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Question: The binding energy of deuteron \({}_{1}^{2}H\)is 1.112 MeV per nucleon and an \(\alpha -\)particle \...

The binding energy of deuteron 12H{}_{1}^{2}His 1.112 MeV per nucleon and an α−\alpha -particle 24He{}_{2}^{4}Hehas a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction12H+12H→24He+Q{}_{1}^{2}H +_{1}^{2}H \rightarrow_{2}^{4}He + Q, the energy Q released is.

A

1 MeV

B

11.9 MeV

C

23.8 MeV

D

931 MeV

Answer

23.8 MeV

Explanation

Solution

Mass of 1H2=2.01478 6mua.m.u.1H^{2} = 2.01478\mspace{6mu} a.m.u.

Mass of 2He4=4.00388 6mua.m.u.2He^{4} = 4.00388\mspace{6mu} a.m.u.

Mass of two deuterium

=2×2.01478=4.02956= 2 \times 2.01478 = 4.02956

Energy equivalent to 21H22_{1}H^{2}

=4.02956×1.112 6muMeV=4.48 6muMeV= 4.02956 \times 1.112\mspace{6mu} MeV = 4.48\mspace{6mu} MeV

Energy equivalent to 2H4{ } _ { 2 } H ^ { 4 }

=4.00388×7.047 6muMeV=28.21 6muMeV= 4.00388 \times 7.047\mspace{6mu} MeV = 28.21\mspace{6mu} MeV

Energy released=28.21−4.48=23.73 6muMeV= 28.21 - 4.48 = 23.73\mspace{6mu} MeV = 24 MeV