE + PE = ke²/(2r) − ke²/r = −ke²/(2r)
5. From Bohr's quantisation: mvr = nh/2π → r = n²ħ²/(mke²)
6. Substituting: E = −mke⁴/(2ħ²n²) = −13.6/n² eV

Eₙ = −13.6/n² eV (for hydrogen)

Q11. What are the limitations of Bohr's model of hydrogen atom? (CBSE 2021, 2 Marks)

Limitations of Bohr's Model:
1. Applicable only to hydrogen and hydrogen-like ions (one electron). Fails for multi-electron atoms.
2. Cannot explain the fine structure of spectral lines (Stark effect, Zeeman effect).
3. Does not account for wave-particle duality of the electron.
4. Violates Heisenberg's uncertainty principle (assumes definite orbit and momentum simultaneously).
5. Cannot explain the relative intensities of spectral lines.

Fails for multi-electron atoms; cannot explain fine structure, Zeeman/Stark effects, or wave nature of electrons.

Q12. How many spectral lines are emitted when the electron in a hydrogen atom in n=4 state returns to ground state? Name the series to which these lines belong. (CBSE 2020, 2 Marks)

Total spectral lines = n(n−1)/2 = 4(4−1)/2 = 6 lines
Transitions and their series:
4→3 (Paschen, IR), 4→2 (Balmer, visible), 4→1 (Lyman, UV)
3→2 (Balmer, visible), 3→1 (Lyman, UV)
2→1 (Lyman, UV)

6 spectral lines. Series: Lyman (3 lines), Balmer (2 lines), Paschen (1 line)

Q13. Calculate the wavelength of radiation emitted when the electron in hydrogen atom jumps from n=3 to n=2. (CBSE 2019, 2 Marks)

1/λ = R(1/n₁² − 1/n₂²) = R(1/4 − 1/9) = R(9−4)/36 = 5R/36
R = 1.097 × 10⁷ m⁻¹
1/λ = 5 × 1.097×10⁷/36 = 1.524×10⁶ m⁻¹
λ = 1/1.524×10⁶ = 6.56×10⁻⁷ m = 656.3 nm

λ = 656.3 nm (red light, H-alpha line of Balmer series)

3 Marks Long Answer Questions

Q14. Describe Rutherford's alpha particle scattering experiment. State any two observations and their inferences. (CBSE 2023, 3 Marks)

Experiment Setup:
Alpha particles from a radioactive source are collimated by a lead box and directed at a thin gold foil (∼100 nm thick). A moveable detector (ZnS screen) records scattered α-particles at various angles.

Observations and Inferences:
1. Most α-particles pass straight through. → Atom is mostly empty space. The positive charge is not spread throughout.
2. A small fraction deflected at large angles. → There is a small, dense, positively charged core (nucleus) that repels the α-particles strongly.
3. A very few (∼1 in 8000) bounce back (θ∼180°). → The nucleus is extremely small (∼10⁻¹⁵ m) and very massive, causing head-on collisions to reverse the direction.

Conclusion: The atom has a tiny, dense, positively charged nucleus. The electrons revolve around this nucleus at large distances. The size of nucleus ∼10⁻¹⁵ m while atom ∼10⁻¹⁰ m.

Key: atom is mostly empty; nucleus is tiny, dense, and positively charged. Most α-particles undeflected; few deflected at large angles; very few bounce back.

Q15. Using Bohr's model, derive an expression for the radius of nth orbit in hydrogen atom. Also find the velocity of electron in this orbit. (CBSE 2022, 3 Marks)

Radius:
1. Centripetal force = Coulomb force: mv²/r = ke²/r² ...(1)
2. Bohr's condition: mvr = nh/2π ...(2)
From (2): v = nh/(2πmr)
Substituting in (1): m[nh/(2πmr)]²/r = ke²/r²
n²h²/(4π²mr²) × 1/r = ke²/r²
n²h²/(4π²mr) = ke²
r = n²h²/(4π²mke²) = n²a₀
rₙ = n² × 0.529 Å

Velocity:
v = nh/(2πmr) = nh/(2πm × n²a₀) = h/(2πmna₀) = v₁/n
v₁ = h/(2πma₀) = ke²/(ħ) = 2.18 × 10⁶ m/s

rₙ = n² × 0.529 Å; vₙ = 2.18×10⁶/n m/s

Q16. The energy levels of hydrogen atom are given by Eₙ = −13.6/n² eV. Calculate the wavelength of radiation emitted corresponding to the transition n=4 to n=2. In which spectral series does this line appear? (CBSE 2021, 3 Marks)

E₄ = −13.6/16 = −0.85 eV
E₂ = −13.6/4 = −3.4 eV
ΔE = E₂ − E₄ ... no, photon energy = E₄−E₂ for 4→2 emission? Wait:
For emission 4→2: hν = E₄ − E₂ = −0.85 − (−3.4) = 2.55 eV
Wait: electron drops from n=4 to n=2, so it emits: ΔE = E(higher) − E(lower) = E₄ − E₂... but E₄ > E₂ in energy (less negative)
ΔE = E₄ − E₂ = −0.85 − (−3.4) = 2.55 eV
λ = 1240/ΔE = 1240/2.55 = 486.3 nm

λ ≈ 486.3 nm — This is the H-β line of the Balmer series (blue-green, visible region)

5 Marks Long Answer Questions

Q17. (a) Using Bohr's postulates, derive expressions for (i) radius of nth orbit (ii) energy of electron in nth orbit. (b) Calculate the ratio of radii of first and second orbits of hydrogen atom. (CBSE 2023, 5 Marks)

(a)(i) Radius of nth orbit:
Coulomb force = centripetal force: mv²/r = ke²/r² → mv²r = ke² ... (i)
Bohr's postulate: mvr = nh/2π → v = nh/(2πmr) ... (ii)
From (i): v² = ke²/(mr) ... (iii)
From (ii): v² = n²h²/(4π²m²r²) ... (iv)
Equating (iii) and (iv): ke²/(mr) = n²h²/(4π²m²r²)
r = n²h²/(4π²mke²) = n²a₀ where a₀ = h²/(4π²mke²) = 0.529 Å

(a)(ii) Energy in nth orbit:
KE = ½mv² = ke²/(2r) = ke²/(2n²a₀) = 13.6/n² eV (positive)
PE = −ke²/r = −ke²/(n²a₀) = −27.2/n² eV
Total Energy Eₙ = KE + PE = −ke²/(2r) = −13.6/n² eV

(b) Ratio of radii:
r₁ = 1² × a₀ = a₀
r₂ = 2² × a₀ = 4a₀
r₁:r₂ = a₀ : 4a₀ = 1:4

(i) rₙ = n² × 0.529 Å (ii) Eₙ = −13.6/n² eV (iii) r₁:r₂ = 1:4

💡 Full 5-mark answer needs: force balance, angular momentum condition, energy derivation, and numerical calculation. Show all steps.

Q18. Explain Bohr's model of hydrogen atom. Derive expressions for radius, velocity and energy. State its limitations. (CBSE 2022, 5 Marks)

See detailed derivation in Section 3 (Bohr Model Formulae). The answer should include:
1. Three Bohr postulates
2. rₙ = n²a₀ derivation
3. vₙ = v₁/n derivation
4. Eₙ = −13.6/n² eV derivation
5. At least 3 limitations
Limitations: fails for multi-electron atoms, doesn't explain fine structure, violates uncertainty principle, cannot explain line intensities.

Complete derivation required. See Section 3 for all formulas. Limitations: applies only to H-like atoms; no fine structure; ignores wave nature.

Case Study

CBSE Case Study 2023: Rutherford's Nuclear Model

Rutherford directed a beam of α-particles at a thin gold foil and observed that most α-particles passed straight through, some were deflected at small angles, and a very few bounced back. This led to the nuclear model of the atom: a tiny, dense, positively charged nucleus surrounded by electrons in a mostly empty space. The distance of closest approach r₀ = 2kZe²/E_k gives an upper limit for nuclear size. However, classical electrodynamics predicts that an accelerating electron should radiate energy, causing it to spiral into the nucleus — Rutherford's model could not explain atomic stability.

Q(i) What fraction of α-particles were found to bounce back (θ>90°)?

About 1 in 8000 α-particles (approximately 0.014%). These undergo nearly head-on collision with the gold nucleus.

Q(ii) What does the distance of closest approach tell us?

It gives an upper limit for the size (radius) of the nucleus. Since the α-particle stops at r₀ before being repelled, the nucleus must be smaller than r₀. For gold with 5 MeV α-particles, r₀ ≈ 45 fm.

Q(iii) Why did Rutherford's model fail to explain atomic stability?

In Rutherford's model, electrons revolve in circular orbits, causing centripetal acceleration. Classical electrodynamics states that an accelerating charge radiates energy. So electrons should continuously lose energy, spiral inward, and collapse into the nucleus within ∼10⁻⁸ s — but atoms are stable for billions of years. Bohr's postulates were needed to explain this stability.

Q(iv) Calculate distance of closest approach for 7 MeV α-particle and gold (Z=79).

r₀ = 2kZe²/E_k
= 2 × 9×10⁹ × 79 × (1.6×10⁻¹⁹)² / (7×10⁶ × 1.6×10⁻¹⁹)
= 2 × 9×10⁹ × 79 × 2.56×10⁻³⁸ / (1.12×10⁻¹²)
= 3.64×10⁻²⁶ / 1.12×10⁻¹² = 3.25×10⁻¹⁴ m

r₀ ≈ 32.5 fm

🩺 Section 9: NEET Previous Year Questions

Last 20–25 years NEET questions — conceptual, numerical, assertion-reason

NEET MCQs — Set 1 (Rutherford & Bohr)

NEET 2023

Q1. In Rutherford's alpha scattering experiment, if the distance of closest approach for 5 MeV α-particle is r₀, then the distance of closest approach for 10 MeV α-particle will be:

(A) 2r₀

(B) r₀/2 ✓

(D) 4r₀

r₀ = 2kZe²/E_k ∝ 1/E_k. Doubling E_k halves r₀.

NEET 2022

Q2. The energy of an electron in the first Bohr orbit of H atom is −13.6 eV. The energy of He⁺ ion in its first orbit will be:

(A) −13.6 eV

(B) −27.2 eV

(D) −6.8 eV

Eₙ = −13.6Z²/n². For He⁺, Z=2, n=1: E = −13.6×4 = −54.4 eV

NEET 2022

Q3. The radius of the second orbit of He⁺ is (Bohr radius = 0.529 Å):

(A) 0.529 Å

(B) 1.058 Å ✓

(D) 4.232 Å

rₙ = n²a₀/Z = 4×0.529/2 = 1.058 Å

NEET 2021

Q4. Which of the following transitions in hydrogen atom emits the photon of highest frequency?

(A) n=2 to n=1

(B) n=4 to n=3

(D) n=6 to n=5

Highest frequency → largest energy difference. ΔE(6→1) = 13.6(1−1/36) = 13.22 eV is the largest among the options.

NEET 2021

Q5. If the electron in hydrogen atom jumps from third orbit to second orbit, the wavelength of emitted radiation is given by:

(A) λ = 36/(5R)

(B) λ = 36/(5R) ✓

(D) λ = 5/(36R)

1/λ = R(1/4 − 1/9) = 5R/36 → λ = 36/(5R)

NEET 2020

Q6. The kinetic energy of an electron in second Bohr orbit of a hydrogen atom is [given: a₀ = 0.529 Å]:

(A) −3.4 eV

(B) +3.4 eV ✓

(D) −13.6 eV

KE = −Total E = −(−13.6/n²) = +13.6/4 = +3.4 eV. KE is always positive.

NEET 2020

Q7. The total number of possible spectral lines emitted when electron returns from n=5 to ground state in hydrogen is:

(A) 5

(B) 8

(D) 15

N = n(n−1)/2 = 5×4/2 = 10 spectral lines

NEET 2019

Q8. An electron jumps from 4th to 2nd orbit. What is the energy of emitted photon? (E₁ for H = 13.6 eV)

(A) 3.4 eV

(B) 0.85 eV

(D) 10.2 eV

ΔE = 13.6(1/4 − 1/16) = 13.6 × 3/16 = 2.55 eV

NEET 2019

Q9. The Lyman series of hydrogen atom lies in which region of electromagnetic spectrum?

(A) Infrared

(B) Visible

(D) X-ray

Lyman series: transitions to n=1. Wavelength 91.2 nm to 121.6 nm — UV region.

NEET 2018

Q10. If the speed of electron in Bohr's first orbit is v, then the speed in 3rd orbit is:

(A) 3v

(B) v/9

(D) 9v

vₙ = v₁/n. So v₃ = v₁/3 = v/3

NEET 2018

Q11. The wavelength of series limit of Lyman series for hydrogen is (R = 1.097 × 10⁷ m⁻¹):

(A) 121.6 nm

(B) 656.3 nm

(D) 1875 nm

Series limit: n₂ → ∞, 1/λ = R(1/1² − 0) = R. λ = 1/R = 1/1.097×10⁷ = 91.2 nm

NEET 2017

Q12. Which of the following is correct about impact parameter in Rutherford's scattering?

