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Nuclear Fusion

Calculate the...

Question

Calculate the energy released in the following reaction.
$_{3} {L i}^{6} +_{0} n^{1} \to_{2} {H e}^{4} +_{1} H^{3}$
Given :
Mass of $_{3} {L i}^{6}$ nucleus $= 6.015126 a m u$
Mass of $_{1} H^{3}$ nucleus $= 3.016049 a m u$
Mass of $_{2} {H e}^{4}$ nucleus $= 4.002604 a m u$
Mass of $_{0} n^{1}$ $= 1.008665 a m u$

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Solution

Given data :
Mass of $_{3} {L i}^{6}$ nucleus $= 6.015126 a m u$
Mass of $_{1} H^{3}$ nucleus $= 3.016049 a m u$
Mass of $_{2} {H e}^{4}$ nucleus $= 4.002604 a m u$
Mass of $_{0} n^{1}$ $= 1.008665 a m u$
Solution:
Mass of the reactant $=$ mass of $_{3} {L i}^{6} +$ mass of neutron
$= (6.015126 + 1.008665) = 7.023791 a m u$
Mass of the product $=$ mass of $_{2} {H e}^{4} +$ mass of $_{1} H^{3}$
$= (3.016049 + 4.002604) = 7.018653 a m u$
Mass of defect $Δ m = (7.023791 - 7.018653)$
$= 0.005138 a m u$
B.E. $= (0.005138 \times 931) M e V = 4.783 M e V$
Energy released $= 4.783 M e V$

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