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In the fusion reaction $_{1}^{2} H +_{1}^{2} H {\to_{}^{}}_{2}^{3} H e +_{0}^{1} n,$ the masses of deuteron helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1kg of deuterium undergoes complete fusion, the amount of total energy released (1amu = 931.5 $M e V / c^{2}$ ) is $p \times 10^{13} J$ . Find $p$ .
(Round off to nearest integer)

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answer is 9.

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Detailed Solution

$Δ m = 2 (2.015) - (3.017 + 1.009) = 0.004 a m u$

Energy released = $(0.004 \times 931.5) M e V = 3.726 M e V$

Number of deuterons in $1 k g = \frac{6.02 \times 10^{26}}{2} = 3.01 \times 10^{26}$

Energy released per kg of deuterium fusion = $(3.01 \times 10^{26} \times 1.863) = 5.6 \times 10^{26} M e V = 9.0 \times 10^{13} J$

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