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The binding energy of deuteron is 2.2 MeV and that of $^{4}_{2}He$ is 28 MeV. If two deuterons are fused to form one $^{4}_{2}He$ then the energy released is :-
A
25.8 MeV
B
23.6 MeV
C
19.2 MeV
D
30.2 MeV
Detailed Solution
Reaction: $^{2}_{1}H + ^{2}_{1}H \rightarrow ^{4}_{2}He + Q$
Energy released = total binding energy of the product $-$ total binding energy of the reactants.
Total binding energy of two deuterons $= 2 \times 2.2 = 4.4$ MeV
Binding energy of the helium nucleus $= 28$ MeV
Energy released $= 28 - 2 \times 2.2 = 28 - 4.4$
$= 23.6$ MeV
Energy released = total binding energy of the product $-$ total binding energy of the reactants.
Total binding energy of two deuterons $= 2 \times 2.2 = 4.4$ MeV
Binding energy of the helium nucleus $= 28$ MeV
Energy released $= 28 - 2 \times 2.2 = 28 - 4.4$
$= 23.6$ MeV