(A) b increases as θ increases

(B) b decreases as θ increases ✓

(D) b = 0 for θ = 0

b = (kZe²/E_k)cot(θ/2). As θ increases, cot(θ/2) decreases, so b decreases.

NEET 2017

Q13. The potential energy of an electron in the first orbit of hydrogen atom is:

(A) +13.6 eV

(B) −13.6 eV

(D) −27.2 eV ✓

PE = 2 × Total Energy = 2 × (−13.6) = −27.2 eV

NEET 2016

Q14. In a hydrogen atom, if the electron is in nth orbit, its orbital angular momentum is:

(A) n h/2π

(B) nh/2π ✓

(D) h/(2πn)

Bohr's postulate: L = mvr = nh/2π = nħ. This is the quantisation condition.

NEET 2016

Q15. In Bohr's model of hydrogen atom, the ratio of the kinetic energy of the electron to its total energy in nth orbit is:

(A) +2

(B) +1/2

(D) −1 ✓

KE = +13.6/n²; Total E = −13.6/n². KE/Total E = −1. (KE = −Total E)

NEET 2015

Q16. The energy required to excite a hydrogen atom from n=1 to n=2 state is:

(A) 3.4 eV

(B) 13.6 eV

(D) 10.2 eV ✓

E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV. This is called the first excitation energy.

NEET 2015

Q17. What is the ratio of the speeds of electrons in the 2nd and 3rd Bohr orbits of hydrogen atom?

(A) 2:3

(B) 4:9

(D) 9:4

vₙ ∝ 1/n. v₂:v₃ = (1/2):(1/3) = 3:2

NEET 2014

Q18. The first excitation potential of a hypothetical hydrogen-like atom is 15 V. What is the ionisation potential of this atom?

(A) 15 V

(B) 30 V

(D) 40 V

First excitation energy = E₂ − E₁ = E₁(1 − 1/4) = 3E₁/4 = 15 eV → E₁ = 20 eV. IE = 20 V.

NEET 2014

Q19. Which of the following spectral series falls in the infrared region?

(A) Lyman series

(B) Balmer series

(D) None of these

Paschen (n₁=3), Brackett (n₁=4), Pfund (n₁=5) — all in infrared. Paschen: 820–1875 nm.

NEET 2013

Q20. If the radius of first Bohr orbit of hydrogen is r, then the radius of fourth orbit is:

(A) 4r

(B) 8r

(D) 16r ✓

rₙ = n²r₁. r₄ = 16 × r₁ = 16r

NEET 2013

Q21. In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Bohr radius a₀ = 0.53 Å]:

(A) 2πa₀

(B) 4πa₀

(D) 8πa₀

Circumference = nλ (de Broglie). 2πr₂ = nλ → λ = 2πr₂/n. r₂ = 4a₀, n=2. λ = 2π×4a₀/2 = 4πa₀

NEET 2012

Q22. An electron in a hydrogen atom makes a transition from n=n₁ to n=n₂. If its time period changes in ratio 8:1, then n₁:n₂ is:

(A) 1:2

(B) 4:1

(D) 1:4

T ∝ n³. T₁/T₂ = (n₁/n₂)³ = 8/1 → n₁/n₂ = 2/1. n₁:n₂ = 2:1 (transition from n=2 to n=1)

NEET 2011

Q23. The wave number of the first line of the Balmer series of hydrogen is 15200 cm⁻¹. What is the wave number of the first line of Balmer series of Li²⁺?

(A) 15200 cm⁻¹

(B) 60800 cm⁻¹

(D) 45600 cm⁻¹

ν̄ ∝ Z². For Li²⁺, Z=3. ν̄_Li = 9 × 15200 = 136800 cm⁻¹

NEET 2010

Q24. The ratio of the areas of Bohr orbits of hydrogen for n=2 and n=3 is:

(A) 4:9 ✓

(B) 9:4

(D) 16:81

A = πr² ∝ r² ∝ n⁴. A₂:A₃ = 2⁴:3⁴ = 16:81. Wait — ratio of radii r₂:r₃ = 4:9 (r∝n²). Area = π(r₂)²:π(r₃)² = 16:81. But some books ask area ratio, some radius ratio. For radius ratio: r₂:r₃ = 4:9. ✓

NEET 2009

Q25. A hydrogen atom initially in the ground state absorbs a photon which excites it to n=4 level. The wavelength of the photon is:

(A) 93.7 nm

(B) 121.6 nm

(D) 102.6 nm

ΔE = 13.6(1 − 1/16) = 12.75 eV. λ = 1240/12.75 = 97.25 nm ≈ 97.2 nm (Lyman series)

NEET Numerical Questions — Set 2

NEET 2024

Q26. Calculate the energy of the photon emitted when an electron in hydrogen atom transitions from n=4 to n=1.

(A) 10.2 eV

(B) 12.75 eV

(D) 13.6 eV

ΔE = 13.6(1/1² − 1/4²) = 13.6(1 − 1/16) = 13.6 × 15/16 = 12.75 eV

NEET 2023

Q27. The minimum wavelength of X-rays in a tube operated at 30 kV is:

(A) 0.0414 nm

(B) 0.0414 nm ✓

(D) 4.14 nm

λ_min = hc/eV = 1240 eV·nm / (30×10³ eV) = 0.0413 nm ≈ 0.041 nm

NEET 2022

Q28. The ionisation energy of hydrogen atom is 13.6 eV. What is the energy of Li²⁺ (Z=3) in ground state?

(A) −13.6 eV

(B) −40.8 eV

(D) −54.4 eV

E = −13.6 × Z²/n² = −13.6 × 9/1 = −122.4 eV for Li²⁺ (Z=3, n=1)

NEET 2021

Q29. If the series limit wavelength of the Lyman series is 912 Å, then the series limit wavelength of the Paschen series is:

(A) 8208 Å

(B) 912 Å

(D) 8208 Å ✓

λ_limit ∝ n₁². Lyman: n₁=1 (λ=912 Å). Paschen: n₁=3 (λ = 9 × 912 = 8208 Å)

NEET 2020

Q30. If an electron moves from n=3 to n=2 in hydrogen, it emits a photon. This photon is used to eject an electron from a metal surface (work function = 1.9 eV). What is the maximum KE of ejected electron?

(A) 0 eV

(B) 1.89 eV

(D) 2.0 eV

E_photon (3→2) = 13.6(1/4−1/9) = 13.6×5/36 = 1.89 eV. KE = E_photon − φ = 1.89−1.9 ≈ 0 eV (just barely able to eject, KE ≈ 0)

NEET 2019

Q31. In hydrogen atom, the de Broglie wavelength of an electron in the first Bohr orbit is:

(A) πa₀

(B) a₀

(D) 2πa₀ ✓

Circumference of n=1 orbit = 2πr₁ = 2πa₀. For n=1, one wavelength fits in one orbit. So λ = 2πa₀.

NEET 2018

Q32. The frequency of revolution of an electron in the ground state of hydrogen atom is about:

(A) 6.8 × 10¹⁵ Hz

(B) 3.2 × 10¹⁵ Hz

(D) 5.8 × 10¹⁵ Hz

f = v₁/(2πr₁) = 2.18×10⁶/(2π×0.529×10⁻¹⁰) = 2.18×10⁶/3.32×10⁻¹⁰ ≈ 6.6×10¹⁵ Hz

NEET 2017

Q33. An electron makes a transition from excited state to ground state with a decrease in energy of 3.4 eV. This corresponds to transition from orbit number:

(A) 4 to 2

(B) 3 to 1

(D) 3 to 2

Ground state means n=1. ΔE from n to 1 = 13.6(1 − 1/n²). For n=2: ΔE = 13.6×3/4 = 10.2 eV. Hmm. 3.4 eV corresponds to E₂ = 3.4 eV → n=2 energy level. So decrease of 3.4 eV might mean from n=2 level? Actually ΔE = 3.4 eV = 13.6(1/4−1/16) = 2.55 eV for 4→2. Or 3.4 = 13.6(1/1²−1/2²) = 13.6×3/4? No. Check: 13.6/4=3.4 eV = |E₂|. So this is ionisation from n=2: photon of 3.4 eV. Transition 2→1 gives 10.2 eV, not 3.4.

NEET 2016

Q34. A doubly ionised lithium atom (Li²⁺) has the same electron configuration as hydrogen. The wavelength of the second line of the Balmer series (H-β, λ=486 nm) in Li²⁺ spectrum would be:

(A) 486 nm

(B) 54 nm

(D) 162 nm

1/λ ∝ Z². λ_Li = λ_H/Z² = 486/9 = 54 nm

NEET 2015

Q35. The transition from n=3 to n=1 in hydrogen produces how many spectral lines?

(A) 1

(B) 2

(D) 4

N = n(n−1)/2 = 3(3−1)/2 = 3. Transitions: 3→1, 3→2, 2→1

NEET Graph-Based & Conceptual Questions — Set 3

NEET 2024

Q36. In Bohr's model, if the electron's energy is plotted against n (orbit number), the graph shape is:

(A) Linear (straight line)

(B) Parabolic

(D) Exponential

Eₙ = −13.6/n² eV. This is a 1/n² function — hyperbolic decrease in magnitude as n increases, approaching 0 from below.

NEET 2023

Q37. How does the total energy of an electron vary with principal quantum number n (for fixed Z)?

(A) E ∝ 1/n² (increases, becomes less negative) ✓

(B) E ∝ n

(D) E ∝ −n

Eₙ = −13.6/n² eV. As n increases, |Eₙ| decreases, total energy becomes less negative (increases towards 0).

NEET 2022

Q38. According to Bohr's model, the angular momentum of an electron in the 4th orbit is:

(A) h/π

(B) h/2π

(D) 2h/π ✓

L = nh/2π = 4h/2π = 2h/π

NEET 2021

Q39. In which transition is a Balmer series photon NOT produced?

(A) 4 → 2

(B) 5 → 2

(D) 5 → 3 ✓

Balmer series: transitions that END at n=2. 5→3 ends at n=3 (Paschen series), not Balmer.

NEET 2020

Q40. The ratio of the longest wavelength in Lyman series to the longest wavelength in Balmer series is:

(A) 5/27

(B) 27/5

(D) 5/27 ✓

Lyman longest (2→1): 1/λ_L = R(1−1/4) = 3R/4 → λ_L = 4/(3R)
Balmer longest (3→2): 1/λ_B = R(1/4−1/9) = 5R/36 → λ_B = 36/(5R)
λ_L/λ_B = [4/(3R)]/[36/(5R)] = (4×5)/(3×36) = 20/108 = 5/27

NEET 2019

Q41. The ground state energy of hydrogen atom is −13.6 eV. If an electron makes a transition from energy level −0.85 eV to −1.51 eV, calculate the wavelength of spectral line emitted.

(A) 1.46 μm

(B) 2.0 μm

(D) 0.97 μm

ΔE = |−0.85 − (−1.51)| = 0.66 eV. λ = 1240/0.66 nm = 1878.8 nm ≈ 1.88 μm (Paschen series, 5→3... Actually n=4 (E=−0.85) to n=3 (E=−1.51): Paschen series)

NEET 2018

Q42. An α-particle travels in a circular path of radius 0.45 m in a magnetic field B = 1.2 Wb/m². The kinetic energy is (mass of α = 6.4×10⁻²⁷ kg):

(A) 4 MeV

(B) 6 MeV

(D) 2 MeV

r = mv/qB → v = qBr/m. KE = ½mv² = q²B²r²/(2m) = (3.2×10⁻¹⁹)²×1.44×0.2025/(2×6.4×10⁻²⁷) ≈ 5.2×10⁻¹³ J ≈ 3.2 MeV

NEET 2017

Q43. The ratio of kinetic energy of electron in 2nd orbit to 3rd orbit of hydrogen atom is:

(A) 4:9

(B) 2:3

(D) 3:2

KE ∝ 1/n². KE₂/KE₃ = (1/4)/(1/9) = 9/4

NEET 2016

Q44. An electron in hydrogen atom makes a transition n₁ → n₂ (n₁ > n₂). The time period T₁ and T₂ satisfy T₁/T₂ = 27/1. If n₂ = 2, find n₁.

(A) 3

(B) 4

(D) 8

T ∝ n³. T₁/T₂ = (n₁/n₂)³ = 27. (n₁/n₂)³ = 27 → n₁/n₂ = 3. n₂ = 2 → n₁ = 6.

NEET 2015

Q45. In Bohr's model, the magnetic field at the centre of the atom due to the motion of electron in nth orbit is proportional to:

(A) n³

(B) 1/n³

(D) 1/n

B = μ₀I/(2r). I = ev/2πr ∝ (v/n²) ∝ (1/n)/n² = 1/n³. B ∝ I/r ∝ (1/n³)/(n²) = 1/n⁵

NEET 2014

Q46. Which of the following graphs correctly represents the variation of electron speed with orbit number in hydrogen atom?

(A) v increases linearly with n

(B) v increases as n²

(D) v is constant

vₙ = v₁/n. Speed decreases hyperbolically as n increases.

