Number of deuterium atoms is 2 kg
= \(\frac{6.023\times10^{23}}{2}\times2000\) = 6.023 × 1026
Energy released when 6.023 × 1023 nuclei of deuterium fuse together
= \(\frac{3.2}{2}\times6.023\times10^{26}MeV\)
= \(\frac{3.2\times6.023\times10^{26}}{2}\times1.6\times10^{-13}J\)
= 15.42 × 1013 J = 15.42 Ws
Power of lamp = 100 W
If the lamp glows for time t, then electric energy consumed = 100 t
∴ 100 t = 15.42 × 1033
∴ t = 0.1542 × 1013 s
= \(\frac{0.1542\times10^{13}}{365\times86400}y\)
or t = 4.0 × 104 y
