JEE Physics Nuclei MCQs Set 30

Answer: (b) 8 and 6

Explanation:
1. \( \alpha \)-particles: \( \dfrac{238-206}4=8 \).
2. Charge: \( 92-16+N_\beta=82\Rightarrow N_\beta=6 \).

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) 3.86

Explanation:
1. \( N_U=N_0\left(\tfrac12\right)^{1.5/4.5}=N_0\,2^{-1/3} \) and \( N_{Pb}=N_0-N_U \).
2. \( \dfrac{N_U}{N_{Pb}}=\dfrac1{2^{1/3}-1}=\dfrac1{0.259}=3.86 \).

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) 7 s

Explanation:
1. Counts in successive equal intervals fall by a constant factor \( e^{-2\lambda}=0.75 \).
2. \( 2\lambda=\ln\dfrac43=2\ln2-\ln3=0.2876\Rightarrow\lambda=0.1438\ \text s^{-1} \), so \( \tau=\dfrac1\lambda=6.95\approx7 \) s.

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) \( 2.32\times10^{19} \)

Explanation:
1. \( N_X=10^{20}e^{-0.1\times16.48}=1.92\times10^{19} \). \( N_Y=\dfrac{10^{20}\times0.1}{0.0667}\left[e^{-0.549}-e^{-1.648}\right]=5.77\times10^{19} \).
2. \( N_Z=10^{20}-N_X-N_Y=2.32\times10^{19} \).

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) 14.43 s

Explanation:
1. 25% to 12.5% in 10 s means the half-life is 10 s.
2. \( \tau=\dfrac{T_{1/2}}{\ln2}=\dfrac{10}{0.693}=14.43 \) s.

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) 40 s

Explanation:
1. Reducing to 6.25% \( =\tfrac1{16} \) of the 12.5% value needs 4 half-lives.
2. \( t=4\times10=40 \) s.

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) \( 3.96\times10^{-6} \)

Explanation:
1. \( v=\sqrt{\dfrac{2K}{m}}=\sqrt{\dfrac{2\times0.0327\times1.6\times10^{-19}}{1.675\times10^{-27}}}=2.5\times10^3 \) m/s, so the time of flight is \( t=\dfrac{10}{2500}=4\times10^{-3} \) s.
2. Fraction decayed \( =1-e^{-\lambda t}\approx\dfrac{0.693\times4\times10^{-3}}{700}=3.96\times10^{-6} \).

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) 91 and 234

Explanation:
1. After \( \alpha \)-emission: \( Z=90 \), \( A=234 \).
2. After \( \beta^- \)-emission: \( Z=91 \), \( A=234 \).

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) \( \dfrac1{2\lambda} \)

Explanation:
1. \( \dfrac{N_A}{N_B}=e^{-(5\lambda-\lambda)t}=e^{-4\lambda t} \).
2. Equal to \( e^{-2} \) gives \( t=\dfrac1{2\lambda} \).

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) 1

Explanation:
1. Nuclear radius \( R=R_0A^{1/3} \), so the volume is proportional to A.
2. Density \( \propto\dfrac AA \) is the same for all nuclei: ratio 1.

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) \( x=n,\ y=n,\ K_{Sr}=129 \) MeV, \( K_{Xe}=86 \) MeV

Explanation:
1. Charge and mass number: \( 92=54+38 \) and \( 236=140+94+2 \), so x and y are two neutrons.
2. Energy released \( =234\times8.5-236\times7.5=219 \) MeV; the two neutrons take 4 MeV, leaving 215 MeV. Equal momenta give \( K\propto1/m \): \( K_{Sr}=215\times\dfrac{140}{234}=129 \) MeV, \( K_{Xe}=86 \) MeV.

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) \( E\left({}^{236}_{92}\text U\right)\gt E\left({}^{137}_{53}\text I\right)+E\left({}^{97}_{39}\text Y\right)+2E(n) \)

Explanation:
1. Fission releases energy, so the rest mass of the parent exceeds the total rest mass of the products.
2. Check the mass numbers: 137 + 97 + 2 = 236, so the first option is a balanced fission and the inequality must point the way energy is released.

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) 5.4 MeV

Explanation:
1. Momentum conservation gives \( K_\alpha=Q\dfrac{A-4}{A} \).
2. \( K_\alpha=5.5\times\dfrac{216}{220}=5.4 \) MeV.

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) \( 10^{17}\ \text{kg/m}^3 \)

Explanation:
1. \( R=1.2\times10^{-15}A^{1/3} \) m, so \( \rho=\dfrac{A\,m_p}{\frac43\pi R_0^3A}=\dfrac{1.67\times10^{-27}}{\frac43\pi(1.2\times10^{-15})^3} \).
2. \( \rho\approx2.3\times10^{17} \) kg/m\(^3\).

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) \( W\rightarrow2Y \)

Explanation:
1. Energy is released when the products have a higher total binding energy than the parent.
2. \( W\rightarrow2Y \): before \( 120\times7.5=900 \) MeV, after \( 2\times60\times8.5=1020 \) MeV, a gain of 120 MeV.

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) \( 10^{12}\ \text s \)

Explanation:
1. Net: \( 3\,^2\text H\rightarrow{}^4\text{He}+n+p \). \( \Delta m=3\times2.014-(4.001+1.008+1.007)=0.026 \) amu \( =24.2 \) MeV per 3 deuterons.
2. Total energy \( =\dfrac{10^{40}}{3}\times24.2\times1.6\times10^{-13}=1.3\times10^{28} \) J, so \( t=\dfrac{1.3\times10^{28}}{10^{16}}\approx10^{12} \) s.

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) two light nuclei combine to give a heavier nucleus and possibly other products

Explanation:
1. Fusion joins light nuclei (such as deuterium) into a heavier one with release of energy.
2. The other statements describe radioactive decay or fission.

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) A deuteron and an alpha particle can undergo complete fusion

Explanation:
1. \( M(\text H^2)+M(\text{He}^4)=6.016705 \) u, which is greater than \( M(\text{Li}^6)=6.01513 \) u.
2. So the fusion \( d+\alpha\rightarrow{}^6\text{Li} \) is exothermic and allowed, while Li-6 cannot split into \( d+\alpha \).

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.