JEE Chemistry Nuclear Chemistry MCQs Set 01

Answer: (b) \( {}^{8}_{5}B \)

Explanation:
1. Positron emission occurs when the n/p ratio is below the stability belt, i.e. the nuclide is neutron deficient.
2. B-8 (n/p = 0.6) is neutron deficient. Ca-49, Al-30 and Kr-94 are heavier than the stable isotopes of their elements (neutron rich) and emit beta particles.

Teacher's Note:
1. Compare the mass number with that of the stable isotopes of the same element. Lighter means positron or electron capture, heavier means beta.

Answer: (a) \( {}^{114}_{49}In \)

Explanation:
1. Cd-114 and Sn-114 have even numbers of protons and neutrons, which makes them stable.
2. In-114 has an odd number of protons (49) and odd number of neutrons (65), an odd-odd nucleus, so it is unstable.

Teacher's Note:
1. Odd-odd nuclei are rarely stable.

Answer: (d) \( {}^1_0n \)

Explanation:
1. Mass number: 9 + 4 = 13 = 12 + A, so A = 1.
2. Charge: 4 + 2 = 6 = 6 + Z, so Z = 0. The particle is a neutron.

Teacher's Note:
1. This is the reaction by which Chadwick discovered the neutron.

Answer: (c) \( {}^6_3Li \)

Explanation:
1. \( {}^9_4Be + {}^1_1H \to {}^4_2He + {}^A_ZX \).
2. A = 9 + 1 - 4 = 6 and Z = 4 + 1 - 2 = 3, so X is lithium-6.

Teacher's Note:
1. In (p, alpha) notation the first symbol is the projectile and the second is the emitted particle.

Answer: (b) \( 2.08 \) h\(^{-1}\)

Explanation:
1. \( \lambda = \dfrac{2.303}{t}\log\dfrac{A_0}{A} = \dfrac{2.303}{2}\log64 = \dfrac{4.159}{2} = 2.08 \) h\(^{-1}\).

Teacher's Note:
1. 1/64 means 6 half lives, so half life is 20 minutes.

Answer: (a) \( 5.6\times10^{5} \)

Explanation:
1. Moles of water \( = 10/18 = 0.556 \); atoms of H \( = 2\times0.556\times6.022\times10^{23} = 6.69\times10^{23} \).
2. Tritium atoms at start \( = 8\times10^{-18}\times6.69\times10^{23} = 5.35\times10^6 \).
3. After 40 y: \( 5.35\times10^6\times2^{-40/12.3} = 5.6\times10^5 \).

Teacher's Note:
1. Count hydrogen atoms (2 per molecule) not molecules.

Answer: (d) 6.25%

Explanation:
1. 240 days is 4 half lives.
2. Fraction left \( = (1/2)^4 = 1/16 = 6.25\% \).

Teacher's Note:
1. Divide the elapsed time by the half life first.

Answer: (c) \( 2.7\times10^{5} \) dps

Explanation:
1. \( \lambda = \dfrac{0.693}{30\times24\times3600} = 2.67\times10^{-7} \) s\(^{-1}\).
2. \( A = \lambda N = 2.67\times10^{-7}\times10^{12} = 2.67\times10^{5} \) dps.

Teacher's Note:
1. Convert half life to seconds to get activity in dps.

Answer: (b) 33.7 years

Explanation:
1. \( t = \dfrac{2.303}{\lambda}\log\dfrac{100}{15} = \dfrac{12.3}{0.693}\times1.897 = 33.7 \) years.

Teacher's Note:
1. Age = t half x log2 (N0/N).

Answer: (a) \( 2.8\times10^{9} \) years

Explanation:
1. Ar atoms have all come from K. If K is 1 part, Ar is 3, then originally 4 parts of K existed.
2. K remaining is 1/4 of the original, i.e. two half lives.
3. Age \( = 2\times1.4\times10^9 = 2.8\times10^9 \) years.

Teacher's Note:
1. Original amount = parent now + daughter now.

Answer: (d) 14.25 MeV

Explanation:
1. \( \Delta m = 2\times2.0200-(3.0160+1.0087) = 0.0153 \) amu.
2. Energy \( = 0.0153\times931.5 = 14.25 \) MeV.

