Atoms — Class 12 MCQs with Answers

Atoms — important questions & MCQs with answers (Class 12 Physics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Rutherford's α-particle experiment showed:

   A.Atom has a small dense nucleus✓ Correct

   B.Atom is solid

   C.No nucleus

   D.All electrons in centre

   Answer: A. Atom has a small dense nucleus

   Explanation: Most α-particles passed straight through, but a few rebounded through large angles, meaning positive charge and mass are concentrated in a tiny, dense nucleus, not spread through the atom.

2. Q2Easy

   Bohr's model: angular momentum is:

   A.Zero

   B.Continuous

   C.Quantised in units of ℏ✓ Correct

   D.Infinite

   Answer: C. Quantised in units of ℏ

   Explanation: Bohr postulated angular momentum is quantised: L = nh/2π = nℏ, allowing only discrete orbits (n=1,2,3…) — not a continuous range as classical mechanics would allow.

3. Q3Easy

   Rutherford used which foil as target?

   A.Aluminium

   B.Silver

   C.Copper

   D.Gold✓ Correct

   Answer: D. Gold

   Explanation: Rutherford used gold because it can be beaten into an extremely thin foil (few atoms thick), minimising multiple scattering so deflections reflect single encounters with one nucleus.

4. Q4Easy

   In a stationary orbit, an electron:

   A.Has zero velocity

   B.Radiates continuously

   C.Absorbs energy

   D.Does not radiate energy✓ Correct

   Answer: D. Does not radiate energy

   Explanation: Bohr's second postulate states that in stationary orbits an electron does not radiate energy even though it's accelerating, avoiding the classical prediction of a decaying orbit.

5. Q5Easy

   Order of size of nucleus is approximately:

   A.10⁻¹⁰ m

   B.10⁻¹⁵ m✓ Correct

   C.10⁻⁵ m

   D.10⁻²⁰ m

   Answer: B. 10⁻¹⁵ m

   Explanation: The closest-approach calculation from large-angle scattering gives nuclear radius ≈ 10⁻¹⁵ m (a few fermi) — five orders of magnitude smaller than the atom itself.

6. Q6Easy

   Ionisation energy of hydrogen atom from ground state is:

   A.1.51 eV

   B.3.4 eV

   C.13.6 eV✓ Correct

   D.27.2 eV

   Answer: C. 13.6 eV

   Explanation: Ionisation removes the electron from n=1 to n=∞, so the energy needed is E∞ − E₁ = 0 − (−13.6) = 13.6 eV, the full ground-state binding energy.

7. Q7Medium

   Distance of closest approach for α-particle of KE K to nucleus of charge Ze:

   A.2kZe²/K✓ Correct

   B.kZe²/K

   C.4kZe²/K

   D.kZe/K

   Answer: A. 2kZe²/K

   Explanation: At closest approach all kinetic energy converts to electrostatic potential energy: K = k(2e)(Ze)/r₀, so r₀ = 2kZe²/K — note the factor 2 from the α-particle's own +2e charge.

8. Q8Medium

   Energy of nth Bohr orbit (H atom):

   A.−13.6·n² eV

   B.+13.6 eV

   C.−13.6/n² eV✓ Correct

   D.0

   Answer: C. −13.6/n² eV

   Explanation: Eₙ = −13.6/n² eV — negative because the electron is bound; as n grows the energy rises toward zero, so higher orbits are less negative, not larger in magnitude.

9. Q9Medium

   Number of α-particles scattered at angle θ varies as:

   A.sin²(θ)

   B.1/sin⁴(θ/2)✓ Correct

   C.cos²(θ/2)

   D.tan(θ)

   Answer: B. 1/sin⁴(θ/2)

   Explanation: Rutherford's scattering formula gives the count of deflected particles varying as 1/sin⁴(θ/2), so a modest rise in angle sharply cuts the number scattered — why large-angle events were rare.

10. Q10Medium

    Photon emitted when electron jumps from higher to lower orbit. Frequency:

    A.hf = E_low − E_high

    B.hf = E_high − E_low✓ Correct

    C.hf = 0

    D.hf = E_total

    Answer: B. hf = E_high − E_low

    Explanation: Energy conservation: the photon's energy equals the drop in atomic energy, hf = E_high − E_low, always positive since the electron falls to a lower, more negative level.

11. Q11Medium

    Energy of He⁺ (Z=2) in ground state is:

    A.−6.8 eV

    B.−13.6 eV

    C.−27.2 eV

    D.−54.4 eV✓ Correct

    Answer: D. −54.4 eV

    Explanation: Eₙ scales as −13.6Z²/n² eV; for He⁺ (Z=2, n=1): −13.6×4 = −54.4 eV — squaring Z is the step often skipped, giving −27.2 eV instead.

12. Q12Hard

    Limitation of Rutherford model:

    A.No limitation

    B.Predicts unstable atom (electrons should radiate and spiral in)✓ Correct

    C.Predicts strong force

    D.Wrong nucleus charge

    Answer: B. Predicts unstable atom (electrons should radiate and spiral in)

    Explanation: Classical electromagnetism says an accelerating orbiting electron must radiate energy continuously and spiral into the nucleus — a fatal flaw Bohr's quantisation postulate fixed.

13. Q13Hard

    Lyman series corresponds to transitions to:

    A.n = 3

    B.n = 2

    C.n = 1✓ Correct

    D.n = ∞

    Answer: C. n = 1

    Explanation: Lyman series is defined by transitions ending at n=1, the ground state, giving the largest energy gaps and shortest (UV) wavelengths of any hydrogen series.

14. Q14Hard

    α-particle of KE 5 MeV approaches gold nucleus (Z=79). Distance of closest approach is approximately:

    A.4.55 × 10⁻¹⁰ m

    B.4.55 × 10⁻¹⁴ m✓ Correct

    C.7.9 × 10⁻¹⁵ m

    D.1.6 × 10⁻¹⁹ m

    Answer: B. 4.55 × 10⁻¹⁴ m

    Explanation: At closest approach, r₀ = 2kZe²/K; plugging K=5 MeV=8×10⁻¹³ J and Z=79 gives r₀ ≈ 4.55×10⁻¹⁴ m — femtometre scale, not the much larger atomic radius (10⁻¹⁰ m).

Atoms — FAQs

More Class 12 Physics chapters

• 01Electric Charges and Fields
• 02Electrostatic Potential and Capacitance
• 03Current Electricity
• 04Moving Charges and Magnetism
• 05Magnetism and Matter
• 06Electromagnetic Induction

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