JEE Physics Nuclei MCQs Set 07

Answer: (d) 4α, 1β

Explanation:
1. α-particles = (238 − 222)/4 = 4
2. Z after 4 α-decays = 90 − 8 = 82; final Z = 83, so one β⁻ particle.

Teacher's Note:
1. Check: 90 − 2(4) + 1 = 83.
2. Mass number change comes only from α-particles.

Answer: (c) 172, 69

Explanation:
1. After α: (176, 70); after β⁻: (176, 71); after α: (172, 69); γ changes nothing
2. X₄ has mass number 172 and atomic number 69.

Teacher's Note:
1. Total change: A decreases by 8 (two α) and Z by 4 − 1 = 3.
2. γ-decay does not change A or Z.

Answer: (b) \(\frac{\ln 2}{\lambda}\) and \(\frac{1}{\lambda}\)

Explanation:
1. Half-life T½ = ln2/λ
2. Mean life τ = 1/λ.

Teacher's Note:
1. Mean life is the average lifetime of a nucleus.
2. τ = T½/ln2 = 1.44 T½.

Answer: (b) about 2/3 of the substance disintegrates

Explanation:
1. After one mean life N = N₀/e = 0.37N₀
2. So 63%, about 2/3, has disintegrated.

Teacher's Note:
1. In one half-life 50% decays; in one mean life 63% decays.
2. In 5 mean lives more than 99% has decayed.

Answer: (c) 125 dps

Explanation:
1. 3 s = 3 half-lives
2. Activity = 1000 × (1/2)³ = 125 dps.

Teacher's Note:
1. Activity halves in each half-life.
2. dps means disintegrations per second.

Answer: (a) 1.35 g

Explanation:
1. m = m₀e^(−λt) with t = 2/λ, so m = 10 e⁻² = 10 × 0.1353
2. m ≈ 1.35 g.

Teacher's Note:
1. Mean life τ = 1/λ, so two mean lives give e⁻².
2. e⁻² = 0.135.

Answer: (c) β-rays

Explanation:
1. β-rays are streams of fast electrons (particles)
2. X-rays, γ-rays and heat (infra-red) rays are electromagnetic waves.

Teacher's Note:
1. Both α and β rays are particle streams.
2. γ-rays have the shortest wavelength of the EM spectrum.

Answer: (c) 10

Explanation:
1. 1/16 = (1/2)⁴, so 4 half-lives = 40 days
2. T½ = 40/4 = 10 days.

Teacher's Note:
1. Write the fraction as a power of 1/2.
2. T½ = t/n.

Answer: (a) β-decay

Explanation:
1. Emission of an electron with an anti-neutrino, with Z increasing by 1 and A unchanged, is β⁻ decay.

Teacher's Note:
1. In β⁻ decay, a neutron changes into a proton.
2. Parent and daughter are isobars.

Answer: (d) 93.7%

Explanation:
1. 4 hours = 240 min = 4 half-lives
2. Remaining = (1/2)⁴ = 1/16 = 6.25%, so decayed = 93.75% ≈ 93.7%.

Teacher's Note:
1. Percentage decayed = 100 − percentage remaining.
2. After 4 half-lives, 15/16 has decayed.

Answer: (a) 1/8

Explanation:
1. 60 min/20 min = 3 half-lives
2. Fraction left = (1/2)³ = 1/8.

Teacher's Note:
1. Convert both times to the same unit.
2. Three half-lives leave one-eighth.

Answer: (d) 100 disintegrations/s

Explanation:
1. 3 s = 3 half-lives
2. Rate = 800/2³ = 100 disintegrations per second.

Teacher's Note:
1. Activity halves with every half-life.
2. The decay rate is the activity.

Answer: (a) 1/8

Explanation:
1. 72000/24000 = 3 half-lives
2. Fraction left = (1/2)³ = 1/8.

Teacher's Note:
1. Plutonium waste must be stored for hundreds of thousands of years.
2. Count the half-lives first.

Answer: (c) β-rays

Explanation:
1. A magnetic field deflects only charged particles; β-rays (electrons) are charged, while X-rays, γ-rays and radio waves are neutral electromagnetic waves.

Teacher's Note:
1. The deflection force is F = qv × B.
2. α and β deflect in opposite directions.

Answer: (a) 1/1024

Explanation:
1. 125/12.5 = 10 half-lives
2. Fraction left = (1/2)¹⁰ = 1/1024.

Teacher's Note:
1. 2¹⁰ = 1024 is worth remembering.
2. Tritium is a radioactive isotope of hydrogen.

Answer: (c) zinc sulphide screen

Explanation:
1. α-particles produce tiny flashes (scintillations) on a zinc sulphide screen, which was used in the alpha-scattering experiment.