Teacher's Note:
1. 1 amu corresponds to 931.5 MeV.

Answer: (c) 8 alpha and 6 beta

Explanation:
1. Alpha particles \( = (238-206)/4 = 8 \).
2. Charge: \( 92-16 = 76 \); to reach 82 needs 6 beta particles.

Teacher's Note:
1. Use mass number to count alpha particles, then correct the charge with beta particles.

Answer: (b) \( {}^{206}_{82}Pb \)

Explanation:
1. Mass number \( = 234-7\times4 = 206 \).
2. Atomic number \( = 90-7\times2+6 = 82 \), which is lead.

Teacher's Note:
1. Each alpha lowers Z by 2 and A by 4; each beta raises Z by 1.

Answer: (a) 0.61 microgram

Explanation:
1. Fraction \( = 2^{-20/28.1} = 2^{-0.712} = 0.61 \).

Teacher's Note:
1. 20 years is less than one half life, so more than half remains.

Answer: (d) 31.2 cm\(^3\)

Explanation:
1. 69.2 days is half a half life; fraction decayed \( = 1-2^{-1/2} = 0.293 \).
2. Moles decayed \( = 0.293/210 = 1.39\times10^{-3} \) and each gives one He atom.
3. Volume \( = 1.39\times10^{-3}\times22400 = 31.2 \) cm\(^3\).

Teacher's Note:
1. Do not assume that 50% decays: 69.2 days is only half of the half life.

Answer: (c) \( 4.5\times10^{9} \) years

Explanation:
1. Moles of U = 23.8/238 = 0.1 and moles of Pb = 20.6/206 = 0.1.
2. Initial U = 0.2 mol, of which 0.1 mol remains: one half life.

Teacher's Note:
1. Equal moles of parent and daughter mean exactly one half life.

Answer: (b) \( 3.09\times10^{-2} \) yr\(^{-1}\)

Explanation:
1. \( \lambda = 0.693/22 = 3.15\times10^{-2} \) yr\(^{-1}\).
2. \( \lambda_{Fr} = 0.98\lambda = 3.09\times10^{-2} \) yr\(^{-1}\) and \( \lambda_{Th} = 6.3\times10^{-4} \) yr\(^{-1}\).

Teacher's Note:
1. The partial constants add up to the overall constant.

Answer: (a) 6 litres

Explanation:
1. Initial activity \( = 3.7\times10^4 \) dps \( = 2.22\times10^6 \) dpm.
2. After 5 h: \( 2.22\times10^6\times2^{-5/15} = 1.76\times10^6 \) dpm.
3. Volume \( = 1.76\times10^6/296 \approx 5950 \) cm\(^3\), about 6 L.

Teacher's Note:
1. Activity per cm3 times volume equals total activity.

Answer: (d) \( 7.1\times10^{8} \) years

Explanation:
1. Moles U \( = 1/238 = 4.20\times10^{-3} \); moles Pb \( = 0.1/206 = 4.85\times10^{-4} \).
2. \( N_0/N = (4.20+0.485)/4.20 = 1.1155 \).
3. \( t = \dfrac{4.5\times10^9}{0.693}\ln1.1155 = 7.1\times10^8 \) years.

Teacher's Note:
1. Convert grams to moles of the same nucleus-count basis, then add parent and daughter.

Answer: (c) 4.12 min

Explanation:
1. \( t_{max} = \dfrac{\ln(\lambda_1/\lambda_2)}{\lambda_1-\lambda_2} \) with \( \lambda_1 = 0.693/3.05 \) and \( \lambda_2 = 0.693/2.68 \).
2. \( = \dfrac{\ln(2.68/3.05)}{0.693(1/3.05-1/2.68)} = \dfrac{-0.1292}{-0.03137} = 4.12 \) min.

Teacher's Note:
1. The daughter is maximum when its production rate equals its decay rate.

Answer: (b) 67.4 h

Explanation:
1. Partial \( \lambda = 0.19\lambda \), so partial \( t_{1/2} = 12.8/0.19 = 67.4 \) h.

Teacher's Note:
1. The partial half life is the half life divided by the fraction of that mode.