How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as ${ }_1^{2}H+{}_1^{2}H\to {}_2^{3}He+n+3.27MeV$

Given reaction,

${ }_1^{2}H+{}_1^{2}H\to {}_2^{3}He+n+3.27MeV$

Mass of deuterium = $2kg=2000g$

Now,

$n=\frac{mass of {}_1{}^{2}H}{atomic weight of {}_1{}^{2}H}\times Avogadro number$

$n=\frac{2000}{1}\times 6.02\times {10}^{23}$

$n=6.02\times {10}^{26}$

3.27 MEV of energy is released in each fusion process.
There are two deuterium nuclei involved in a fusion reaction.
Hence, the energy produced from each deuterium nucleus,

$=\frac{3.27}{2}=1.635MeV$

the energy produced when 2 kg of deuterium fused

$E=1.635\times 6.02\times {10}^{23}=9.843\times {10}^{26}\times 1.6\times {10}^{-13}=1.575\times {10}^{14}J$

Duration of the light's illumination,

$P=100W$

$t=\frac{1.576\times {10}^{14}}{100}$

$t=\frac{1.576\times {10}^{14}s}{100\times 60\times 60\times 24\times 365}\approx 4.9\times {10}^{4}years$

Hence the correct answer is $4.9\times {10}^4 years$.