NEET 2013

Q47. For hydrogen atom, the wavelength of the spectral line of the Lyman series for the transition n=3 → n=1 is:

(A) 102.5 nm ✓

(B) 121.6 nm

(D) 656 nm

1/λ = R(1/1−1/9) = 8R/9. λ = 9/(8×1.097×10⁷) = 9/(8.776×10⁷) = 1.026×10⁻⁷ m = 102.6 nm

NEET 2012

Q48. The binding energy of an electron in the ground state of He atom is 24.6 eV. The energy required to remove both electrons is:

(A) 24.6 eV

(B) 49.2 eV

(D) 79 eV ✓

First electron: 24.6 eV (from neutral He). After first electron removed, He⁺ is like H with Z=2: E = −54.4 eV. So second electron needs 54.4 eV. Total = 24.6 + 54.4 = 79 eV.

NEET 2011

Q49. The ionisation energy of the hydrogen atom is 13.6 eV. An electron with energy 12.5 eV can excite a ground state hydrogen atom to:

(A) n = 2 only

(B) n = 4

(D) n = 5

E(n=3) − E(n=1) = −1.51−(−13.6) = 12.09 eV < 12.5 eV ✓. E(n=4)−E(1) = 12.75 eV > 12.5 eV ✗. So max excitation is n=3.

NEET 2010

Q50. Which series of hydrogen atom spectrum is in the UV region?

(A) Balmer

(B) Paschen

(D) Brackett

Lyman series (n₁=1, n₂=2,3,...): wavelengths 91.2–121.6 nm — ultraviolet region.

🎯 Section 10: JEE Main Previous Year Questions

Single correct, integer type and numerical value questions

JEE Main — Single Correct

JEE Main 2024 (Jan)

Q1. If the kinetic energy of an electron in the first Bohr orbit of hydrogen is K, the potential energy in the second orbit is:

(A) K/4

(B) −K/4

(D) −K/2 ✓

KE₁ = K (n=1). KE₂ = K/4. PE₂ = 2×E₂ = 2×(−KE₂) = −K/2

JEE Main 2024 (Apr)

Q2. The wavelength of the first line of the Lyman series for hydrogen is identical to that of the second line of the Balmer series for some hydrogen-like atom X. Find the atomic number of X.

(A) 1

(B) 2

(D) 4

Lyman 1st (H): 1/λ = R(1−1/4) = 3R/4
Balmer 2nd (X, Z): 1/λ = RZ²(1/4−1/16) = 3RZ²/16
Setting equal: 3R/4 = 3RZ²/16 → Z² = 4 → Z = 2. Wait: 3R/4 = 3RZ²/16 → Z² = 4 → Z = 2. Hmm. Some versions: Lyman 1st H, Balmer 2nd atom X (transition 4→2). 3R/4 = Z²×3R/16 → Z²=4 → Z=2. Answer should be 2 for He⁺, but let me check: for Z=3, 3R/4 =? 3×9R/16 = 27R/16 ≠. Z=2: 3×4R/16 = 3R/4 ✓. Z=2.

JEE Main 2023 (Jan)

Q3. In hydrogen atom, the orbital speed of electron in nth orbit is proportional to:

(A) n

(B) n²

(D) 1/n ✓

vₙ = v₁/n ∝ 1/n

JEE Main 2023 (Apr)

Q4. The graph of the de Broglie wavelength of an electron versus its energy E in Bohr orbit of hydrogen is:

(A) Linear

(B) Exponential

(D) Parabolic

λ = h/p = h/√(2mE). So λ ∝ 1/√E — inverse square root relationship.

JEE Main 2022 (Jun)

Q5. In an hydrogen-like atom, when an electron jump from M shell to L shell, the wavelength of emitted radiation is λ. If we consider the nuclear charge to be doubled (Z→2Z), what will be the new wavelength?

(A) λ

(B) 2λ

(D) λ/2

1/λ ∝ Z². If Z doubles, 1/λ becomes 4 times larger, so λ becomes λ/4.

JEE Main 2022 (Jul)

Q6. Hydrogen atom is excited from ground state to n=4 state. How many spectral lines are possible in emission spectrum?

(A) 3

(B) 5

(D) 4

N = n(n−1)/2 = 4×3/2 = 6

JEE Main 2021 (Feb)

Q7. The shortest wavelength in Paschen series of hydrogen is: (R_H = 1.097 × 10⁷ m⁻¹)

(A) 8.2 × 10⁻⁷ m

(B) 12.8 × 10⁻⁷ m

(D) 18.75 × 10⁻⁷ m

Shortest λ in Paschen: n₂→∞, n₁=3. 1/λ = R/9. λ = 9/R = 9/1.097×10⁷ = 8.2×10⁻⁷ m

JEE Main 2021 (Mar)

Q8. A 12.09 eV photon is absorbed by hydrogen atom. Calculate the orbital angular momentum of the excited electron.

(A) h/π

(B) h/2π

(D) h/π ✓

12.09 eV = excitation to n=3 (E₃−E₁ = −1.51+13.6 = 12.09 eV). L = nħ = 3h/2π = h/π × 3/2... For n=3: L = 3h/2π. Option (C) = 3h/2π ✓

JEE Main 2020 (Jan)

Q9. Consider an electron in the nth orbit of hydrogen atom in Bohr model. The circumference of the orbit can be expressed in terms of the de Broglie wavelength λ of the electron as:

(A) λ/n

(B) nλ/2

(D) √n × λ

2πrₙ = nλ (de Broglie condition for standing waves). Circumference = nλ.

JEE Main 2020 (Sep)

Q10. An alpha particle is accelerated through a potential difference of 1 MV. Its kinetic energy is:

(A) 1 MeV

(B) 0.5 MeV

(D) 4 MeV

KE = qV = 2e × 1MV = 2 MeV (α-particle has charge 2e)

JEE Main 2019 (Jan)

Q11. In Bohr's model of hydrogen, the speed of the electron in the 4th orbit is approximately:

(A) 5.45 × 10⁵ m/s

(B) 2.18 × 10⁶ m/s

(D) 5.45 × 10⁵ m/s ✓

v₄ = v₁/4 = 2.18×10⁶/4 = 0.545×10⁶ = 5.45×10⁵ m/s

JEE Main 2019 (Apr)

Q12. The energy required to ionise a hydrogen atom from n=2 state is:

(A) 13.6 eV

(B) 10.2 eV

(D) 3.4 eV ✓

|E₂| = 13.6/4 = 3.4 eV. This is the ionisation energy from n=2 state.

JEE Main 2018

Q13. A particle of mass m moves in a circular orbit of radius r. Its angular momentum is L. The kinetic energy is:

(A) L²/(2mr²)

(B) L²/(2mr²) ✓

(D) 2L²/mr²

L = mvr → v = L/(mr). KE = ½mv² = ½m × L²/(m²r²) = L²/(2mr²)

JEE Main 2018

Q14. The electron of hydrogen atom is in 5th orbit. How many maximum spectral lines can be emitted?

(A) 5

(B) 8

(D) 15

N = n(n−1)/2 = 5×4/2 = 10

JEE Main 2017

Q15. If the wavelength of photon emitted due to transition of electron from third orbit to first orbit in a hydrogen atom is λ, then the wavelength of photon emitted due to transition from fourth orbit to second orbit is:

(A) 208λ/27

(B) 16λ/3

(D) 27λ/208

3→1: 1/λ = R(1−1/9) = 8R/9 → λ = 9/(8R)
4→2: 1/λ' = R(1/4−1/16) = 3R/16 → λ' = 16/(3R)
λ'/λ = [16/(3R)]/[9/(8R)] = 128/27. Hmm. λ' = 128λ/27? Let me recalculate: 16×8/(3×9) = 128/27. So λ'= 128λ/27. None of options listed... checking option C: 36λ/7 = 36/7 × λ. 128/27 ≈ 4.74; 36/7 ≈ 5.14. Not matching. This problem has different versions.

JEE Main 2016

Q16. In a hydrogen-like atom, if the radii of successive orbits are in ratio 1:4, the atom must be:

(A) Any H-like atom

(B) H atom only

(D) Cannot determine

r₁:r₂ = 1²:2² = 1:4 for any hydrogen-like atom (ratio depends only on n, not Z). So all H-like atoms satisfy this for n=1 and n=2.

JEE Main 2015

Q17. In the Bohr model, the ratio of the period of revolution of the electron in n=2 to n=1 orbit is:

(A) 2

(B) 4

(D) 8 ✓

T ∝ n³. T₂/T₁ = 2³ = 8

JEE Main 2015

Q18. The angular momentum of an electron in 3rd orbit of He⁺ is:

(A) h/2π

(B) 2h/π

(D) 6h/2π

L = nh/2π = 3h/2π. Angular momentum does NOT depend on Z — same formula for all H-like atoms.

JEE Main 2014

Q19. The ratio of minimum wavelength of Lyman series to that of Paschen series is:

(A) 7/8

(B) 1/9

(D) 4/9

λ_min(Lyman) = 1/R; λ_min(Paschen) = 9/R. Ratio = (1/R)/(9/R) = 1/9

JEE Main 2014

Q20. Hydrogen atom is bombarded with 10.2 eV electrons. It will emit:

(A) Only Lyman series photons

(B) Lyman + Balmer series photons

(D) No photons

10.2 eV exactly matches E₂−E₁ (excitation to n=2). Atom goes from n=1 to n=2, then falls back to n=1 emitting 121.6 nm (Lyman-α only). Cannot reach n=3 (needs 12.09 eV).

JEE Main — Integer Type / Numerical Value

Q21. The sum of orbital angular momenta of electrons in 1st, 2nd, and 3rd Bohr orbits of hydrogen in units of h/2π is: (JEE Main 2023)

L₁ = 1×h/2π, L₂ = 2×h/2π, L₃ = 3×h/2π
Sum = (1+2+3)×h/2π = 6h/2π

Sum = 6 units of ħ (i.e., 6ħ = 6h/2π)

Q22. The number of spectral lines emitted by hydrogen when electrons drop from n=6 to ground state: (JEE Main 2022)

N = n(n−1)/2 = 6×5/2 = 15

15 spectral lines

Q23. If the energy difference between ground state and first excited state of hydrogen is 10.2 eV, what is the energy difference between first and second excited states? (JEE Main 2021)

E₂−E₁ = −3.4−(−13.6) = 10.2 eV ✓
E₃−E₂ = −1.51−(−3.4) = 1.89 eV

1.89 eV

Q24. In hydrogen atom, the wavelength (in nm) of photon emitted in the transition from n=5 to n=3 is: (JEE Main 2020) (Use 1/R = 91.2 nm)

1/λ = R(1/9 − 1/25) = R × 16/225
λ = 225/(16R) = 225×91.2/16 = 20520/16 = 1282.5 nm

λ ≈ 1282 nm (Paschen series, infrared)

Q25. An electron in hydrogen atom is in the third excited state. How many spectral lines are possible from this state? (JEE Main 2019)

Third excited state = n=4 (ground=1, 1st excited=2, 2nd=3, 3rd=4).
N = 4(4−1)/2 = 6

6 spectral lines

🏆 Section 11: JEE Advanced Previous Year Questions

Single correct, multiple correct, paragraph and matrix match questions

JEE Advanced — Single Correct

JEE Advanced 2023

Q1. In Bohr model, for a hydrogen-like atom with atomic number Z, the minimum energy required to ionise the atom from its first excited state is E. The value of E in terms of E₀ = 13.6 eV is:

(A) Z²E₀/4

(B) Z²E₀/4 ✓

(D) ZE₀/4

First excited state: n=2. IE from n=2: |E₂| = 13.6Z²/4 = Z²E₀/4

JEE Advanced 2022

Q2. A proton is fired toward a gold nucleus (Z=79) from far away with kinetic energy K. The distance of closest approach is d. If the proton were replaced by an α-particle of the same kinetic energy K, the distance of closest approach would be:

(A) d/2

(B) d

(D) 4d

For proton: d = kZ(e)(e)/K = kZe²/K
For α-particle (charge 2e): d' = kZ(e)(2e)/K = 2kZe²/K = 2d

JEE Advanced 2021

Q3. In the Bohr model, the electron in nth orbit has kinetic energy Kₙ. Match the following for hydrogen: K₁, K₂, total energy E₁, E₂.

(A) K₁ > K₂, E₁ < E₂

(B) K₁ > K₂, E₁ < E₂ ✓

(D) K₁ = K₂

KE ∝ 1/n²: K₁ = 13.6 eV > K₂ = 3.4 eV. Total E ∝ −1/n²: E₁ = −13.6 < E₂ = −3.4 (more negative = lower energy)

JEE Advanced 2020

Q4. In a hydrogen atom, the transition n=4 to n=2 occurs. Using Rydberg formula, what is the value of 1/λ in terms of R?

(A) 3R/16

(B) 3R/16 ✓

(D) 5R/36

1/λ = R(1/4 − 1/16) = R(4−1)/16 = 3R/16

JEE Advanced 2019

Q5. An electron in a hydrogen atom jumps from n=1 to n=4, and then immediately falls back to n=1 emitting a photon. The process repeats. Which of the following spectral series can be observed from a large collection of hydrogen atoms excited to n=4?

(A) Lyman only

(B) Balmer only

(D) Lyman + Balmer + Paschen ✓

From n=4, all possible transitions: 4→3 (Paschen), 4→2 (Balmer), 4→1 (Lyman), 3→2 (Balmer), 3→1 (Lyman), 2→1 (Lyman). All three series observed.

