Question

If all the atoms of $1kg$ of deuterium undergo fusion approximately how much energy could be released?

• A. $7.87\times {10}^{13}J$
• B. $9\times {10}^{11}$ calories
• C. $9\times {10}^{12}kWh$
• D. $79\times {10}^{7}kWh$

Answer 1

The correct option is A $7.87\times {10}^{13}J$

Fusion reaction of deuterium is given by

${}_1^{2}H{+}_1^{2}H{\to }_2^{3}He{+}_0^{1}n+3.27MeV$

No of atoms present in 2g of deuterium is given by

$N_a=6.02\times {10}^{23}2g=6.02\times {10}^{23}1000g=\frac{6.02\times {10}^{23}\times {10}^3}{2}=\frac{6.02\times {10}^{26}}{2}$

Energy releases in the fusion of 2 deuterium nuclei is 3.27MeV

Energy released in the fusion of $\frac{6.02\times {10}^{26}}{2}$

$=\frac{6.02\times {10}^{26}\times 3.27}{2\times 2}=7.87\times {10}^{13}J$