Question

$A {}^{7} Li$ target is bombarded with a proton beam current of magnitude $10^{-4} A$ for an hour to produce ${ }^{7} Be$ of activity $1.8 \times 10^{8}$ disintegrations per second. If we assume that one radioactive nucleus of beryllium is produced by bombarding $1000$ protons, then its half-life is close to

• A. $100$ days
• B. $110$ days
• C. $150$ days
• D. $80$ days

Answer 1

Answer: (A)

At time $t$, let say there are $N$ atoms of ${ }^{7} Be$ (radioactive).
Then the net rate of formation of ${ }^{7} Be$ nuclei at this instant is,
$\frac{d N}{d t}=\frac{10^{-4}}{1.6 \times 10^{-19} \times 1000}-\lambda N$
or $\frac{d N}{d t}=6.25 \times 10^{11}-\lambda N$
or $\int\limits_{0}^{N_{0}} \frac{d N}{6.25 \times 10^{11}-\lambda N}=\int\limits_{0}^{3600} d t$
where $N_{0}$ are the number of nuclei at $t=1$ hour or $3600\, s$.
$\therefore \quad-\frac{1}{\lambda} \ln \left(\frac{6.25 \times 10^{11}-\lambda N_0}{6.25 \times 10^{11}}\right)=3600$
$\lambda N_{0}=$ activity of ${}^{7} Be$ at $t=1$ hour $=1.8 \times 10^{8}$
disintegrations per second
$\therefore -\frac{1}{\lambda} \ln \left(\frac{6.25 \times 10^{11}-1.8 \times 10^{8}}{6.25 \times 10^{11}}\right)=3600$
$\therefore \lambda=8.0 \times 10^{-8} s ^{-1}$
Therefore, half-life $\begin{aligned} t_{1 / 2} & =\frac{0.693}{8.0 \times 10^{-8}}=8.66 \times 10^6 \mathrm{~s} \\ & =100.26 \text { days }\end{aligned}$