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In the fusion reaction $_{1} H^{2} +_{1} H^{2} \to_{2} H e^{3} +_{0} n^{1}$ the masses of deuteron, helium and neutron expressed in amu are 2.015u,3.017u and 1.009u respectively. It 1kg deuteron undergoes complete fusion, the amount of energy released (in $10^{13} J)$

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answer is 9.

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Detailed Solution

$Δm = 2 (2.015) - (3.017 + 1.009) = 0.004 amu$

$E = Δm \times 931.5 = 3.726 MeV$

No.of deuterons in $1 kg = \frac{6.02 \times 10^{26}}{2} = 3.01 \times 10^{26}$

Energy released per 1kg deuteron

$= 3.01 \times 10^{26} \times \frac{3.726}{2} = 5.6 \times 10^{26} MeV = 9 \times 10^{13} J$

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