$\mathrm{A}^7 \mathrm{Li}$ target is bombarded with a proton beam current of $10^{-4} \mathrm{~A}$ for one hour to produce ${ }^7 \mathrm{Be}$ of activity $1.8 \times 10^8$ disintegrations per second. Assuming that one ${ }^7$ Be radioactive nuclei is produced by bombarding 1000 protons, its half-life is

• Correct answer
  A

  $0.87 \times 10^7 \mathrm{~s}$
• B

  $0.2 \times 10^7 \mathrm{~s}$
• C

  $0.67 \times 10^8 \mathrm{~s}$
• D

  $0.87 \times 10^6 \mathrm{~s}$

Solution

Given, $i=10^{-4} \mathrm{~A}, t=1 \mathrm{~h}=3600 \mathrm{~s}$ We know that, $q=$ it $=10^{-4} \times 1 \times 3600=0.36 \mathrm{C}$ $\therefore$ Number of protons $=\frac{q}{e}=\frac{0.36}{1.6 \times 10^{-19}}$ $\left[\because e=1.6 \times 10^{-19} \mathrm{C}\right]$ $=2.25 \times 10^8$ Number of ${ }^7$ Be nuclei produced $=\frac{2.25 \times 10^{18}}{1000}=2.25 \times 10^{15}$ Activity of ${ }^7 \mathrm{Be}=1.8 \times 10^8$ disintegration $/ \mathrm{s}$ $\begin{aligned} \therefore \text { Activity, } A & =\lambda N \\ & =\frac{\ln 2}{t_{1 / 2}} N\end{aligned}$ $\left[\because \lambda=\frac{\ln 2}{t_{1 / 2}}\right]$ $1.8 \times 10^8=\frac{0.693}{t_{1 / 2}} \times 2.25 \times 10^{15}$ $\therefore \quad t_{1 / 2}=\frac{0.693 \times 2.25 \times 10^{15}}{1.8 \times 10^8}=0.87 \times 10^7 \mathrm{~s}$

Hence, the correct answer is option: 'A'.