JEE Physics Nuclei MCQs Set 19

Answer: (d) \( \frac53R_{Al} \)

Explanation:
1. \( R=R_0A^{1/3}\Rightarrow\frac{R_{Te}}{R_{Al}}=\left(\frac{125}{27}\right)^{1/3}=\frac53 \)

Teacher's Note:
1. Nuclear radius varies as the cube root of the mass number.

Answer: (d) 17.3 MeV

Explanation:
1. BE of Li-7 \( =7\times5.60=39.2\ \text{MeV} \); BE of each He-4 \( =4\times7.06=28.24\ \text{MeV} \).
2. \( Q=2\times28.24-39.2=17.28\approx17.3\ \text{MeV} \)

Teacher's Note:
1. Energy released equals the increase in total binding energy.

Answer: (c) 4.8

Explanation:
1. \( R_{Cu}=3.6\times\left(\frac{64}{27}\right)^{1/3}=3.6\times\frac43=4.8\ \text{fermi} \)

Teacher's Note:
1. Use R proportional to A to the one third.

Answer: (b) 50 days

Explanation:
1. \( N=N_0e^{-\lambda t} \): at \( t_1 \) \( N=\frac23N_0 \), at \( t_2 \) \( N=\frac13N_0 \).
2. \( t_2-t_1=\frac1\lambda\ln\frac{2/3}{1/3}=\frac{\ln2}{\lambda}=T_{1/2}=50\ \text{days} \)

Teacher's Note:
1. The activity halves between those two stages, which takes one half-life.

Answer: (c) \( \frac{9N_0}{2} \)

Explanation:
1. \( 4N_0\,2^{-t}=N_0\,2^{-t/2}\Rightarrow t=4\ \text{min} \).
2. P left \( \frac{N_0}{4} \), Q left \( \frac{N_0}{4} \).
3. R formed \( =(4N_0-\tfrac{N_0}4)+(N_0-\tfrac{N_0}4)=\frac{9N_0}2 \)

Teacher's Note:
1. Find the time when the two populations are equal, then count what has decayed.

Answer: (d) \( 5\log_e2 \)

Explanation:
1. \( A=A_0e^{-\lambda t}\Rightarrow\frac1e=e^{-5\lambda}\Rightarrow\lambda=\frac15\ \text{min}^{-1} \)
2. \( t_{1/2}=\frac{\ln2}\lambda=5\ln2 \)

Teacher's Note:
1. The decay constant follows from the 1/e time.

Answer: (c) \( M(A,Z)=ZM_p+(A-Z)M_n-BE/c^2 \)

Explanation:
1. The mass defect equals the binding energy divided by \( c^2 \): nuclear mass is smaller than the sum of its parts.

Teacher's Note:
1. A bound nucleus weighs less than its separated nucleons.

Answer: (a) \( \frac1{4\lambda} \)

Explanation:
1. \( \frac{N_1}{N_2}=e^{-5\lambda t}/e^{-\lambda t}=e^{-4\lambda t}=e^{-1}\Rightarrow t=\frac1{4\lambda} \)

Teacher's Note:
1. Divide the two exponentials and equate to 1/e.

Answer: (d) 5 : 4

Explanation:
1. A: 4 half-lives, decayed fraction \( 1-\frac1{16}=\frac{15}{16} \).
2. B: 2 half-lives, decayed fraction \( 1-\frac14=\frac34 \).
3. Ratio \( \frac{15/16}{3/4}=\frac54 \)

Teacher's Note:
1. Convert the elapsed time to half-lives for each element.

Answer: (b) \( 7.5\times10^{21} \)

Explanation:
1. Nuclei in 1 g: \( \frac{6.023\times10^{23}}{24}=2.51\times10^{22} \).
2. Decayed fraction in half a half-life: \( 1-2^{-1/2}=0.293 \).
3. \( N_\beta=0.293\times2.51\times10^{22}\approx7.4\times10^{21} \)

Teacher's Note:
1. Each decay of Na-24 emits one beta particle.

Answer: (b) 13

Explanation:
1. \( N=8\times10^{14}\times2^{-24/7}=8\times10^{14}\times0.0929=7.43\times10^{13} \), so n = 13.

Teacher's Note:
1. Apply N = N0 times 2 to the power of minus t over T.

Answer: (b) 1

Explanation:
1. \( \frac{2}{0.25}=8=2^3 \): three half-lives in 3 h, so \( T_{1/2}=1\ \text{h} \)

Teacher's Note:
1. The material has dropped to 1/8 of its original amount.

Answer: (c) 6

Explanation:
1. Rate \( R=0.2\times10^{-6}\times3.7\times10^{10}=7400\ \text{Bq} \); \( \lambda=\frac{0.693}{122}=5.68\times10^{-3}\ \text{s}^{-1} \).
2. \( N=\frac R\lambda=1.30\times10^6 \), so n = 6.

Teacher's Note:
1. The number of nuclei equals activity divided by decay constant.

Answer: (c) 98

Explanation:
1. Nuclei \( =\frac{0.80\times10^{-6}}{131}\times6.02\times10^{23}=3.68\times10^{15} \); \( \lambda=\frac{0.693}{8.1\times86400}=9.9\times10^{-7}\ \text{s}^{-1} \).
2. Activity \( =3.6\times10^9\ \text{Bq}=\frac{3.6\times10^9}{3.7\times10^{7}}\ \text{mCi}\approx98\ \text{mCi} \)

Teacher's Note:
1. Convert the activity from becquerel to curie with 1 mCi = 3.7 x 10^7 Bq.

Answer: (b) 2

Explanation:
1. \( t=\frac{T_{1/2}}{\ln2}\ln\frac1{0.048}=\frac{5700}{0.693}\times3.04\approx2.50\times10^4 \) years, so \( 2n=4 \), n = 2.

Teacher's Note:
1. Age follows from the fraction of activity remaining.

Answer: (b) 2.8

Explanation:
1. Initial rate \( =24\times10^{-6}\times3.7\times10^{10}=8.9\times10^5\ \text{Bq} \).
2. Since \( \lambda t\ll1 \), decays in one year \( \approx8.9\times10^5\times3.16\times10^7=2.8\times10^{13} \)

Teacher's Note:
1. Over a time much shorter than the half-life the activity is practically constant.