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,
$\frac{dN}{dt}=\frac{10^{-4}}{1.6\times 10^{-19}\times 1000}-\lambda N$
or $\frac{dN}{dt}=6.25\times 10^{11}-\lambda N$
or $\underset{0}{\overset{N_0}{\int }}\frac{dN}{6.25\times 10^{11}-\lambda N}=\underset{0}{\overset{3600}{\int }}dt$
where $N_0$ are the number of nuclei at $t=1$ hour or $3600s$.
$\therefore -\frac{1}{\lambda }\ln \left(\frac{6.25\times 10^{11}-\lambda N_0}{6.25\times 10^{11}}\right)=3600$
$\lambda N_0=$ 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 $\begin{array}{rl}{t}_{1/2}& =\frac{0.693}{8.0\times 10^{-8}}=8.66\times 10^6\mathrm{s}\\ & =100.26\text{ days }\end{array}$