JEE Physics Nuclei MCQs Set 29

Answer: (b) 200

Explanation:
1. 1600 to 100 is a factor 16 = \( 2^4 \) in 8 s, so the half-life is 2 s.
2. At 6 s (3 half-lives): \( \dfrac{1600}{2^3}=200 \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) 20 days and 5 days

Explanation:
1. \( \lambda_AN_A=10 \), \( \lambda_BN_B=20 \) with \( N_A=2N_B \): \( \dfrac{\lambda_A}{\lambda_B}=\dfrac{10/2}{20}=\dfrac14 \).
2. So \( T_A=4T_B \): 20 days and 5 days.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) \( t=T\dfrac{\log1.3}{\log2} \)

Explanation:
1. \( N_A=N_0e^{-\lambda t} \), \( N_B=N_0(1-e^{-\lambda t}) \), so \( \dfrac{N_B}{N_A}=e^{\lambda t}-1=0.3 \).
2. \( \lambda t=\ln1.3\Rightarrow t=\dfrac{T}{\ln2}\ln1.3=T\dfrac{\log1.3}{\log2} \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) 108

Explanation:
1. A reduction by a factor of 64 needs \( 64=2^6 \), i.e. 6 half-lives.
2. \( t=6\times18=108 \) days.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) 20 years and 5 years, respectively

Explanation:
1. Activity \( =\lambda N \): \( \lambda_1(2N)=5 \), \( \lambda_2N=10 \), so \( \dfrac{\lambda_1}{\lambda_2}=\dfrac14 \).
2. \( T_1=4T_2 \): 20 years and 5 years.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) 24000

Explanation:
1. The activity halves in 140 days, so the half-life is 140 days.
2. 280 days is two half-lives: \( A_0=6000\times4=24000 \) dps.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) \( {}^A X_Z\rightarrow{}^A X_Z+f \)

Explanation:
1. In gamma decay a nucleus de-excites by emitting a photon; A and Z do not change.
2. The other processes change Z or A and so are not simple gamma emission.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) the decay of a neutron in a nucleus

Explanation:
1. A neutron converts as \( n\rightarrow p+e^-+\bar\nu \).
2. The electron is created at the moment of decay; it is not stored in the nucleus.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) y will decay at a faster rate than x

Explanation:
1. \( T_x=\tau_y\Rightarrow\dfrac{\ln2}{\lambda_x}=\dfrac1{\lambda_y}\Rightarrow\lambda_x=0.693\lambda_y \).
2. So \( \lambda_y\gt \lambda_x \): y decays faster.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) Beta rays are the same as cathode rays

Explanation:
1. Both beta rays and cathode rays are streams of electrons.
2. Gamma rays are photons, alpha particles are doubly ionised helium nuclei, and the neutron is slightly heavier than the proton.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) Cu\(^{64}\) is radioactive, decaying to Zn\(^{64}\) through \( \beta \)-decay

Explanation:
1. Cu-64 is heavier, so it can decay to the lighter Zn-64 with release of energy.
2. Z increases by 1 at constant A, which is \( \beta^- \) decay (not \( \gamma \)-decay).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) \( \alpha,\beta,\gamma \)

Explanation:
1. Alpha particles are heavily ionising and stopped by paper; beta particles by a thin sheet of metal; gamma rays are the most penetrating.
2. Order: \( \alpha\lt \beta\lt \gamma \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) 12 h

Explanation:
1. A factor 64 means \( 2^6 \): 6 half-lives.
2. \( t=6\times2=12 \) h.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) a neutron in the nucleus decays emitting an electron

Explanation:
1. \( n\rightarrow p+e^-+\bar\nu \) takes place inside the nucleus.
2. The electron is newly created, not a pre-existing nuclear or atomic electron.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (d) (i) and (iii)

Explanation:
1. Mass number: \( 232-212=20=4N_\alpha\Rightarrow N_\alpha=5 \).
2. Charge: \( 90-2\times5+N_\beta=82\Rightarrow N_\beta=2 \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (b) 4

Explanation:
1. \( \lambda=\dfrac{0.693}{1386}=5\times10^{-4}\ \text s^{-1} \), so \( \lambda t=0.04 \).
2. Fraction decayed \( =1-e^{-0.04}=0.039\approx4\% \).

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (c) 1

Explanation:
1. \( A=\lambda N\Rightarrow N=A\tau=10^{10}\times10^9=10^{19} \) atoms.
2. Mass \( =10^{19}\times10^{-25}=10^{-6} \) kg \( =1 \) mg.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.

Answer: (a) lithium, with mass number 7

Explanation:
1. \( ^{10}_5\text B+{}^1_0n\rightarrow{}^4_2\text{He}+{}^A_ZX \).
2. \( A=10+1-4=7 \), \( Z=5-2=3 \): lithium-7.

Teacher's Note:
1. Decide whether the question needs the decay law, the half-life relation or the mass defect, then write that relation.
2. Check the result with a simple case such as one or two half-lives.