Question

In fusion reaction ${}_1^{2}H{+}_1^{2}H{\to }_2^{3}He{+}_0^{1}n,$ the masses of deuteron, helium and neutron expressed in $\text{amu}$ are $2.015,3.017$ and $1.009$ respectively. If $1\text{kg}$ of deuterium undergoes complete fusion, find the amount of total energy released $(1\text{amu}{c}^2=931.5\text{MeV})$.

• A. $9\times {10}^{14}\text{J}$
• B. $9\times {10}^{15}\text{J}$
• C. $9\times {10}^{16}\text{J}$
• D. $9\times {10}^{13}\text{J}$

Answer 1

The correct option is D $9\times {10}^{13}\text{J}$

Given equation:

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

Mass defect:

$\Delta m=\left[\right(2.015\times 2)-(3.017+1.009\left)\right]=4\times {10}^{-3}\text{amu}$

$\text{Equivalent energy}=\Delta m{c}^2=4\times {10}^{-3}\times 931.5$

$=3.726\text{MeV}=3.726\times \left({10}^6\right)\times (1.6\times {10}^{-19}))=5.96\times {10}^{-13}\text{J}$

$\text{Number of atoms in 1 kg of}{}_1^{2}H\text{are}$,

$N_a=\text{Number of moles}\times (6.023\times {10}^{23})=\frac{1000\text{g}}{2\text{g}}\times (6.023\times {10}^{23})$

Total energy released:

$E=\frac{N_a}{2}\times E$

$=\frac{1}{2}(\frac{1000}{2}\times 6.023\times {10}^{23})\times 5.96\times {10}^{-13}=9\times {10}^{13}\text{J}$

Hence, $\left(D\right)$ is the correct answer.