JEE Advanced — Multiple Correct

JEE Advanced 2023

Q6. Which of the following statements about the Bohr model of hydrogen atom are CORRECT? (Choose all that apply)

(A) Radius of nth orbit ∝ n² ✓

(B) Speed of electron ∝ 1/n ✓

(D) Angular momentum is quantised ✓

A: rₙ ∝ n² ✓. B: vₙ ∝ 1/n ✓. C: Eₙ ∝ −1/n² (not n) ✗. D: L = nh/2π is Bohr's postulate ✓.

JEE Advanced 2022

Q7. For a hydrogen atom in the ground state, which of the following are correct?

(A) Total energy = −13.6 eV ✓

(B) KE = +13.6 eV ✓

(D) KE = 2|Total energy|

A: E₁ = −13.6 eV ✓. B: KE = −E₁ = +13.6 eV ✓. C: PE = −27.2 eV = 2×(−13.6), so |PE| = 2|E| = 2|KE| ✓. D: KE = |Total energy|, not 2× ✗.

JEE Advanced 2021

Q8. Which of the following transitions in hydrogen will produce photons belonging to the Balmer series?

(A) n = 3 → n = 2 ✓

(B) n = 5 → n = 2 ✓

(D) n = 4 → n = 2 ✓

Balmer series: lower level n₁ = 2. So any transition ending at n=2 is Balmer. A, B, D all end at n=2 ✓. C ends at n=1 → Lyman ✗.

JEE Advanced — Paragraph Based

Paragraph (JEE Advanced 2022): Hydrogen Spectrum

The emission spectrum of hydrogen contains series of spectral lines. In the Balmer series, the lines are in the visible region. The wavelengths of visible Balmer lines correspond to transitions from higher energy levels (n ≥ 3) down to n = 2. The Rydberg constant for hydrogen is R_H = 1.097 × 10⁷ m⁻¹. The speed of light c = 3 × 10⁸ m/s and h = 6.626 × 10⁻³⁴ J·s.

Q(i) Calculate the wavelength of the first line of the Balmer series.

1/λ = R_H(1/4 − 1/9) = R_H × 5/36
λ = 36/(5R_H) = 36/(5 × 1.097×10⁷) = 36/5.485×10⁷ = 6.56×10⁻⁷ m

λ = 656.3 nm (H-α, red line)

Q(ii) What is the series limit of the Balmer series?

n₂→∞: 1/λ = R_H/4. λ = 4/R_H = 4/1.097×10⁷ = 3.646×10⁻⁷ m

λ_limit = 364.6 nm

Q(iii) In what region does the series limit of Balmer series lie?

364.6 nm lies in the UV (near ultraviolet) region, just below visible light (400–700 nm).

Paragraph (JEE Advanced 2021): Bohr Model Calculations

Consider the Bohr model for a hydrogen-like atom with atomic number Z. The electron's energy in nth orbit is Eₙ = −13.6Z²/n² eV and radius is rₙ = 0.529n²/Z Å.

Q(i) For what value of Z does the first orbit radius of the hydrogen-like atom equal the second orbit radius of hydrogen?

r₁(Z-atom) = r₂(H): 0.529×1/Z = 0.529×4/1 → 1/Z = 4 → Z = 1/4?
Wait: r₁ = 0.529/Z Å. r₂(H) = 0.529×4 = 2.116 Å. 0.529/Z = 2.116 → Z = 0.529/2.116 = 0.25? That's not an integer.
Alternatively: Which Z gives r₁ = 2a₀? 0.529×1²/Z = 2×0.529 → Z = 1/2? Not physical.
Interpretation: Which atom's 2nd orbit = H's 1st orbit? r₂(Z)/r₁(H) = 1: 4/Z = 1 → Z = 4 (Beryllium Be³⁺)

Z = 4 (Be³⁺): r₂(Be³⁺) = 0.529×4/4 = 0.529 Å = r₁(H)

Q(ii) Find the ground state energy of He⁺ (Z=2).

E₁(He⁺) = −13.6 × Z²/n² = −13.6 × 4/1 = −54.4 eV

E₁(He⁺) = −54.4 eV; IE = 54.4 eV

JEE Advanced — Numerical Answer Type

Q9. In a hydrogen atom, the electron moves with speed 2.18×10⁶ m/s. Find the magnetic moment of this orbiting electron (in units of 9.27×10⁻²⁴ A·m²). (JEE Advanced 2022)

For n=1: r = 0.529×10⁻¹⁰ m, v = 2.18×10⁶ m/s
Current I = ev/(2πr) = 1.6×10⁻¹⁹ × 2.18×10⁶ / (2π × 0.529×10⁻¹⁰)
I = 3.488×10⁻¹³ / (3.32×10⁻¹⁰) = 1.05×10⁻³ A
Magnetic moment μ = I × πr² = 1.05×10⁻³ × π × (0.529×10⁻¹⁰)²
μ = 1.05×10⁻³ × 8.8×10⁻²¹ = 9.24×10⁻²⁴ A·m²

μ ≈ 9.27×10⁻²⁴ A·m² = 1 Bohr magneton (μ_B)

💡 This is the Bohr Magneton: μ_B = eħ/(2m) = 9.274×10⁻²⁴ J/T

Q10. Find the ratio of minimum wavelength in Lyman series to maximum wavelength in Paschen series. (JEE Advanced 2021)

Min λ in Lyman (n₂→∞, n₁=1): 1/λ_L = R. λ_L = 1/R
Max λ in Paschen (n₂=4, n₁=3): 1/λ_P = R(1/9−1/16) = 7R/144. λ_P = 144/(7R)
Ratio = λ_L/λ_P = (1/R)/(144/(7R)) = 7/144

λ_L(min)/λ_P(max) = 7/144

🌍 Section 12: IB Physics Questions

IB SL and HL questions on atomic structure and spectra

IB SL Questions

IB SL 2023

Q1. State what is meant by the ground state of an atom.

✓ The ground state is the lowest energy state of an atom, in which the electron occupies the innermost allowed orbit (n=1). The atom is most stable in this state.

IB SL 2023

Q2. An electron in a hydrogen atom makes a transition from the n=3 level to the n=1 level. Calculate the wavelength of the emitted photon. [Use R_H = 1.10×10⁷ m⁻¹]

✓ 1/λ = R(1/1 − 1/9) = 8R/9 = 8×1.10×10⁷/9 = 9.78×10⁶ m⁻¹
λ = 1/9.78×10⁶ = 1.02×10⁻⁷ m = 102 nm (UV, Lyman series)

IB SL 2022

Q3. Explain why atoms emit only discrete spectral lines (line spectra) rather than a continuous spectrum.

✓ Electrons in atoms can only exist in discrete energy levels (quantised energies). When an electron transitions from a higher to a lower energy level, it emits a photon of energy equal to the energy difference: E = hf. Since only discrete energy differences are possible, only discrete frequencies (wavelengths) of photons are emitted → line spectrum.

IB SL 2022

Q4. The ionisation energy of hydrogen is 13.6 eV. Calculate the minimum frequency of radiation that can ionise a ground-state hydrogen atom.

✓ E = hf → f = E/h = (13.6 × 1.6×10⁻¹⁹) / (6.63×10⁻³⁴) = 2.176×10⁻¹⁸/6.63×10⁻³⁴ = 3.28×10¹⁵ Hz

IB SL 2021

Q5. State the significance of the negative sign in the expression for the energy of an electron in a Bohr orbit.

✓ The negative sign indicates that the electron is bound to the nucleus. Energy must be supplied to remove the electron. The total energy of the bound electron is less than that of a free electron at rest (E = 0 at infinity). The atom is in a stable bound state.

IB SL 2021

Q6. An atom absorbs a photon and an electron is excited from n=2 to n=4. Calculate the energy of this photon for hydrogen.

✓ ΔE = 13.6(1/4 − 1/16) = 13.6 × 3/16 = 2.55 eV = 2.55 × 1.6×10⁻¹⁹ = 4.08×10⁻¹⁹ J

IB SL 2020

Q7. Outline Rutherford's model of the atom and state one piece of experimental evidence that supports this model.

✓ Rutherford model: The atom consists of a tiny (∼10⁻¹⁵ m), massive, positively charged nucleus at the centre, surrounded by electrons moving in orbits at large distances. The atom is mostly empty space (∼10⁻¹⁰ m radius).
Evidence: The observation that a small fraction (∼1 in 8000) of α-particles were deflected through angles greater than 90°, and some even bounced back, showed that there must be a massive, concentrated, positively charged region — the nucleus — within the atom.

IB SL 2020

Q8. Hydrogen atoms can be excited to the n=3 level. Determine the number of possible spectral lines that can be emitted as atoms return to the ground state.

✓ N = n(n-1)/2 = 3(3-1)/2 = 3. Lines: 3→2 (Balmer), 3→1 (Lyman), 2→1 (Lyman)

IB SL 2019

Q9. Explain the emission spectrum of atomic hydrogen in terms of electron energy transitions.

✓ Hydrogen atoms are excited by heat or electrical discharge. Electrons are promoted to higher energy levels. When they fall back to lower levels, photons are emitted. Photon energy = difference between energy levels: E_photon = E₂ − E₁ = hf. Since energy levels are discrete, photon frequencies are discrete → line spectrum. Different transitions → different series (Lyman, Balmer, Paschen).

IB SL 2019

Q10. The first line of the Balmer series has wavelength 656 nm. Calculate the frequency of this radiation and identify the series.

✓ f = c/λ = 3×10⁸/656×10⁻⁹ = 4.57×10¹⁴ Hz. This is in the visible red region. This line (H-α) is the first line of the Balmer series (n=3→n=2 transition).

IB HL Questions

Q11. IB HL 2023: Derive the Rydberg formula using Bohr model. (Long answer)

1. Bohr condition: mvr = nh/2π
2. Coulomb = Centripetal: mv²/r = ke²/r² → r = n²h²/(4π²mke²) = n²a₀
3. Energy: Eₙ = −ke²/2r = −me⁴k²/(2h²n²) = −13.6/n² eV
4. Photon from transition n₂→n₁: hν = Eₙ₂ − Eₙ₁ = 13.6(1/n₁² − 1/n₂²) eV
5. ν/c = 1/λ = 13.6/hc × (1/n₁² − 1/n₂²) = R_H(1/n₁² − 1/n₂²)
where R_H = me⁴k²/(4πħ³c) = 1.097×10⁷ m⁻¹

1/λ = R_H(1/n₁² − 1/n₂²) where R_H = 1.097×10⁷ m⁻¹

Q12. IB HL 2022: Using data analysis, the series limit of Balmer series is 365 nm and first line is 656 nm. Verify Rydberg formula for both.

Series limit (n₂→∞): 1/λ = R/4 → λ = 4/R = 4/1.097×10⁷ = 3.646×10⁻⁷ m = 364.6 nm ≈ 365 nm ✓
First line (n=3→2): 1/λ = R(1/4−1/9) = 5R/36 → λ = 36/(5R) = 36/(5×1.097×10⁷) = 6.56×10⁻⁷ m = 656 nm ✓

Both values verified. Rydberg formula is consistent with experimental data.

Q13. IB HL 2021: Why does the Bohr model succeed for hydrogen but fail for multi-electron atoms?

Success for H: Only one electron — no electron-electron repulsion. The only forces are the nucleus-electron Coulomb attraction. Bohr's circular orbit model with quantised angular momentum gives exact energy levels for H.
Failure for multi-electron atoms: In multi-electron atoms, electrons repel each other (electron-electron interaction). This makes the problem much more complex — each electron's orbit is affected by all others. Bohr's model cannot handle these many-body interactions. Also, electron shielding (inner electrons screen the nuclear charge for outer electrons) requires quantum mechanical treatment.

Bohr model ignores electron-electron repulsion — valid only for one-electron systems (H, He⁺, Li²⁺).

Q14. IB HL 2020: An electron in H makes transition from n=4 to n=2. (a) State the series. (b) Calculate λ. (c) Explain why the atom is said to undergo de-excitation.

(a) n=4→n=2: Balmer series (visible region)
(b) 1/λ = R(1/4−1/16) = 3R/16. λ = 16/(3R) = 16/(3×1.097×10⁷) = 4.86×10⁻⁷ m = 486 nm (H-β, blue-green)
(c) De-excitation: the electron moves from a higher energy level to a lower energy level, releasing energy in the form of a photon. The atom's total energy decreases — it becomes less energetic, returning toward lower energy states.

(a) Balmer (b) λ = 486 nm (c) Electron moves to lower level, losing energy as emitted photon

Q15. IB HL 2019: The electron in ground state H has speed 2.18×10⁶ m/s and r=5.3×10⁻¹¹ m. Find (a) the KE, (b) the PE, (c) show that total energy = −13.6 eV.

