Question

How long can an electric lamp of 1000 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:

\[{}_{1}^{2}\mathrm{H}+{}_{1}^{2}\mathrm{H}\rightarrow{}_{2}^{3}\mathrm{He}+\mathrm{n}+3.27\mathrm{MeV}\]

Options

• 5.0 × 10⁴ years

• 5.0 × 10¹⁴ years

• 4.5 × 10⁴ years

• 4.5 × 10¹⁴ years

Answer 1

5.0 × 10⁴ years

Explanation:

Fusion reaction is:

\[_1^2\mathrm{H+}_1^2\mathrm{H\to}_2^3\mathrm{He+n+3.27MeV}\]

Given that, Amount of deuterium, m = 2 kg

So, 1 mole, i.e., 2 g of deuterium contains 6.023 × 10²³ atoms.

2.0 kg of deuterium contains

\[=\frac{6.023\times10^{23}}{2}\times2000\]

= 6.023 × 10²⁶ atoms

Given that when two atoms of deuterium fuse, 3.27 MeV energy is released.

∴ Total energy released in the fusion reaction:

\[E=\frac{[3.27\times6.023\times10^{26}]}{2}\mathrm{MeV}\]

\[E=\frac{3.27}{2}\times6.023\times10^{26}\times1.6\times10^{-19}\times10^{6}\]

= 1.576 × 10¹⁴ J

Power of the electric lamp, P = 100 W = 100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow

\[=\frac{1.576\times10^{14}}{100}s\]

\[=\frac{1.576\times10^{14}}{100\times60\times60\times24\times365}=5.0\times10^{4}\mathrm{years}\]