JEE Physics Nuclei MCQs Set 11

Answer: (c) \(\frac1{\sqrt2}\)

Explanation:
1. \(N/N_0=(1/2)^{t/T}\)
2. For \(t=T/2\): \((1/2)^{1/2}=\frac1{\sqrt2}\)

Teacher's Note:
1. Use the formula \((1/2)^{t/T}\) for any time, not only whole numbers of half-lives.

Answer: (b) 12 hours

Explanation:
1. The intensity must fall to \(1/64=(1/2)^6\) of its value
2. That needs 6 half-lives
3. Time = \(6\times2=12\) hours

Teacher's Note:
1. Convert the factor into a power of 2 to get the number of half-lives directly.

Answer: (a) \(4.5\times10^{9}\) yr

Explanation:
1. Equal numbers of U and Pb mean half of the uranium has decayed
2. That takes exactly one half-life
3. Age = \(4.5\times10^9\) years

Teacher's Note:
1. Each decayed uranium nucleus becomes a lead nucleus, so a 1 : 1 ratio is always one half-life.

Answer: (b) 2 g

Explanation:
1. Energy per day = \(10^6\times86400=8.64\times10^{10}\) J
2. Energy per fission = \(100\times1.6\times10^{-13}=1.6\times10^{-11}\) J, so number of fissions = \(5.4\times10^{21}\)
3. Mass = \(\frac{5.4\times10^{21}}{6.02\times10^{23}}\times235\approx2.1\) g

Teacher's Note:
1. Always convert MeV to joule and use Avogadro's number to turn the number of nuclei into mass. About 1 MW needs only a few grams of uranium a day.

Answer: (d) 1.862 MeV

Explanation:
1. Mass before = \(1.002+1.004=2.006\) u; mass after = \(1.001+1.003=2.004\) u
2. Mass defect = 0.002 u
3. \(Q=0.002\times931=1.862\) MeV

Teacher's Note:
1. 1 u of mass corresponds to 931 MeV of energy. Subtract the product masses from the reactant masses.

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

Explanation:
1. Activity \(A=\lambda N\)
2. \(\frac{A_1}{A_2}=\frac{\lambda_1N_1}{\lambda_2N_2}=\frac{\lambda_1}{\lambda_2}\times2=\frac{5}{10}\), so \(\lambda_1/\lambda_2=\tfrac14\)
3. Half-life is inversely proportional to \(\lambda\), so \(T_1/T_2=4\), which matches 20 years and 5 years

Teacher's Note:
1. Activity depends on both the decay constant and the number of nuclei. A smaller activity with more nuclei means a smaller decay constant, hence a longer half-life.