(a) KE = ½mv² = ½×9.11×10⁻³¹×(2.18×10⁶)² = ½×9.11×10⁻³¹×4.75×10¹² = 2.16×10⁻¹⁸ J = 13.5 eV ≈ 13.6 eV
(b) PE = −ke²/r = −9×10⁹×(1.6×10⁻¹⁹)²/(5.3×10⁻¹¹) = −9×10⁹×2.56×10⁻³⁸/5.3×10⁻¹¹ = −4.35×10⁻¹⁸ J = −27.2 eV
(c) Total E = KE + PE = 13.6 + (−27.2) = −13.6 eV ✓

(a) 13.6 eV (b) −27.2 eV (c) Total = −13.6 eV ✓

📚 Section 13: IGCSE Questions

Cambridge IGCSE Physics questions on atomic structure

IGCSE MCQs

Cambridge IGCSE 2023

Q1. An atom emits a photon when:

(A) An electron gains energy

(B) An electron moves from a higher to lower energy level ✓

(D) A proton changes orbit

When an electron moves from a higher to a lower energy level, it releases energy as a photon: E_photon = E₂ − E₁ = hf

Cambridge IGCSE 2023

Q2. Which statement about the nucleus of an atom is correct?

(A) It contains most of the atom's volume

(B) It is negatively charged

(D) Its size is the same as the atom

The nucleus is tiny (∼10⁻¹⁵ m) but contains protons and neutrons — nearly all of the atom's mass. The atom's volume is mostly empty space.

Cambridge IGCSE 2022

Q3. What was the key evidence from Rutherford's experiment that showed the nucleus is small?

(A) Most α-particles were deflected

(B) All α-particles passed straight through

(D) No α-particles reached the detector

Very few large-angle deflections (∼1 in 8000) means the nucleus is tiny — most α-particles never come close to it. This proves the nucleus is very small compared to the atom.

Cambridge IGCSE 2022

Q4. A hydrogen atom absorbs a photon of energy 10.2 eV. This causes the electron to move from:

(A) n=2 to n=3

(B) n=3 to n=1

(D) n=2 to n=1

E₂−E₁ = 10.2 eV. Absorption of 10.2 eV photon excites electron from n=1 to n=2.

Cambridge IGCSE 2021

Q5. What is meant by the ionisation of an atom?

(A) An electron moves to a higher orbit

(B) The nucleus splits

(D) A photon is absorbed

Ionisation = complete removal of an electron from the atom. The atom becomes an ion. Energy required = ionisation energy (13.6 eV for hydrogen ground state).

IGCSE Structured Questions

Q6. [IGCSE 2023] Describe Rutherford's alpha particle scattering experiment and explain the model of the atom it supports. (6 marks)

Description (3 marks):
• A thin gold foil is bombarded with alpha (α) particles from a radioactive source.
• The α-particles are collimated into a narrow beam by a lead plate with a small hole.
• A moveable detector (zinc sulphide screen) records where scattered α-particles land.

Observations and model (3 marks):
• Most α-particles pass straight through (undeflected) → atom is mostly empty space.
• A small number are deflected at various angles → there is a positively charged region inside the atom that repels α-particles.
• Very few (∼1 in 8000) bounce back at ∼180° → the positive charge is concentrated in a very small, dense region (the nucleus).

Nuclear Model: The atom has a tiny, dense, positively charged nucleus at the centre, surrounded by electrons orbiting at large distances. The atom is mostly empty space.

Nuclear model: tiny, dense positive nucleus at centre; electrons in orbit; mostly empty space.

Q7. [IGCSE 2022] A hydrogen atom emits a photon of wavelength 656 nm. (a) Calculate the energy of this photon in eV. (b) Identify the transition.

(a) E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸) / (656×10⁻⁹)
= 1.989×10⁻²⁵ / 6.56×10⁻⁷ = 3.03×10⁻¹⁹ J
E(eV) = 3.03×10⁻¹⁹ / 1.6×10⁻¹⁹ = 1.89 eV

(b) ΔE = 1.89 eV = E₃ − E₂ = −1.51 − (−3.4) = 1.89 eV ✓
This is the transition n=3 → n=2 (first line of Balmer series, H-α)

(a) E = 1.89 eV (b) n=3 → n=2, Balmer series (visible red)

Q8. [IGCSE 2021] Explain why a hydrogen gas discharge lamp emits light of specific colours only.

1. In a gas discharge lamp, electrical energy excites hydrogen atoms — electrons absorb energy and jump to higher energy levels.
2. Electrons are unstable in these higher levels and fall back to lower levels.
3. When an electron falls from a higher level to a lower level, it emits a photon.
4. The photon's energy = difference between the two energy levels: E = hf.
5. Since energy levels in hydrogen are discrete (quantised), only specific energy differences (and thus specific frequencies/colours) are possible.
6. This produces a line spectrum with specific wavelengths — e.g., red (656 nm), blue-green (486 nm), violet (434 nm) in the visible Balmer series.

Discrete energy levels → discrete photon energies → specific frequencies/colours only (line spectrum)

Q9. [IGCSE 2020] The ground state energy of hydrogen is −13.6 eV. Calculate the energy needed to move an electron from n=2 to n=4.

E₂ = −13.6/4 = −3.4 eV; E₄ = −13.6/16 = −0.85 eV
Energy needed = E₄ − E₂ = −0.85 − (−3.4) = 2.55 eV
In Joules: 2.55 × 1.6×10⁻¹⁹ = 4.08×10⁻¹⁹ J

ΔE = 2.55 eV = 4.08×10⁻¹⁹ J (energy absorbed)

Q10. [IGCSE 2020] State two limitations of Rutherford's model of the atom.

1. Cannot explain atomic stability: A classical electron in circular orbit accelerates → should radiate energy → spiral into nucleus within ∼10⁻⁸ s. But atoms are observed to be stable.
2. Cannot explain discrete spectral lines: Classical theory predicts a continuous spectrum as the electron spirals inward with continuously varying frequency. But atomic spectra show sharp, discrete lines.

1. Cannot explain stability (electron should spiral in). 2. Cannot explain line spectra (should emit continuous spectrum).

IGCSE Theory Questions

Q11. [IGCSE 2023] Draw and label a diagram showing the first three energy levels of hydrogen atom and show two possible electron transitions.

Draw three horizontal lines labelled:
n=3: E = −1.51 eV (top)
n=2: E = −3.4 eV (middle)
n=1: E = −13.6 eV (bottom, ground state)
Draw downward arrows for transitions:
Arrow 1: n=3 → n=1 (Lyman, UV) with energy 12.09 eV
Arrow 2: n=2 → n=1 (Lyman, UV, 10.2 eV) or n=3→n=2 (Balmer, visible, 1.89 eV)

See Energy Level Diagram in Section 5. Label energy values and series names for full marks.

Q12. [IGCSE 2022] Distinguish between emission spectrum and absorption spectrum of hydrogen.

Emission spectrum:
• Produced when excited hydrogen atoms emit photons as electrons fall to lower levels.
• Appears as bright coloured lines on a dark background.
• Each line corresponds to a specific electron transition (higher → lower level).

Absorption spectrum:
• Produced when white light passes through hydrogen gas — atoms absorb photons.
• Appears as dark lines on a continuous (rainbow) spectrum background.
• Dark lines occur at the same wavelengths as the emission lines — electrons absorb specific energies to jump to higher levels.
• The two spectra are complementary (same wavelengths, opposite appearance).

Emission: bright lines on dark background (photons emitted). Absorption: dark lines on continuous background (photons absorbed). Same wavelengths.

Q13. [IGCSE 2021] What is the de Broglie hypothesis and how does it justify Bohr's quantisation condition?

de Broglie hypothesis: All matter has wave properties. The de Broglie wavelength is λ = h/p = h/(mv).

Justification of Bohr's condition: For a stable orbit, the electron's wave must form a standing wave around the orbit. This means the circumference must be an integer multiple of the wavelength:
2πr = nλ = n(h/mv)
→ mvr = nh/2π = nħ
This is exactly Bohr's angular momentum quantisation condition! de Broglie's hypothesis gives a physical reason for Bohr's postulate.

Standing wave condition: 2πr = nλ → mvr = nħ (Bohr's condition justified by wave nature of electron)

Q14–Q20. [IGCSE Mixed Questions] (Click to expand each)

Q14. Which has more energy: n=2 or n=3 electron in H? Answer: n=3 (E₃ = −1.51 eV > E₂ = −3.4 eV, less negative = more energy)

Q15. Name the series of spectral lines in visible region. Answer: Balmer series (n₁=2, n₂≥3)

Q16. Why are spectral lines of hydrogen called line spectra? Answer: Because they appear as discrete, narrow lines at specific wavelengths — not a continuous band of colours.

Q17. An electron falls from n=4 to n=3. In which spectral series is this line? Answer: Paschen series (transitions ending at n=3)

Q18. State Bohr's frequency condition. Answer: hν = E₂ − E₁ where ν is the frequency of emitted/absorbed photon, and E₂, E₁ are energies of higher and lower levels respectively.

Q19. What happens to the energy of a photon emitted when n₂ increases (with n₁ fixed)? Answer: Energy increases (as n₂ increases, 1/n₂² decreases, so 1/n₁² − 1/n₂² increases, hence ΔE increases)

Q20. Why is the ionisation energy of He⁺ greater than that of H? Answer: He⁺ has Z=2, so E₁ = −13.6×4 = −54.4 eV. IE(He⁺) = 54.4 eV > IE(H) = 13.6 eV. Greater nuclear charge → stronger binding → more energy needed to remove electron.

🎓 Section 14: A-Level Questions

Cambridge A-Level and Edexcel A-Level structured and calculation questions

Cambridge A-Level Structured Questions

Q1. [Cambridge A-Level 2023] (a) State Bohr's three postulates for the hydrogen atom. (b) Using these postulates, derive the radius of the nth orbit. (c) Calculate the radius of the 3rd orbit. (9 marks)

(a) Bohr's Postulates:
1. Electrons revolve in fixed circular orbits called stationary states without radiating energy.
2. Only orbits in which angular momentum L = mvr = nh/2π are allowed (n = 1, 2, 3...)
3. When an electron transitions between levels, it emits or absorbs a photon of energy hν = |E₂ − E₁|

(b) Derivation of rₙ:
Coulomb force provides centripetal force:
ke²/r² = mv²/r → mv² = ke²/r ... (i)
Angular momentum condition:
mvr = nh/2π → v = nh/(2πmr) ... (ii)
From (i): v² = ke²/(mr) ... (iii)
From (ii): v² = n²h²/(4π²m²r²) ... (iv)
Equating (iii) and (iv):
ke²/(mr) = n²h²/(4π²m²r²)
r = n²h²/(4π²mke²) = n²a₀
where a₀ = h²/(4π²mke²) = 0.529 Å (Bohr radius)

(c) r₃ = 9 × a₀ = 9 × 0.529 Å = 4.76 Å = 4.76 × 10⁻¹⁰ m

(a) Three postulates stated. (b) rₙ = n²a₀ derived. (c) r₃ = 4.76 × 10⁻¹⁰ m

Q2. [Cambridge A-Level 2023] The energy levels of hydrogen are given. Show that the shortest wavelength emitted in the Balmer series is 364.6 nm.

Shortest λ in Balmer: n₂ → ∞, n₁ = 2
1/λ = R_H(1/4 − 0) = R_H/4
λ = 4/R_H = 4/(1.097 × 10⁷) = 3.646 × 10⁻⁷ m = 364.6 nm ✓

λ = 364.6 nm verified using series limit formula for Balmer (n₂→∞)

Q3. [Cambridge A-Level 2022] An electron in hydrogen is in n=4 state. (a) Calculate the energy of the electron. (b) How much energy must be supplied to remove it? (c) What is the longest wavelength photon it can emit?

(a) E₄ = −13.6/16 = −0.85 eV = −1.36 × 10⁻¹⁹ J
(b) IE from n=4: |E₄| = 0.85 eV = 1.36 × 10⁻¹⁹ J
(c) Longest λ → smallest ΔE → transition to next lower level (n=4→n=3):
ΔE = |E₄ − E₃| = |−0.85 − (−1.51)| = 0.66 eV = 1.056×10⁻¹⁹ J
λ = hc/ΔE = (6.63×10⁻³⁴ × 3×10⁸)/(1.056×10⁻¹⁹) = 1884 nm ≈ 1875 nm (Paschen)

(a) −0.85 eV (b) 0.85 eV (c) ≈1875 nm (infrared, Paschen series)

Q4. [Cambridge A-Level 2022] Explain how the line spectrum of hydrogen provides evidence for quantised energy levels in atoms.

1. When a discharge passes through hydrogen gas, the emission spectrum shows sharp, discrete spectral lines — not a continuous spectrum.
2. Each line corresponds to a specific wavelength (and thus specific photon energy).
3. Photon energy E = hf = hc/λ is measured precisely.
4. These discrete photon energies match the differences between specific energy levels: ΔE = hf = E₂ − E₁.
5. Since only specific photon energies are emitted, only specific energy differences exist in the atom.
6. This implies that electron energies are not continuous but take only discrete, allowed values — i.e., energy levels are quantised.
7. Classical physics cannot explain this — it predicts a continuous spectrum. Quantisation is a quantum mechanical effect.

Discrete spectral lines → discrete photon energies → discrete energy differences between levels → quantised energy levels. Classical physics predicts continuous spectra (inconsistent).

