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,
$\frac{dN}{dt}=\frac{10^{-4}}{1\text{.}6\times 10^{-19}\times 1000}-\lambda N$
or $\frac{dN}{dt}=6\text{.}25\times 10^{11}-\lambda N$
or ${\int }_o^{N{}_0}\frac{dN}{6\text{.}25\times 10^{11}-\lambda N}={\int }_0^{3600}dt$
where $N_0$ are the number of nuclei at $t=1hour$ or $3600s$ .
∴ $-\frac{1}{\lambda }\text{ln}\left(\frac{6\text{.}25\times 10^{11}-\lambda {\left(N\right)}_0}{6\text{.}25\times 10^{11}}\right)=3600$
$\lambda N_0$ = activity of $\_{}^{7}Be$ at $t=1hour$ $=1.8\times 10^8$ disintegrations per second
∴ $-\frac{1}{\lambda }\text{ln}\left(\frac{6\text{.}25\times 10^{11}-1\text{.}8\times 10^8}{6\text{.}25\times 10^{11}}\right)=3600$
∴ $\lambda =8.0\times 10^{-8}{s}^{-1}$
Therefore, half-life $t_{1/2}=\frac{0\text{.}693}{8\text{.}0\times 10^{-8}}=8\text{.}66\times 10^{6}s$
$=100.26$ days