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In the fusion reaction ${ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \longrightarrow$ ${ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n}$, the masses of deuteron, helium and neutron expressed in amu are $2.015,3.017$ and 1.009 , respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released. (1 amu $=931.5 \mathrm{MeV})$.
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The correct answer is:
$9.0 \times 10^{13} \mathrm{~J}$
$\Delta \mathrm{m}=2(2.015)-(3.017+1.009)=0.004 \mathrm{amu}$
$\therefore$ Energy released $=(0.004 \times 931.5) \mathrm{MeV}$
$=3.726 \mathrm{MeV}$
Energy released per deuteron
$=\frac{3.726}{2}=1.863 \mathrm{MeV}$
Number of deuterons in 1 kg
$=\frac{6.02 \times 10^{26}}{2}=3.01 \times 10^{26}$
$\therefore$ Energy released per kg of deuterium fusion
$\begin{aligned} & =\left(3.01 \times 10^{26} \times 1.863\right) \\ & =5.6 \times 10^{26} \mathrm{MeV} \approx 9.0 \times 10^{13} \mathrm{~J}\end{aligned}$
$\therefore$ Energy released $=(0.004 \times 931.5) \mathrm{MeV}$
$=3.726 \mathrm{MeV}$
Energy released per deuteron
$=\frac{3.726}{2}=1.863 \mathrm{MeV}$
Number of deuterons in 1 kg
$=\frac{6.02 \times 10^{26}}{2}=3.01 \times 10^{26}$
$\therefore$ Energy released per kg of deuterium fusion
$\begin{aligned} & =\left(3.01 \times 10^{26} \times 1.863\right) \\ & =5.6 \times 10^{26} \mathrm{MeV} \approx 9.0 \times 10^{13} \mathrm{~J}\end{aligned}$
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