Question

The binding energy of deuteron ${ }_{1}^{2} H$ is $1.112\, MeV$ per nucleon and an $\alpha$-particle ${ }_{2}^{4}$ He has a binding energy of $7.047\, MeV$ per nucleon. Then in the fusion reaction ${ }_{1}^{2} H +{ }_{1}^{2} H \rightarrow{ }_{2}^{4} He +Q$, the energy $Q$ released is

• A. 1 MeV
• B. 11.9 MeV
• C. 23.8 MeV
• D. 931 MeV

Answer 1

Answer: (C)

Mass of ${ }_{1} H ^{2}=2.01478$ a.m.u.
Mass of ${ }_{2} He ^{4}=4.00388$ a.m.u.
Mass of two deuterium $=2 \times 2.01478=4.02956$
Energy equivalent to $2_{1} He ^{2}$
$=4.02956 \times 1.112\, MeV =4.48 \,MeV$
Energy equivalent to ${ }_{2} H ^{4}$
$=4.00388 \times 7.047 \,MeV =28.21 \,MeV$
Energy released $=28.21-4.48=23.73 \,MeV$
$=23.8 \,MeV$