Question

How long can an electric lamp of $100\, W$ be kept glowing by fusion of $2.0\, kg$ of deuterium? Take the fusion reaction as $^{2}_{1}H+^{2}_{1}H \rightarrow ^{3}_{2}He +n+3.27\,MeV$

• 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 $2 \,g$ of deuterium $=6.023\times10^{23}$
Total number of atoms present in $2000 \,g$ of deuterium
$=\frac{6.023\times10^{23}\times2000}{2}=6.023\times10^{26}$
Energy released in the fusion of $2$ deuterium atoms
$=3.27\,MeV$
Total energy released in the fusion of $2.0 \,kg$ of deuterium atoms
$E=\frac{3.27}{2}\times6.023\times10^{26}=9.81\times10^{26}\,MeV$
$=15.696\times10^{13}\,J$
Energy consumed by the bulb per second $= 100 \,J$
Time for which the bulb will glow
$t=\frac{15.69\times10^{13}}{100} s$ or $t=\frac{15.69\times10^{11}}{3.15\times10^{7}}$ years
$=4.9\times10^{4} $ years