CBSE Class 12 Physics Atoms MCQs Set 07

Answer: (c) most of the space inside an atom is empty

Explanation:
1. An α-particle is deflected appreciably only when it passes very close to a nucleus.
2. Since most α-particles go straight through, they met nothing in their path, so most of the atom is empty space.

Teacher's Note:
1. Large-angle scattering of a few α-particles is what shows the nucleus is tiny and heavy.
2. Option (a) describes Thomson's model, which the experiment disproved.

Answer: (b) 4.5 × 10⁻¹⁴ m

Explanation:
1. At closest approach K = k \( \frac{(2e)(Ze)}{r_0} \), so r₀ = \( \frac{k \cdot 2Ze^2}{K} \)
2. With K = 5 MeV = 5 × 10⁶ × 1.6 × 10⁻¹⁹ J, one factor of e cancels: r₀ = \( \frac{9 \times 10^9 \times 2 \times 79 \times 1.6 \times 10^{-19}}{5 \times 10^6} \)
3. r₀ = 4.5 × 10⁻¹⁴ m

Teacher's Note:
1. The charge of an α-particle is 2e; forgetting the 2 gives 2.3 × 10⁻¹⁴ m.
2. Writing K in eV lets one factor of e cancel and keeps the arithmetic short.

Answer: (d) half

Explanation:
1. r₀ = \( \frac{k \cdot 2Ze^2}{K} \), so r₀ ∝ \( \frac{1}{K} \)
2. Doubling K makes r₀ half.

Teacher's Note:
1. A more energetic α-particle pushes closer against the repulsion.
2. Do not confuse with speed: doubling speed would make K four times and r₀ one-fourth.

Answer: (b) 180°

Explanation:
1. Zero impact parameter means a head-on approach along the line through the centre of the nucleus.
2. The α-particle stops at the distance of closest approach and is pushed straight back, so θ = 180°.

Teacher's Note:
1. Small impact parameter gives large deflection; large impact parameter gives small deflection.
2. 0° would be the case for a very large impact parameter.

Answer: (b) −0.85 eV

Explanation:
1. E_n = −\( \frac{13.6}{n^2} \) eV
2. E₄ = −\( \frac{13.6}{16} \) = −0.85 eV

Teacher's Note:
1. Multiplying 13.6 by 4 (giving −54.4 eV) is a common slip; energy goes as 1/n².
2. −1.51 eV is the n = 3 value.

Answer: (c) 2.12 Å

Explanation:
1. r_n = n²a₀
2. r₂ = 4 × 0.53 = 2.12 Å

Teacher's Note:
1. Radius grows as n², not as n, so 1.06 Å is wrong.
2. 4.77 Å is the radius of the n = 3 orbit.

Answer: (a) 2 : 1

Explanation:
1. v_n ∝ \( \frac{1}{n} \)
2. v₁ : v₂ = \( \frac{1}{1} : \frac{1}{2} \) = 2 : 1

Teacher's Note:
1. The electron moves fastest in the innermost orbit.
2. 4 : 1 would be the ratio of kinetic energies, not speeds.

Answer: (d) −27.2 eV

Explanation:
1. In any Bohr orbit U = 2E.
2. Ground state E₁ = −13.6 eV, so U = 2 × (−13.6) = −27.2 eV

Teacher's Note:
1. The kinetic energy is +13.6 eV, so K + U = −13.6 eV.
2. −13.6 eV is the total energy, not the potential energy.

Answer: (a) 486 nm

Explanation:
1. ΔE = E₄ − E₂ = −0.85 − (−3.4) = 2.55 eV
2. λ = \( \frac{1240}{2.55} \) ≈ 486 nm (visible, blue-green)

Teacher's Note:
1. Find the energy difference first, then use λ = hc/ΔE.
2. 656 nm belongs to the smaller 3 → 2 jump; a smaller gap gives a longer wavelength.

Answer: (c) 3.4 eV

Explanation:
1. The first excited state is n = 2, where E₂ = −3.4 eV.
2. To free the electron its energy must be raised to 0, so the energy needed is 3.4 eV.

