Question

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

• A. $2.4\times 10^6$ years
• B. $7.4\times 10^4$ years
• C. $1.6\times 10^6$ years
• D. $4.9\times 10^4$ years

Answer 1

Answer: (D)

Number of atoms present in $2g$ of deuterium $=6.023\times 10^{23}$
Total number of atoms present in $2000g$ of deuterium
$=\frac{6.023\times 10^{23}\times 2000}{2}=6.023\times 10^{26}$
Energy released in the fusion of $2$ deuterium atoms
$=3.27MeV$
Total energy released in the fusion of $2.0kg$ of deuterium atoms
$E=\frac{3.27}{2}\times 6.023\times 10^{26}=9.81\times 10^{26}MeV$
$=15.696\times 10^{13}J$
Energy consumed by the bulb per second $=100J$
Time for which the bulb will glow
$t=\frac{15.69\times 10^{13}}{100}s$ or $t=\frac{15.69\times 10^{11}}{3.15\times 10^7}$ years
$=4.9\times 10^4$ years