Deuterium undergoes the following fusion reaction: _1^2 H +_1^2 H →_2^3 He +_0^1 n +3.27 MeV How long an electric bulb of 200 W will glow by using…

Number of atoms in $\displaystyle 2$ g deuterium $\displaystyle =6.023 \times 10^{23}$ Energy released per atom $\displaystyle =\frac{3.27}{2}=1.635 \mathrm{MeV}$ $\displaystyle t=\frac{\text { Total energy released }}{\text { Power }}$ $\displaystyle t=\frac{6.023 \times 10^{23} \times 1.635 \times 1.6 \times 10^{-13}}{200}$ $\displaystyle t=7.88 \times 10^{8} \mathrm{~s}$