Teacher's Note:
1. The first excited state is n = 2, not n = 1.
2. 10.2 eV is the energy needed to reach n = 2 from the ground state, a different question.

Answer: (b) 3.17 × 10⁻³⁴ J s

Explanation:
1. L = \( \frac{nh}{2\pi} \) = \( \frac{3 \times 6.63 \times 10^{-34}}{2 \times 3.14} \)
2. L ≈ 3.17 × 10⁻³⁴ J s

Teacher's Note:
1. Angular momentum grows in steps of h/2π ≈ 1.06 × 10⁻³⁴ J s.
2. Forgetting the 2π gives 1.99 × 10⁻³³ J s.

Answer: (d) 4

Explanation:
1. A stationary orbit holds a standing wave: 2πr_n = nλ.
2. For n = 4 the circumference contains 4 wavelengths.

Teacher's Note:
1. This condition leads directly to mvr = nh/2π.
2. 16 comes from wrongly using n² instead of n.

Answer: (c) C

Explanation:
1. λ = hc/ΔE, so the longest wavelength comes from the smallest energy gap.
2. Gaps: A (2 → 1) = 10.2 eV, B (3 → 2) = 1.89 eV, C (4 → 3) = 0.66 eV, D (4 → 2) = 2.55 eV.
3. The smallest gap is for C, so it gives the longest wavelength.

Teacher's Note:
1. Levels get closer together as n increases, so jumps between high levels give long wavelengths.
2. Transition A, ending on the ground state, gives the shortest wavelength (ultraviolet).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. About 1 in 8000 α-particles was deflected by more than 90°, so the Assertion is true.
2. The nucleus is about 10⁻¹⁵ to 10⁻¹⁴ m across while the atom is about 10⁻¹⁰ m, so the Reason is true.
3. Since the nucleus is so small, very few α-particles come close enough to it to be strongly deflected. The Reason explains the Assertion.

Teacher's Note:
1. Large deflection needs a small impact parameter, which is rare.
2. This is the key conclusion of Rutherford's model.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. An orbiting electron is accelerated, so classically it should radiate and spiral into the nucleus. Rutherford's model cannot prevent this, so the Assertion is true.
2. The Reason is a correct statement of classical theory.
3. The radiation of energy by the accelerated electron is exactly why the model fails, so the Reason explains the Assertion.

Teacher's Note:
1. Bohr solved this by postulating non-radiating stationary orbits.
2. The same argument also predicts a continuous spectrum, which is not observed.

Answer: (d) A is false, but R is true.

Explanation:
1. K_n = 13.6/n² eV decreases as n increases, so the Assertion is false.
2. v_n = 2.18 × 10⁶/n m s⁻¹, so the Reason is true.

Teacher's Note:
1. The total energy increases (becomes less negative) with n, but the kinetic energy decreases.
2. K = −E links the two.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. E₁ = −13.6 eV, so 13.6 eV is needed to free the electron. The Assertion is true.
2. The Bohr radius is 0.53 Å, so the Reason is true.
3. The value of the radius alone does not explain why the ionisation energy is 13.6 eV, so the Reason is not the correct explanation.

Teacher's Note:
1. The ionisation energy follows from the energy of the ground state, E₁ = −13.6 eV.
2. Choose (b) when both statements are true facts but unrelated as cause and effect.

Answer: (c) A is true, but R is false.

Explanation:
1. E₁ = −13.6 eV is the most negative allowed energy, so the Assertion is true.
2. r_n = n²a₀, so the ground state has the smallest radius. The Reason is false.

Teacher's Note:
1. Lowest energy goes with the smallest orbit.
2. Energy rises towards 0 as the orbit gets larger.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. E₂ − E₁ = 10.2 eV, E₃ − E₂ = 1.89 eV, E₄ − E₃ = 0.66 eV, so the Assertion is true.
2. E_n = −13.6/n² eV, so the Reason is true.
3. Because 1/n² changes less and less as n increases, the levels crowd together. The Reason explains the Assertion.