Q5. [Edexcel A-Level 2023] Using the Bohr model, show that the energy of the electron in the nth orbit of hydrogen is Eₙ = −13.6/n² eV.

1. From Coulomb = centripetal: ke²/r² = mv²/r → ke²/(2r) = ½mv² = KE
2. KE = ke²/(2r) > 0
3. PE = −ke²/r (by convention, zero at infinity)
4. Total E = KE + PE = ke²/(2r) − ke²/r = −ke²/(2r)
5. From Bohr's condition and Coulomb: r = n²a₀ where a₀ = 0.529 Å
6. E = −ke²/(2n²a₀) = −ke²/(2a₀) × 1/n²
7. ke²/(2a₀) = 9×10⁹ × (1.6×10⁻¹⁹)² / (2 × 0.529×10⁻¹⁰) = 2.18×10⁻¹⁸ J = 13.6 eV
8. Therefore: Eₙ = −13.6/n² eV ✓

Eₙ = −ke²/(2n²a₀) = −13.6/n² eV (shown)

Q6–Q10. [Mixed A-Level Calculation Questions]

Q6. Find wavelength of photon emitted when Li²⁺ electron falls from n=3 to n=1.
1/λ = R × Z² × (1/1 − 1/9) = 1.097×10⁷ × 9 × 8/9 = 8 × 1.097×10⁷ = 8.776×10⁷
λ = 1/8.776×10⁷ = 11.4 nm (extreme UV)

Q7. The work function of a metal is 2.0 eV. What is the maximum wavelength of light that can eject electrons?
λ = hc/φ = 1240/2.0 nm = 620 nm (red light)

Q8. For hydrogen, calculate the ratio of wavelengths of Lyman limit to Balmer limit.
λ_Lyman_limit = 1/R; λ_Balmer_limit = 4/R. Ratio = 1:4

Q9. An electron in H makes transition n=5 to n=2. Calculate λ.
1/λ = R(1/4 − 1/25) = R × 21/100. λ = 100/(21R) = 100/(21×1.097×10⁷) = 434.1 nm (H-γ, violet, Balmer)

Q10. For He⁺, what is the first excitation energy?
E₁(He⁺) = −54.4 eV. E₂(He⁺) = −54.4/4 = −13.6 eV. ΔE = 54.4 − 13.6 = 40.8 eV

A-Level Extended Questions

Q11. [A-Level 2022] Compare and contrast Bohr's model of the atom with Rutherford's model. Include experimental evidence and limitations. (10 marks)

Rutherford's Model:
• Based on α-scattering experiment (1911). Found that atom has tiny, dense, positive nucleus surrounded by electrons in mostly empty space.
• Nuclear size ∼10⁻¹⁵ m; atomic size ∼10⁻¹⁰ m.
• No explanation for where electrons are or their stability.
• Predicts classical collapse (radiating, spiralling electron).
• Cannot explain line spectra.

Bohr's Model (1913):
• Keeps Rutherford's nuclear idea but adds quantisation rules.
• Postulate 1: Electrons in stationary orbits don't radiate.
• Postulate 2: Angular momentum quantised: L = nħ.
• Postulate 3: Photon emitted/absorbed during level transitions: hν = ΔE.
• Successfully explains H spectrum: predicts exact wavelengths.
• Derives ionisation energy, orbit radii, electron speeds.

Limitations of Bohr's model:
• Fails for multi-electron atoms.
• Cannot explain fine structure (Zeeman, Stark effect).
• Violates Heisenberg's uncertainty principle.
• Doesn't use wave mechanics — superseded by Schrödinger's model (1926).

Bohr improved on Rutherford by adding quantisation. Both agree on nuclear structure. Bohr explains H spectrum; Rutherford cannot. Both fail for multi-electron atoms. Modern quantum mechanics (Schrödinger) supersedes both.

Q12–Q15. [A-Level Miscellaneous]

Q12. Derive the Rydberg formula from Bohr's model. Show that R = me⁴k²/(4πħ³c).
Eₙ = −13.6/n² eV → hν = 13.6(1/n₁² − 1/n₂²) eV → 1/λ = 13.6/(hc) × (1/n₁² − 1/n₂²) = R_H(1/n₁² − 1/n₂²). Numerically R_H = 1.097×10⁷ m⁻¹ ✓

Q13. A photon of 97.2 nm ionises a hydrogen atom. Was the atom in the ground state?
E = 1240/97.2 = 12.76 eV. Ground state needs 13.6 eV. Since 12.76 < 13.6, it cannot ionise from ground state. It could ionise from n=2 (needs 3.4 eV) — with excess energy — but 12.76 matches 1→4 transition (12.75 eV). So atom in ground state absorbs it to reach n=4, not ionised.

Q14. In Bohr model, as n→∞, what happens to adjacent energy levels?
ΔE = 13.6(1/n² − 1/(n+1)²) ≈ 13.6 × 2/n³ → 0 as n→∞. Energy levels become continuous → classical limit (Correspondence Principle).

Q15. Calculate the momentum of a photon emitted in Lyman-α transition.
λ = 121.6 nm. p = h/λ = 6.63×10⁻³⁴/121.6×10⁻⁹ = 5.45×10⁻²⁷ kg·m/s

💡 Section 15: Assertion Reason Questions

50 Assertion-Reason questions on the Atoms chapter

Instructions: Each question has an Assertion (A) and a Reason (R). Choose:
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is NOT the correct explanation
(c) A is true, R is false
(d) A is false, R is true (or both false)

Q1.

Assertion (A): The ground state of hydrogen atom is the most stable state.

Reason (R): In the ground state, the electron has the minimum energy (most negative total energy).

✓ Answer: (a) — Both A and R are true, and R correctly explains A. Minimum energy state is the most stable configuration.

Q2.

Assertion (A): The radius of the Bohr orbit increases with increasing quantum number n.

Reason (R): rₙ = n² × a₀, so radius is proportional to n².

✓ Answer: (a) — Both A and R are true, and R correctly explains A. rₙ ∝ n².

Q3.

Assertion (A): The speed of the electron increases as n increases in Bohr model.

Reason (R): vₙ = v₁/n, so speed decreases with n.

✓ Answer: (d) — A is false (speed decreases with n), R is true (vₙ = v₁/n is the correct formula).

Q4.

Assertion (A): Lyman series of hydrogen lies in the ultraviolet region.

Reason (R): Lyman series involves transitions to n=1, which have large energy differences producing high-frequency UV photons.

✓ Answer: (a) — Both true, R correctly explains A. Transitions to n=1 involve ΔE ≥ 10.2 eV → UV wavelengths (91–122 nm).

Q5.

Assertion (A): Bohr's model cannot explain the spectrum of helium atom.

Reason (R): Bohr's model does not account for electron-electron interactions in multi-electron atoms.

✓ Answer: (a) — Both A and R are true, and R correctly explains A.

Q6.

Assertion (A): The kinetic energy of the electron in a Bohr orbit is always positive.

Reason (R): KE = ½mv² is always positive for any moving object.

✓ Answer: (a) — Both true, R correctly explains A. KE = +13.6/n² eV > 0 always.

Q7.

Assertion (A): The total energy of the electron in a hydrogen atom is negative.

Reason (R): Negative total energy indicates the electron is in a bound state (cannot escape the atom).

✓ Answer: (a) — Both true, R correctly explains A. E = −13.6/n² eV < 0 means bound state.

Q8.

Assertion (A): No back-scattering of α-particles is observed when they bombard hydrogen gas.

Reason (R): The proton (hydrogen nucleus) is lighter than the α-particle, so the α-particle cannot bounce back in a collision.

✓ Answer: (a) — Both true, R correctly explains A. Heavier α (4 amu) cannot be deflected backwards by lighter proton (1 amu) in a collision.

Q9.

Assertion (A): The ionisation energy of He⁺ is four times that of hydrogen atom.

Reason (R): IE = 13.6 Z²/n² eV. For He⁺ (Z=2, n=1), IE = 4 × 13.6 = 54.4 eV.

✓ Answer: (a) — Both true, R correctly explains A. IE ∝ Z².

Q10.

Assertion (A): Rutherford's model of the atom predicted atomic instability.

Reason (R): According to classical electrodynamics, an orbiting electron radiates energy and should spiral into the nucleus.

✓ Answer: (a) — Both true, R correctly explains A. This is the famous "collapse problem" of Rutherford's model.

Q11.

Assertion (A): The Balmer series appears in the visible region of the electromagnetic spectrum.

Reason (R): Balmer series corresponds to transitions from n ≥ 3 to n = 2.

✓ Answer: (b) — Both true, but R alone doesn't explain WHY it's visible — the energy differences in Balmer correspond to visible photon wavelengths (400–700 nm). R gives the transition, not the reason for visible region.

Q12.

Assertion (A): The potential energy of the electron in hydrogen atom is twice its total energy.

Reason (R): PE = −ke²/r and Total E = −ke²/(2r), so PE = 2E_total.

✓ Answer: (a) — Both true, R correctly explains A. |PE| = 2|KE| = 2|Total E| (Virial theorem).

Q13.

Assertion (A): An atom can absorb photons of only specific energies.

Reason (R): Energy levels of atoms are discrete, so only photons matching exact energy differences can be absorbed.

✓ Answer: (a) — Both true, R correctly explains A. This is why absorption spectra show dark lines at the same wavelengths as emission lines.

Q14.

Assertion (A): The angular momentum of an electron in the nth orbit of hydrogen is nħ.

Reason (R): Bohr's second postulate states L = mvr = nh/2π = nħ.

✓ Answer: (a) — Both true, R correctly explains A. This is Bohr's quantisation condition.

Q15.

Assertion (A): The shortest wavelength in each spectral series is called the series limit.

Reason (R): At the series limit, n₂ → ∞, giving maximum energy difference and minimum wavelength.

✓ Answer: (a) — Both true, R correctly explains A. Series limit: n₂→∞, so 1/λ = R/n₁² (maximum 1/λ → minimum λ).

Q16.

Assertion (A): Hydrogen atom emits only UV radiation in the ground state.

Reason (R): An atom in the ground state cannot emit any radiation without first being excited.

✓ Answer: (d) — A is false (ground state atom cannot emit — it has no lower level to fall to). R is true (must be excited first).

Q17.

Assertion (A): The Paschen series lies in the infrared region.

Reason (R): Paschen series corresponds to transitions ending at n=3, with wavelengths 820–1875 nm.

✓ Answer: (a) — Both true, R correctly explains A. Wavelengths 820–1875 nm are all in the infrared region (>700 nm).

Q18.

Assertion (A): The energy of the photon emitted in the transition 2→1 in hydrogen is 10.2 eV.

Reason (R): E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV.

✓ Answer: (a) — Both true, R correctly explains A. This is the Lyman-α photon energy.

Q19.

Assertion (A): Distance of closest approach is inversely proportional to kinetic energy of α-particle.

Reason (R): r₀ = 2kZe²/E_k ∝ 1/E_k.

✓ Answer: (a) — Both true, R correctly explains A. Higher KE → deeper penetration → smaller r₀.

Q20.

Assertion (A): The maximum number of spectral lines from n=4 state is 6.

Reason (R): N = n(n−1)/2 = 4×3/2 = 6.

✓ Answer: (a) — Both true, R correctly explains A.

Q21.

Assertion (A): The impact parameter for 180° scattering is zero.

Reason (R): b = (kZe²/E_k)cot(90°) = 0, since cot(90°) = 0.

✓ Answer: (c) — A is true (b=0 for 180°). R is incorrect in formula: for θ=180°, b = cot(180°/2) = cot(90°) = 0. So A is true and R is also true! Actually both are true and R explains A. → Answer: (a). b = cot(θ/2) evaluated at θ=180°: cot(90°) = 0 ✓.

Q22.

Assertion (A): Emission and absorption spectra of hydrogen have the same set of wavelengths.

Reason (R): Both emission and absorption involve the same energy level transitions.

✓ Answer: (a) — Both true, R correctly explains A. The same transitions that emit photons can also absorb photons of identical energy/wavelength (Kirchhoff's law).

Q23.

Assertion (A): The de Broglie wavelength of the electron decreases as n increases in hydrogen atom.

Reason (R): λ = h/(mv). As n increases, v decreases (v ∝ 1/n), so λ increases (not decreases).

✓ Answer: (d) — A is false (λ increases with n: λ = 2πrₙ/n = 2πna₀ ∝ n). R is true (gives the reason correctly: lower v means larger λ). Actually λ = h/p = h/(mv), v = v₁/n, so λ = hn/(mv₁) ∝ n → increases. A is false, R is true. → (d)

Q24.

Assertion (A): Bohr's postulate of quantisation of angular momentum was justified by de Broglie's hypothesis.

Reason (R): de Broglie showed that for a standing wave in the orbit: 2πr = nλ → mvr = nħ (Bohr's condition).

✓ Answer: (a) — Both true, R correctly explains A. de Broglie's standing wave condition gives the same result as Bohr's postulate.

Q25.

Assertion (A): Most of the mass of the atom is concentrated in the nucleus.

Reason (R): Proton mass (1836 mₑ) and neutron mass (1839 mₑ) are much greater than electron mass (mₑ).

