(a)

Differentiate between nuclear fission and nuclear fusion.

(b)

Deuterium undergoes fusion as per the reaction: \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}+3 \cdot 27 \mathrm{MeV} \] Find the duration for which an electric bulb of $\displaystyle 500$ W can be kept glowing by the fusion of $\displaystyle 100$ g of deuterium.

\begin{aligned} t & =\frac{E}{P} & =\frac{7 \cdot 878 \times 10^{12} \mathrm{~J}}{500 \mathrm{~J} / \mathrm{s}}=1.5756 \times 10^{10} \mathrm{~s} & =\frac{1.5756 \times 10^{10}}{3 \cdot 15 \times 10^{7}} \simeq 500 y \end{aligned}Alternatively :\begin{aligned} & E=\frac{M Q}{2 m_{d}} & =\frac{(0 \cdot 1 \mathrm{~kg}) \times(3 \cdot 27 \mathrm{MeV})}{2(2 \cdot 04) \times\left(1 \cdot 66 \times 10^{-27} \mathrm{~kg} / 4\right)} & =0 \cdot 0492 \times 10^{27}=4 \cdot 92 \times 10^{25} \mathrm{MeV} & t=\frac{E}{P} & =\frac{\left(4 \cdot 92 \times 10^{25}\right) \times\left(1 \cdot 6 \times 10^{-13}\right)}{500}=1.5756 \times 10^{10} \mathrm{~s} & =\frac{1.5756 \times 10^{10}}{3 \cdot 15 \times 10^{7}} \simeq 500 y \end{aligned} $$