Teacher's Note:
1. The levels converge towards E = 0 at n = ∞.
2. This is why the energy-level diagram looks crowded near the top.

Answer: (c) 5.7 × 10⁻¹⁴ m

Explanation:
1. r₀ = \( \frac{k \cdot 2Ze^2}{K} \) = \( \frac{9 \times 10^9 \times 2 \times 79 \times 1.6 \times 10^{-19}}{4 \times 10^6} \) (K in eV)
2. r₀ ≈ 5.7 × 10⁻¹⁴ m

Teacher's Note:
1. Use 2e for the α-particle and 79e for the gold nucleus.
2. The result is far smaller than the 10⁻¹⁰ m size of an atom.

Answer: (b) zero kinetic energy; all its energy is electrostatic potential energy

Explanation:
1. As the α-particle moves towards the nucleus, the repulsion slows it down and kinetic energy turns into potential energy.
2. At the closest point it stops momentarily, so K = 0 and all its energy is potential energy, equal to 4 MeV.

Teacher's Note:
1. The total energy stays 4 MeV throughout (energy conservation).
2. This is the condition used to find r₀.

Answer: (d) had a very small impact parameter

Explanation:
1. A large scattering angle needs a strong repulsive force, which acts only close to the nucleus.
2. So this α-particle approached with a very small impact parameter.

Teacher's Note:
1. Electrons are far too light to turn an α-particle back.
2. Few α-particles have small impact parameters, so large-angle events are rare.

Answer: (a) 2.8 × 10⁻¹⁴ m

Explanation:
1. r₀ ∝ 1/K, so doubling K halves r₀.
2. r₀ = 5.7 × 10⁻¹⁴/2 ≈ 2.8 × 10⁻¹⁴ m

Teacher's Note:
1. Use proportion rather than recalculating from scratch.
2. 1.4 × 10⁻¹⁴ m would need a fourfold increase in energy.

Answer: (b) less than r₀

Explanation:
1. The α-particle turns back at r₀ without touching the nucleus, because the scattering follows Coulomb's law.
2. So the nucleus must be smaller than this distance: its radius is less than r₀.

Teacher's Note:
1. r₀ gives an upper limit to the size of the nucleus.
2. 10⁻¹⁰ m is the size of the whole atom.

Answer: (b) −1.51 eV

Explanation:
1. E₃ = −\( \frac{13.6}{9} \) = −1.51 eV

Teacher's Note:
1. Divide by n², not by n (−4.53 eV comes from dividing by 3).
2. The energy is negative because the electron is still bound.

Answer: (d) 12.09 eV

Explanation:
1. hν = E₃ − E₁ = −1.51 − (−13.6)
2. hν = 12.09 eV

Teacher's Note:
1. This is also the excitation energy needed to raise the atom from n = 1 to n = 3.
2. 1.89 eV is the 3 → 2 jump and 10.2 eV is the 2 → 1 jump.

Answer: (a) 3

Explanation:
1. Possible jumps: 3 → 2, 3 → 1 and 2 → 1.
2. Number of lines = \( \frac{n(n-1)}{2} \) = \( \frac{3 \times 2}{2} \) = 3

Teacher's Note:
1. Many atoms are excited, so all possible routes occur together.
2. 6 would be the answer for atoms excited to n = 4.

Answer: (b) 9 times

Explanation:
1. r_n ∝ n²
2. \( \frac{r_3}{r_1} \) = \( \frac{3^2}{1^2} \) = 9 times

Teacher's Note:
1. r₃ = 9 × 0.53 Å ≈ 4.77 Å.
2. One-ninth is the ratio of energies in magnitude, |E₃|/|E₁|.

Answer: (c) 1.51 eV

Explanation:
1. Ionisation raises the electron's energy from E₃ = −1.51 eV to 0.
2. Energy needed = 1.51 eV

Teacher's Note:
1. An excited atom is much easier to ionise than a ground-state atom.
2. 13.6 eV is needed only from the ground state.