How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:
${}_1{}^2\mathrm{H} + {}_1{}^2\mathrm{H} \to {}_2{}^3\mathrm{He}+\mathrm{n}+3.2 \mathrm{MeV}.$

Power of the electric lamp = 100 W 

When two nuclei of deuterium fuse together, energy released = 3.2 MeV 

Number of deuterium atoms in 2 kg is, 
           $$\begin{gathered} = \frac{6.023 \times {10}^{23}}{2}\times 2000 \\ = 6.023 \times {10}^{26} \end{gathered}$$

Energy released when $6.023 \times {10}^{26}$ nuclei of deuterium fuse together, 

               $$\begin{gathered} = \frac{3.2}{2}\times 6.023\times {10}^{26} \mathrm{MeV} \\ = \frac{3.2}{2}\times 6.023 \times {10}^{26} \times 1.6 \times {10}^{-13}\mathrm{J} \\ = 1.54 \times {10}^{14} \mathrm{J} \mathrm{or} \mathrm{Ws} \end{gathered}$$ 
If the lamp glows for time t, then the electrical energy consumed by the lamp is 100 t, 

$\therefore$ 100 t =  $1.54 \times {10}^{14}$ 

            $\mathrm{t} = 1.54 \times {10}^{12}\mathrm{s}$

             $$\begin{gathered} = \frac{1.54 \times {10}^{12}}{3.154 \times {10}^7}\mathrm{years} \\ = 4.88 \times {10}^4 \mathrm{years}. \end{gathered}$$

which is the life span of an electric lamp.