Teacher's Note:
1. Geiger–Müller counters detect α and β particles.
2. Gold foil was the target in Rutherford's experiment.

Answer: (c) 2300 years

Explanation:
1. τ = T½/ln2 = 1600/0.693 = 2309 years ≈ 2300 years.

Teacher's Note:
1. Mean life = 1.44 × half-life.
2. 1.44 × 1600 = 2304 years.

Answer: (a) a = 89, b = 234

Explanation:
1. After α-emission: Z = 92 − 2 = 90 and A = 238 − 4 = 234
2. After β⁺ emission A is unchanged and Z decreases by 1: Z = 89
3. So a = 89 and b = 234.

Teacher's Note:
1. α lowers A by 4 and Z by 2; β⁺ lowers Z by 1 and leaves A unchanged.
2. Track A and Z step by step.

Answer: (c) a = 110, b = 219, c = 109, d = 215

Explanation:
1. The last step is β⁻, so A stays 215 and Z rises by 1: d = 215 and c + 1 = 110, hence c = 109
2. Before the α-decay: Z = 109 + 2 = 111 and A = 215 + 4 = 219
3. Before the β⁻ decay: Z = 111 − 1 = 110 and A = 219, so a = 110, b = 219.

Teacher's Note:
1. Work backwards from the final nucleus by reversing each decay.
2. Reverse of α: add 4 to A and 2 to Z; reverse of β⁻: subtract 1 from Z.

Answer: (a) n = 8, n′ = 6

Explanation:
1. n = (238 − 206)/4 = 8
2. Charge balance: 92 − 2n + n′ = 82, so n′ = 82 − 92 + 16 = 6.

Teacher's Note:
1. The mass number change fixes the number of α-particles.
2. Then use the atomic number change to find the β⁻ particles.

Answer: (a) 8 α and 8 β

Explanation:
1. α-particles = (232 − 200)/4 = 8
2. Charge: 90 − 2(8) + n′ = 82, so n′ = 8 β⁻ particles.

Teacher's Note:
1. Same two-step method: mass number gives α, charge gives β.
2. The ordinary natural series ends in Pb-208 with 6 α and 4 β; here the target is Pb-200.

Answer: (d) 172 and 69

Explanation:
1. After α: (176, 70)
2. After β⁻: (176, 71)
3. After the second α: (172, 69)
4. γ-emission changes neither A nor Z, so A₄ has mass number 172 and atomic number 69.

Teacher's Note:
1. γ-decay only removes excess energy of the nucleus.
2. Add up the changes: A falls by 8 in total, Z falls by 3.

Answer: (c) \({}^{226}_{88}\mathrm{Ra}\) → \({}^{222}_{86}\mathrm{Rn}\) + \({}^{4}_{2}\mathrm{He}\)

Explanation:
1. α-emission reduces A by 4 and Z by 2: (226 − 4, 88 − 2) = (222, 86)
2. Element with Z = 86 is radon.

Teacher's Note:
1. Check that both the mass number and the charge balance on the two sides.
2. The daughter of α-decay lies two places to the left in the periodic table.

Answer: (c) \({}^{242}_{94}\mathrm{Pu}\) → \({}^{238}_{92}\mathrm{U}\) + \({}^{4}_{2}\mathrm{He}\)

Explanation:
1. A = 242 − 4 = 238 and Z = 94 − 2 = 92
2. The element with Z = 92 is uranium, so the daughter is U-238.

Teacher's Note:
1. In every correct nuclear equation the sums of A and Z are equal on both sides.
2. α-decay is common among heavy nuclei such as Pu.

Answer: (c) \({}^{32}_{15}\mathrm{P}\) → \({}^{32}_{16}\mathrm{S}\) + \({}^{0}_{-1}e\) + antineutrino

Explanation:
1. In β⁻ decay a neutron changes into a proton: Z increases by 1 and A stays the same
2. So Z = 16 (sulphur), A = 32, and an antineutrino is emitted with the electron.

Teacher's Note:
1. β⁻ decay always comes with an antineutrino, β⁺ with a neutrino.
2. The parent and daughter are isobars.

Answer: (c) \({}^{210}_{83}\mathrm{Bi}\) → \({}^{210}_{84}\mathrm{Po}\) + \({}^{0}_{-1}e\) + antineutrino

Explanation:
1. β⁻ decay: Z → Z + 1 = 84 (polonium), A = 210 unchanged
2. An antineutrino accompanies the electron.

Teacher's Note:
1. Count the charge on both sides: 83 = 84 + (−1).
2. β⁻ emitters have too many neutrons.