✓ Answer: (a) — Both true, R correctly explains A. Since proton/neutron masses ≫ electron mass, nucleus (containing protons and neutrons) has nearly all the mass.

Q26.

Assertion (A): Hydrogen and deuterium have identical atomic spectra.

Reason (R): Both have Z=1 (one proton), so their electron energy levels are nearly identical (mass difference of nucleus causes very tiny shifts).

✓ Answer: (b) — Both true, but R only partially explains: the spectra are nearly identical (not perfectly identical) due to reduced mass effect. The reason given is a partial truth. Answer (b) is most appropriate.

Q27.

Assertion (A): In Bohr's model, the orbit of an electron is a physical path along which the electron travels.

Reason (R): The Heisenberg uncertainty principle makes it impossible to define a precise orbit for the electron.

✓ Answer: (d) — A is false (in modern quantum mechanics, we cannot define a precise circular path — only a probability distribution; Bohr's orbits are a simplification). R is true (HUP: Δx·Δp ≥ ħ/2 prevents simultaneous knowledge of position and momentum needed to define a classical orbit).

Q28.

Assertion (A): The first line of the Lyman series has a longer wavelength than the first line of the Balmer series.

Reason (R): Lyman transitions (to n=1) have larger energy differences than Balmer transitions (to n=2), so Lyman photons have shorter wavelengths.

✓ Answer: (d) — A is false (Lyman-α = 121.6 nm is shorter than Balmer-α = 656.3 nm). R is true (Lyman has larger ΔE → shorter λ).

Q29.

Assertion (A): The frequency of revolution of an electron in the nth Bohr orbit is proportional to 1/n³.

Reason (R): f = v/(2πr) ∝ (1/n)/(n²) = 1/n³.

✓ Answer: (a) — Both true, R correctly explains A. f = vₙ/(2πrₙ) = (v₁/n)/(2πn²a₀) = v₁/(2πa₀) × 1/n³ ∝ 1/n³.

Q30.

Assertion (A): The nuclear radius is of the order of 10⁻¹⁵ m.

Reason (R): R = R₀A^(1/3) where R₀ ≈ 1.2 fm = 1.2 × 10⁻¹⁵ m.

✓ Answer: (a) — Both true, R correctly explains A. For typical nuclei (A ∼ 1–238), R ∼ 10⁻¹⁵ m range.

Q31–Q40.

Q31 A: An electron in n=2 state has more energy than in n=1 state.

Q31 R: E₂ = −3.4 eV > E₁ = −13.6 eV (less negative = more energy).

Answer: (a). Both true, R explains A.

Q32 A: The Brackett series is in the infrared region.

Q32 R: Brackett series: transitions to n=4 give λ = 1458–4051 nm, which is far infrared.

Answer: (a). Both true, R explains A.

Q33 A: The energy of the emitted photon equals the kinetic energy lost by the electron.

Q33 R: hν = E₂ − E₁ (total energy difference, not just KE difference).

Answer: (d). A is false (photon energy = total energy difference, not just KE change). R is true.

Q34 A: Rutherford's model is called the nuclear model of the atom.

Q34 R: Rutherford discovered the nucleus through the α-scattering experiment.

Answer: (a). Both true, R explains A.

Q35 A: The wavelength of radiation increases as the electron makes transitions to higher orbits.

Q35 R: Higher orbits have smaller energy differences, so emitted photons have lower energy and longer wavelength.

Answer: (a). Both true, R explains A. (Within a series: higher n → smaller ΔE → longer λ)

Q36 A: Hydrogen atom cannot emit more than one spectral line at a time.

Q36 R: A single electron can make only one transition at a time.

Answer: (a). Correct — a single hydrogen atom emits one photon per transition. Multiple lines come from many atoms making different transitions.

Q37 A: He⁺ has the same spectral lines as hydrogen at the same wavelengths.

Q37 R: He⁺ has Z=2, so its spectral lines are at different wavelengths (shorter, higher energy).

Answer: (d). A is false (He⁺ spectrum is different from H). R is true (Z=2 gives 4× more energy → different λ).

Q38 A: The electron in the nth orbit of hydrogen completes n revolutions per second.

Q38 R: The frequency of revolution fₙ = v₁/(2πn³a₀) ∝ 1/n³, not n.

Answer: (d). A is false. R is true (fₙ ∝ 1/n³).

Q39 A: The radius of the nth Bohr orbit for He⁺ (Z=2) is half that of hydrogen in the same orbit.

Q39 R: rₙ = n²a₀/Z. For He⁺, Z=2, so rₙ = n²a₀/2 = half of hydrogen's rₙ.

Answer: (a). Both true, R explains A. r(He⁺)/r(H) = 1/Z = 1/2.

Q40 A: Atomic nuclei are positively charged.

Q40 R: Nuclei contain protons (positive charge) and neutrons (no charge).

Answer: (a). Both true, R explains A.

Q41–Q50.

Q41 A: The energy of the hydrogen atom in a state characterised by n=∞ is zero.

Q41 R: At n=∞, the electron is free (ionised), with no binding energy.

Answer: (a). Both true, R explains A. E_∞ = −13.6/∞² = 0 eV (reference level for free electron).

Q42 A: Balmer series is the only visible series in the hydrogen spectrum.

Q42 R: Only the Balmer series (n₁=2) produces photons in the 400–700 nm visible range.

Answer: (a). Both true, R correctly explains A.

Q43 A: In Bohr's model, the electron does not radiate energy while in a stationary orbit.

Q43 R: This is Bohr's first postulate — an ad hoc assumption contrary to classical electrodynamics.

Answer: (a). Both true, R explains why it's a postulate (not a derived result).

Q44 A: The Rydberg constant has units of per metre (m⁻¹).

Q44 R: In the formula 1/λ = R_H(1/n₁² − 1/n₂²), λ is in metres, so R_H must be in m⁻¹ for dimensional consistency.

Answer: (a). Both true, R explains A.

Q45 A: For transition 3→1 in hydrogen, the emitted photon belongs to the Lyman series.

Q45 R: Lyman series consists of all transitions ending at n=1.

Answer: (a). Both true, R explains A. 3→1 ends at n=1 → Lyman ✓.

Q46 A: Thomson's model of atom was disproved by Rutherford's experiment.

Q46 R: In Thomson's model, large-angle scattering of α-particles would be impossible; but it was observed.

Answer: (a). Both true, R explains A. Thomson's uniform positive sphere cannot produce the observed large-angle deflections.

Q47 A: The number of spectral lines emitted by hydrogen when electron drops from n=5 is 10.

Q47 R: N = n(n−1)/2 = 5×4/2 = 10.

Answer: (a). Both true, R explains A.

Q48 A: The electron's angular momentum is larger in higher Bohr orbits.

Q48 R: L = nħ. As n increases, L increases proportionally.

Answer: (a). Both true, R explains A.

Q49 A: α-particles are emitted by radioactive nuclei with a range of kinetic energies.

Q49 R: α-particles from a given source have discrete, well-defined energies (not a continuous range).

Answer: (d). A is false (α-particles from a given isotope have discrete, fixed energies). R is true (unlike β-particles which have continuous energy).

Q50 A: The size of the atom is of the order of 10⁻¹⁰ m while nuclear size is 10⁻¹⁵ m.

Q50 R: The nucleus is 10⁵ times smaller than the atom, leaving the rest of the atom as empty space.

Answer: (a). Both true, R explains A. Ratio of sizes ∼10⁻¹⁰/10⁻¹⁵ = 10⁵. Atom is mostly empty space.

📋 Section 16: Case Study Questions

25 Case Studies on Rutherford, Bohr, Hydrogen Spectrum, Energy Levels

Case Study 1: Rutherford's Alpha Scattering Experiment

Geiger and Marsden, under Rutherford's supervision, directed a narrow beam of α-particles at a thin gold foil (2.1×10⁻⁷ m thick). A circular detector surrounded the foil. They found that approximately 1 in 8000 α-particles were deflected through angles greater than 90°. This was completely unexpected based on Thomson's model, which predicted only small deflections. Rutherford concluded that all the positive charge and most of the mass must be concentrated in an extremely small region at the centre of the atom — the nucleus. The distance of closest approach for 5 MeV α-particles with gold (Z=79) is approximately 45 fm, giving the upper limit for the nuclear radius. The actual nuclear radius is given by R = R₀A^(1/3) where R₀ = 1.2 fm.

Q1. Why was 1-in-8000 back-scattering surprising based on Thomson's model?

In Thomson's model, the positive charge is spread uniformly throughout the atom. The maximum electric force on an α-particle is very small and distributed — it should cause only tiny deflections, not 90°+ deflections. A 1-in-8000 back-scattering rate was 100,000× larger than Thomson's model predicted.

Q2. Calculate the nuclear radius of gold (A=197) using R₀ = 1.2 fm.

R = R₀A^(1/3) = 1.2×10⁻¹⁵ × (197)^(1/3) = 1.2×10⁻¹⁵ × 5.82 = 6.98×10⁻¹⁵ m

R(Au) ≈ 7.0 fm = 7.0×10⁻¹⁵ m

Q3. What physical principle is used to derive the distance of closest approach?

Conservation of energy: At the closest approach, the α-particle momentarily stops. All initial kinetic energy is converted to electrostatic potential energy: ½mv² = kZe(2e)/r₀ → r₀ = 2kZe²/E_k.

Q4. Why is r₀ = 45 fm only an UPPER LIMIT for nuclear size?

The distance of closest approach is the distance at which the α-particle turns around — it doesn't actually touch the nucleus. The actual nucleus is somewhere inside this radius. The nuclear surface is smaller than where the α-particle stopped. Hence r₀ is an upper limit (the nucleus is ≤ r₀).

Case Study 2: Bohr Model of Hydrogen Atom

Niels Bohr in 1913 modified Rutherford's model by proposing three postulates. He postulated that electrons occupy fixed, stable orbits without radiating energy. The angular momentum of the electron in any orbit is quantised: L = nh/2π. When an electron transitions between orbits, it emits or absorbs a photon of energy hν = |E₂ − E₁|. Using these postulates, Bohr derived that the radius of the nth orbit is rₙ = n² × 0.529 Å and the energy is Eₙ = −13.6/n² eV for hydrogen. This model successfully explained the hydrogen spectrum, predicting the Balmer series wavelengths with great accuracy. However, the model fails for multi-electron atoms and does not incorporate quantum mechanics properly.

Q1. State Bohr's angular momentum quantisation condition and give its physical significance.

L = mvr = nh/2π (n = 1, 2, 3...). Physical significance: Only certain discrete orbits are allowed — those in which the electron's angular momentum is an integer multiple of ħ. This prevents the electron from spiralling into the nucleus and explains atomic stability.

Q2. Using Bohr's model, calculate the radius and energy of the second orbit in hydrogen.

r₂ = n²a₀ = 4 × 0.529 Å = 2.116 Å
E₂ = −13.6/n² = −13.6/4 = −3.4 eV

r₂ = 2.116 Å; E₂ = −3.4 eV

Q3. What is the first excitation potential of hydrogen?

First excitation potential = E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV. This is the minimum energy needed to excite a ground-state hydrogen atom to its first excited state (n=2).

Q4. Why does the Bohr model fail for helium but work for He⁺?

He⁺ has only one electron (like hydrogen), so Bohr's model applies. Helium atom has two electrons — they interact with each other (electron-electron repulsion), which Bohr's model does not account for. This additional interaction changes the energy levels in ways Bohr cannot predict.

Case Study 3: Hydrogen Spectrum and Spectral Series

When hydrogen gas is excited electrically, it emits light that, when passed through a prism, produces a characteristic line spectrum. Johann Balmer in 1885 found that the wavelengths of the visible hydrogen spectrum fit the formula 1/λ = R(1/4 − 1/n²) for n = 3, 4, 5... Later, Lyman found the UV series (transitions to n=1) and Paschen found the IR series (transitions to n=3). The Rydberg constant R_H = 1.097 × 10⁷ m⁻¹. The H-α line (656.3 nm) is red, H-β (486.1 nm) is blue-green, H-γ (434.1 nm) is violet, and H-δ (410.2 nm) is violet. The series limit represents the transition from n=∞ to the base level of the series.

Q1. Calculate the wavelength of H-α (3→2) and verify it is 656.3 nm.

1/λ = R(1/4 − 1/9) = R×5/36 = 1.097×10⁷×5/36 = 1.524×10⁶ m⁻¹
λ = 1/1.524×10⁶ = 6.563×10⁻⁷ m = 656.3 nm ✓

λ = 656.3 nm verified

Q2. Why is the H-α line the longest wavelength (and lowest energy) in the Balmer series?

H-α corresponds to the 3→2 transition — consecutive levels. This gives the smallest energy difference in the Balmer series (ΔE = 1.89 eV), hence the largest wavelength (656.3 nm). Larger n transitions give larger ΔE and shorter wavelengths.

Q3. Find the series limit wavelength of the Paschen series.

Series limit: n₂→∞, n₁=3. 1/λ = R/9. λ = 9/R = 9/1.097×10⁷ = 8.2×10⁻⁷ m = 820 nm

λ_limit(Paschen) = 820 nm

Q4. How many series of spectral lines are produced when hydrogen is excited to n=5?

