In the fusion reaction ${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}H + {}_{0}^{1}n,$ 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, the amount of total energy released $____\times 10^{13} J$
$(1 a m u c^{2} = 931.5 M e V)$

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Detailed Solution

$\begin{array}{l} Δ M = 2 (2.015) - 3.017 + (1.009) \\ Δ M = 0.004 a m u \\ Δ E = (Δ M) 931.5 M e V \\ Δ E = 3.726 M e V \end{array}$

Energy release per duetron is $\frac{3.726}{2} = 1.863 M e V$
No. of duetrons in 1 kg
$\frac{6.02 \times 10^{26}}{2} = 3.01 \times 10^{26}$

Energy released per 1 kg of deuterium fusion is

$\begin{array}{l} E = (3.01 \times 10^{26} \times 1.863) \\ = 5.6 \times 10^{26} M e V \\ = 9 \times 10^{13} j o u l e \end{array}$

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