Atoms: Class 12 Important Questions with Answers
72 important questions for Class 12 Physics Chapter 12, Atoms, the way they come in the board exam: MCQs, assertion–reason, case-based and short and long answer questions, with answers. Taken from our Class 12 Physics notes for 2026-27.
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Jump to: Practice set: test yourself20 MCQsAssertion–ReasonAnswers to the practice set
Practice set: test yourself
Try every question before looking at the answers below. 64 or more out of 80 means you are ready for the board.
A. Fill in the blanks (1 mark each)
- The Bohr radius is ______ Å.
- The ionisation energy of hydrogen is ______ eV.
- Angular momentum in the nth orbit = ______.
- The Balmer series lies in the ______ region.
- For head-on collision the impact parameter is ______.
- Total energy E and kinetic energy K are related by ______.
- Time period T ∝ ______.
- 2πrₙ = ______ (de Broglie).
B. True or false? (1 mark each)
- Most α-particles pass through the gold foil undeflected.
- In Thomson's model the positive charge is in a nucleus.
- The electron's speed increases with n.
- The total energy of a bound electron is negative.
- Lyman lines are infrared.
- Bohr's model works for helium.
- Energy levels get closer as n increases.
- Rutherford's atom should be unstable classically.
C. Very short answer (1 mark each)
- Define impact parameter.
- Define ionisation energy.
- Energy of H in n = 3?
- Who explained Bohr's second postulate?
- Name the series ending on n = 3.
- What does the distance of closest approach estimate?
- Value of PE in the ground state of H?
- How does the radius depend on n?
- Why is the energy negative?
- Give one limitation of Bohr's model.
D. Short answer I (2 marks each)
- Write two conclusions of the α-scattering experiment.
- Derive d for an α-particle of KE K.
- Find the wavelength of the 4 → 2 transition.
- Show T ∝ n³.
- Explain how de Broglie's idea gives L = nh/2π.
- Find the energy needed to take an electron from n = 2 to n = 5.
- Why does Rutherford's model fail? Give two reasons.
- Find the speed and radius in n = 3.
| Section | A /8 | B /8 | C /10 | D /16 | E /18 | F /20 | G (bonus) | Total /80 |
|---|---|---|---|---|---|---|---|---|
| My marks |
Rough work
E. Short answer II (3 marks each)
- State Bohr's postulates and derive the radius of the nth orbit.
- Derive the total energy of the electron in the nth orbit; explain the negative sign.
- Draw the energy level diagram of H and mark the Lyman, Balmer and Paschen series.
- An electron in H falls from n = 3 to n = 2. Find (a) the photon energy, (b) λ, (c) the change in the radius.
- Describe the Geiger–Marsden experiment with a diagram.
- Show that the energy levels crowd together at large n.
F. Long answer (5 marks each)
- Describe Rutherford's α-scattering experiment. Draw the N–θ graph and the trajectories for different impact parameters. What conclusions follow? Find the closest approach of a 7.7 MeV α-particle to gold.
- State Bohr's postulates. Derive expressions for rₙ, vₙ and Eₙ. Find the wavelength of the first line of the Balmer series. Give two limitations of the model.
G. HOTS (higher-order thinking)
- Why is gold used in the α-scattering experiment?
- Why can't an electron in H have an energy of −5 eV?
- An atom absorbs 10.2 eV. Which level does it reach and how many lines can it then emit?
- If the electron in H were replaced by a particle 200 times heavier with the same charge, how would a₀ change?
(i) Which series do these belong to? (ii) Find the photon energy of the 656 nm line. (iii) Which transition gives 486 nm? (iv) Why does each gas give its own set of lines?
20 MCQs
- Bohr radius(a) 5.3 Å (b) 1.06 Å (c) 0.53 Å (d) 0.053 Å
- Energy of H in n = 2(a) −3.4 eV (b) −13.6 eV (c) −1.51 eV (d) −6.8 eV
- rₙ ∝(a) n (b) n³ (c) 1/n (d) n²
- vₙ ∝(a) n (b) 1/n (c) n² (d) 1/n²
- T ∝(a) n (b) n³ (c) n² (d) 1/n³
- Lyman series region(a) IR (b) visible (c) microwave (d) UV
- Head-on collision: θ =(a) 180° (b) 90° (c) 0° (d) 45°
- Ionisation energy of H(a) 10.2 eV (b) 3.4 eV (c) 13.6 eV (d) 27.2 eV
- In the ground state, PE(a) 13.6 eV (b) 27.2 eV (c) −13.6 eV (d) −27.2 eV
- Angular momentum in n = 3(a) h/2π (b) 3h/2π (c) 3h (d) h/3π
- Most of the atom is(a) nucleus (b) electrons (c) empty space (d) protons
- Second Balmer line(a) 4 → 2 (b) 3 → 2 (c) 4 → 3 (d) 2 → 1
- d ∝ (for fixed Z)(a) K (b) 1/K (c) K² (d) √K
- Gold foil experiment by(a) Thomson (b) Bohr (c) Geiger and Marsden (d) de Broglie
- Number of lines from n = 4 to 1(a) 3 (b) 4 (c) 10 (d) 6
- Energy to go from n = 1 to 2(a) 10.2 eV (b) 13.6 eV (c) 3.4 eV (d) 12.09 eV
- 2πrₙ equals(a) nλ (b) λ/n (c) n²λ (d) λ
- K = ? in terms of E(a) E (b) −2E (c) 2E (d) −E
- Paschen series ends on(a) n = 1 (b) n = 3 (c) n = 2 (d) n = 4
- Thomson's model is called(a) planetary (b) nuclear (c) plum pudding (d) wave
1 (c) · 2 (a) · 3 (d) · 4 (b) · 5 (b) · 6 (d) · 7 (a) · 8 (c) · 9 (d) · 10 (b) · 11 (c) · 12 (a) · 13 (b) · 14 (c) · 15 (d) · 16 (a) · 17 (a) · 18 (d) · 19 (b) · 20 (c)
Assertion–Reason (6)
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
- A: Very few α-particles bounce back from gold foil. R: The nucleus is very small compared with the atom.
- A: The total energy of the electron in H is negative. R: The electron is bound to the nucleus.
- A: The speed of the electron is greatest in n = 1. R: vₙ ∝ 1/n.
- A: Bohr's model explains the spectrum of helium. R: Bohr's model ignores electron–electron repulsion.
- A: Energy levels of H are equally spaced. R: Eₙ ∝ 1/n².
- A: Balmer lines are visible. R: Balmer transitions end on n = 2.
Key: 1 (a) · 2 (a) · 3 (a) · 4 (d) · 5 (d) · 6 (a)
Answers to the practice set
A. 1. 0.53 2. 13.6 3. nh/2π 4. visible 5. zero 6. K = −E 7. n³ 8. nλ
B. 1. T 2. F 3. F 4. T 5. F 6. F 7. T 8. T
C. 1. perpendicular distance of the α velocity line from the nucleus 2. energy to free the electron from the ground state 3. −1.51 eV 4. de Broglie 5. Paschen 6. an upper limit to the nuclear size 7. −27.2 eV 8. r ∝ n² 9. the electron is bound 10. only one-electron atoms / no line intensities
D. 1. p. 2 2. p. 3 3. 2.55 eV: 486 nm 4. p. 6 5. p. 8 6. 3.4 − 0.544 = 2.86 eV 7. accelerated charge radiates → collapse; continuous spectrum 8. 7.3 × 10⁵ m/s, 4.77 Å
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