Possible transitions from n=5: to n=4 (Brackett), n=3 (Paschen), n=2 (Balmer), n=1 (Lyman). From n=4: to n=3, n=2, n=1. From n=3: to n=2, n=1. From n=2: to n=1. All 4 series (Lyman, Balmer, Paschen, Brackett) are represented. Total lines: N = 5(4)/2 = 10.

Case Study 4: Energy Levels of Hydrogen

The energy levels of hydrogen are given by Eₙ = −13.6/n² eV (n = 1, 2, 3...). The ground state (n=1) has energy −13.6 eV. The ionisation energy (energy to remove electron from ground state) is 13.6 eV. When 12.5 eV electrons bombard hydrogen atoms in the ground state, atoms can be excited to n=3 (needing 12.09 eV) but not n=4 (needing 12.75 eV). An atom excited to n=3 can return to ground state via n=3→2→1 or n=3→1, emitting photons from Lyman and Balmer series.

Q1. Identify which spectral series are emitted when hydrogen is bombarded with 12.5 eV electrons.

Max excitation to n=3 (12.09 eV < 12.5 eV < 12.75 eV for n=4). Transitions: 3→1 and 2→1 (Lyman series, UV) and 3→2 (Balmer series, visible). Paschen series NOT possible (needs n=4 or higher).

Q2. Calculate the energies of photons emitted in all three transitions.

3→1: ΔE = 13.6(1−1/9) = 12.09 eV; λ = 1240/12.09 = 102.6 nm (UV)
2→1: ΔE = 13.6(1−1/4) = 10.2 eV; λ = 1240/10.2 = 121.6 nm (UV)
3→2: ΔE = 13.6(1/4−1/9) = 1.89 eV; λ = 1240/1.89 = 656 nm (red, visible)

12.09 eV (UV), 10.2 eV (UV), 1.89 eV (red)

Q3. What happens if 14 eV electrons are used instead of 12.5 eV?

14 eV > ionisation energy (13.6 eV). The electron would be removed from the atom — ionisation occurs. Remaining energy = 14 − 13.6 = 0.4 eV appears as kinetic energy of the liberated electron. No photon is emitted from an excited state (atom is ionised, not excited).

Q4. Why is the ionisation energy exactly 13.6 eV?

Ionisation energy = |E₁| = 13.6 eV. From Bohr's formula: E₁ = −me⁴k²/(2ħ²) = −13.6 eV. This is derived from the balance between Coulomb force and centripetal force combined with angular momentum quantisation. It represents the binding energy of the electron in the ground state — the energy needed to bring the electron from r = a₀ (Bohr radius) to r = ∞.

Case Study 5: Photon Emission and Absorption

Photons are emitted when an electron transitions from a higher energy level to a lower one, and are absorbed when an electron transitions to a higher level. The photon energy equals the energy difference between the two levels: E = hν = hc/λ. In an emission spectrum, bright lines appear on a dark background. In an absorption spectrum, dark lines (Fraunhofer lines) appear on a continuous background. Both spectra have lines at the same wavelengths — this is Kirchhoff's law. The solar spectrum shows absorption lines corresponding to atoms in the Sun's cooler outer atmosphere absorbing specific wavelengths from the continuous spectrum produced by the hotter interior.

Q1. Explain why absorption and emission spectra of hydrogen have lines at the same wavelengths.

Same energy differences between levels → same photon energies → same wavelengths. Emission: electron falls from level 2 to 1, releasing photon of energy E₂−E₁. Absorption: photon of the same energy E₂−E₁ is needed to excite electron from 1 to 2. Since ΔE is the same in both processes, the wavelength λ = hc/ΔE is the same (Kirchhoff's law of radiation).

Q2. A photon of wavelength 97.2 nm is absorbed by hydrogen. To which level is the electron excited?

E = 1240/97.2 = 12.76 eV. E₁ = −13.6 eV.
E_n = −13.6 + 12.76 = −0.84 eV ≈ −0.85 eV = E₄. So n=4.

Electron excited to n=4 (third excited state)

Q3. What is the significance of Fraunhofer lines in the solar spectrum?

Fraunhofer lines are dark absorption lines in the solar spectrum caused by specific elements in the Sun's cooler outer atmosphere absorbing particular wavelengths. Each element has a characteristic set of absorption wavelengths. By identifying these dark lines, astronomers can determine the chemical composition of the Sun and other stars — a powerful technique of spectroscopy.

Q4. An electron makes transitions from n=5 to n=1. How many photons are emitted in this process?

Only ONE photon is emitted per transition. If the electron goes directly 5→1, one photon of energy 13.6(1−1/25) = 13.06 eV is emitted. If it goes step by step (5→4→3→2→1), four photons are emitted (one per step). Different paths from n=5 give one or more photons, but each individual transition produces exactly one photon.

Case Study 6: Hydrogen-Like Atoms (He⁺, Li²⁺)

Atoms or ions with only one electron are called hydrogen-like. Examples: H (Z=1), He⁺ (Z=2), Li²⁺ (Z=3), Be³⁺ (Z=4). The Bohr model gives exact results for these systems. The radius is rₙ = n²a₀/Z and energy is Eₙ = −13.6Z²/n² eV. For He⁺ (Z=2): E₁ = −54.4 eV, E₂ = −13.6 eV, r₁ = 0.265 Å. The spectrum of He⁺ partially overlaps with the hydrogen spectrum — the Balmer series of He⁺ has the same wavelengths as every other line of the Lyman series of H (Pickering-Fowler lines).

Q1. Calculate the ionisation energy of Li²⁺ (Z=3).

E₁(Li²⁺) = −13.6 × Z²/1 = −13.6 × 9 = −122.4 eV
IE = 122.4 eV

IE(Li²⁺) = 122.4 eV

Q2. For He⁺, find the wavelength of the photon emitted in the 2→1 transition.

ΔE = 13.6 × 4 × (1 − 1/4) = 54.4 × 3/4 = 40.8 eV
λ = 1240/40.8 = 30.4 nm

λ = 30.4 nm (extreme UV)

Q3. For which hydrogen-like atom does the first orbit have the same radius as the second orbit of hydrogen?

r₁(Z) = a₀/Z. r₂(H) = 4a₀. Set equal: a₀/Z = 4a₀ → Z = 1/4? Not physical.
Alternatively: r₂(Z) = 4a₀/Z = r₁(H) = a₀ → Z = 4 (Be³⁺).
Be³⁺ second orbit radius = 4×0.529/4 = 0.529 Å = r₁(H) ✓

Be³⁺ (Z=4): r₂(Be³⁺) = r₁(H) = 0.529 Å

Q4. What is the ratio of ground state energies of He⁺ and Li²⁺?

E₁ ∝ Z². E₁(He⁺)/E₁(Li²⁺) = Z(He⁺)²/Z(Li²⁺)² = 4/9

Ratio = 4:9

🔄 Section 17: Quick Revision Notes

25 most important formulas, concepts, and exam questions — one page revision

🔢 25 Most Important Formulae

rₙ = n²a₀/Z (orbit radius)
a₀ = 0.529 Å (Bohr radius)
Eₙ = −13.6Z²/n² eV
vₙ = 2.18×10⁶/n m/s (H)
KE = +13.6Z²/n² eV
PE = −27.2Z²/n² eV
L = nh/2π = nħ
IE = 13.6Z²/n² eV
1/λ = R(1/n₁² − 1/n₂²)
R_H = 1.097×10⁷ m⁻¹
E_photon = hν = hc/λ
λ(nm) = 1240/E(eV)
r₀ = 2kZe²/E_k
b = kZe²cot(θ/2)/E_k
N = n(n−1)/2 (lines)
T ∝ n³; f ∝ 1/n³
R = R₀A^(1/3)
ΔE = 13.6(1/n₁² − 1/n₂²) eV
cR_H = 3.29×10¹⁵ Hz
v₁ = 2.18×10⁶ m/s
|PE| = 2|KE| = 2|E_total|
μ_B = 9.27×10⁻²⁴ J/T
hc = 1240 eV·nm
Tₙ = n³T₁; T₁ = 1.52×10⁻¹⁶ s
2πrₙ = nλ_dB

💡 25 Most Important Concepts

Rutherford: nucleus is tiny, dense, positive
Most α-particles pass straight through
Bohr Postulate 1: Stationary orbits
Bohr Postulate 2: L = nħ
Bohr Postulate 3: hν = ΔE
Negative E → bound (stable) state
Ground state = lowest energy state
Ionisation = removal of electron
Excitation = jump to higher level
Emission: higher → lower level
Absorption: lower → higher level
Lyman: UV (n₁=1)
Balmer: Visible (n₁=2)
Paschen: IR (n₁=3)
Series limit: n₂→∞ (shortest λ)
Longest λ in series: consecutive levels
de Broglie: 2πr = nλ justifies Bohr
Correspondence Principle: n→∞ = classical
Bohr model valid only for 1-electron atoms
KE = −Total E (always positive)
PE = 2×Total E (always negative)
Nuclear size ∼10⁻¹⁵ m; atomic ∼10⁻¹⁰ m
Emission ↔ absorption: same wavelengths
rₙ ∝ n², Eₙ ∝ 1/n², vₙ ∝ 1/n
IE of H = 13.6 eV, He⁺ = 54.4 eV

❓ 25 Most Important Exam Questions

State Bohr's three postulates
Derive rₙ = n²a₀ from Bohr's model
Derive Eₙ = −13.6/n² eV
Explain Rutherford's α-scattering
Why is Rutherford's model unstable?
Find wavelength of H-α line
Calculate r₂ and r₃ of hydrogen
Find KE and PE in ground state H
Explain line spectrum of hydrogen
Number of spectral lines from n=4
What is ionisation energy of H?
Which series lies in visible region?
Find distance of closest approach
Impact parameter at 90° scattering
Wavelength from 4→2 transition
First excitation energy of H
Why Bohr fails for multi-electron atoms
Series limit of Lyman/Balmer/Paschen
Photon energy for n=1→4 absorption
Explain stability in Bohr model
Energy levels of He⁺ (Z=2)
de Broglie justification of Bohr
Correspondence principle
12.5 eV bombardment: which series?
Ratio of KE to total energy in nth orbit

🩺 NEET Revision Tips

Memorise: E₁=−13.6, E₂=−3.4, E₃=−1.51, E₄=−0.85 eV
Use λ(nm)=1240/E(eV) shortcut always
rₙ:rₙ₊₁ = n²:(n+1)² (ratio trick)
vₙ ∝ Z/n; for H: v₁=2.18×10⁶ m/s
N=n(n-1)/2 for spectral lines
KE is always positive in Bohr orbits
Series: Lyman UV, Balmer Visible, Paschen IR
IE from nth orbit = 13.6/n² eV
First excitation of H = 10.2 eV
Hydrogen-like: scale by Z² for energy
Impact parameter: large b → small θ
Series limit: 1/λ = R/n₁² (n₂→∞)
For graph questions: E vs n is 1/n² curve

🏆 JEE Revision Tips

Derive all formulas — don't just memorise
Rydberg formula: 1/λ = RZ²(1/n₁²-1/n₂²)
Overlap: Balmer He⁺ ↔ Lyman H (for Z comparison)
Series limit wavelengths: λ∝n₁² (ratio questions)
Magnetic moment μ_B from orbital motion
de Broglie: 2πrₙ = nλ → standing waves
Zeeman/Stark effects: Bohr model fails here
Correspondence principle for large n
Ratio problems: always use proportionality
Photoelectric + hydrogen combined problems
Multiple-correct: all properties of Bohr orbits
Integer answer: n from mvr = nħ (NCERT 12.9 type)
Bohr magneton = μ_B = 9.27×10⁻²⁴ J/T

📝 One-Page Key Values

a₀ = 0.529 Å = 5.29×10⁻¹¹ m
R_H = 1.097×10⁷ m⁻¹
h = 6.626×10⁻³⁴ J·s
c = 3×10⁸ m/s; hc = 1240 eV·nm
m_e = 9.11×10⁻³¹ kg
e = 1.6×10⁻¹⁹ C
k = 9×10⁹ N·m²/C²
H-α = 656.3 nm; H-β = 486.1 nm
Lyman-α = 121.6 nm; Ly-limit = 91.2 nm
Balmer limit = 364.6 nm
Paschen limit = 820.4 nm; Pa-α = 1875 nm
IE(H) = 13.6 eV; IE(He⁺) = 54.4 eV
IE(Li²⁺) = 122.4 eV
E₁=−13.6, E₂=−3.4, E₃=−1.51, E₄=−0.85 eV
v₁(H) = 2.18×10⁶ m/s

Still Confused?

If you are searching for a Physics Tutor and still do not understand Rutherford Model, Bohr Model, Hydrogen Spectrum, Spectral Series, Energy Levels, Numericals or PYQs, contact Kumar Sir for personalised one-to-one online Physics classes.

30+ Years Teaching Experience FIITJEE Alumni Aakash Faculty IIT-JEE Expert NEET Expert AP Physics IB Physics IGCSE Physics A-Level Physics British Curriculum

📞 +91-9958461445 📧 kumarsirphysics@gmail.com 🌐 kumarphysicsclasses.com