Q: How long can be electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium ? take the fusion reaction as:

_{1}H^{2} + _{1}H^{2} → _{2}He^{3} + n + 3.27 MeV

Sol: _{1}H^{2} + _{1}H^{2} → _{2}He^{3} + n + 3.27 MeV

No. of atoms in 2 kg of _{1}H^{2} is

$\displaystyle = \frac{2}{2}\times 6.023 \times 10^{26} $

= 6.023 x 10^{26} atoms.

In the above reaction two deuterium nuclei are combined power (P) = W × rate of fusion.

= 3.2 MeV × (Number of atoms)/(Time expended )

100 = 3.27 x 10^{6} x 1.6 x 10^{-19} x (6.023 × 10^{26})/2x

x = (3.27 × 1.6 × 6.023 × 10^{11})/2 = 15.756 x 10^{11} S

= (15.756 × 10^{11})/(365 × 24 × 60 × 60)

= (15.756 × 10^{11})/(3.15 × 10^{7} ) = 5 x 10^{4} years.