(a) Plot of binding energy per nucleon is shown in Figure. From BE/nucleon curve, we note that first B.E. increases rapidly and then decreases slowly and B.E is max i.e. $8.8\mathrm{Mev}$ for ${}^{56}\mathrm{Fe}$ atom. Again by decreasing slowly B.E. become 8.5Mev for uranium atom ${}_{92}^{238}\mathrm{U}$ . This shows that nucleus with mass number are less stable, but some nucleus as ${}^4\mathrm{He},{}^{12}\mathrm{C},{}^{16}\mathrm{O}$ (even-even) nuclei are stable. Thus the nuclei with mass number shows fusion reaction as ${}^2\mathrm{H}$ and ${}^3\mathrm{H}$ have very low $\mathrm{BE}∕$ nucleon in comparison to ${}^4\mathrm{He}$ . Thus when two very light nuclei $(\mathit{A}\le 10$ say) fuse to form a heavy nucleus, the B.E/A of fused heavier nucleus is more than the B.E/A of lighter nuclei. This implies release of energy in nuclear fusion.
Similarly, due to fission of a very heavy nucleus, the B.E/A of the product as daughter nuclei increases which implies the release of huge amount of energy.
Thus for lighter nuclei nuclear fusion and for heavier nuclei nuclear fission takes place and huge amount of energy is released.

(b) Let the initial activity is ${\mathit{R}}_0$ and final activity is $\mathit{R}$ then we have
$\frac{\mathit{R}}{{\mathit{R}}_0}={\left(\frac{1}{2}\right)}^{\mathit{t}∕{\mathit{T}}_1∕2}$
Given $\mathit{R}=3.125\%$ ,

$\mathit{R}=0.03125{\mathit{R}}_0$
$\Rightarrow \frac{\mathit{R}}{{\mathit{R}}_0}=0.03125={\left(\frac{1}{2}\right)}^5$
$\Rightarrow {\left(\frac{1}{2}\right)}^5={\left(\frac{1}{2}\right)}^{\mathit{t}∕1∕1∕2}$

$\mathit{t}=5\times 10$