Question

The binding energy of deutron ${}_1{H}^2$ is $1.112MeV$ per nucleon and an alpha particle ${}_{2}H{e}^4$ has a binding energy of $7.047MeV$ per nucleon. Then, in the fusion rection
${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^4+Q$, the energy $Q$ released is
(take mass of ${}_1{H}^2=2.01478u$;
mass of ${}_{2}H{e}^4=4.00338u$)

• A. $1.8MeV$
• B. $11.9MeV$
• C. $23.8MeV$
• D. $931MeV$

Answer 1

The correct option is C $23.8MeV$

Mass of ${}_1{H}^2=2.01478u$
Mass of 2 deuterium $=2\times 2.01478=4.02956u$
Mass of ${}_{2}H{e}^4=4.00338u$

Energy equivalent to 2 ${}_1{H}^2=4.02956\times 1.112MeV=4.48MeV$
Energy equivalent to 2 ${}_{2}H{e}^4=4.00388\times 7.047MeV=28.21MeV$

Energy released $=28.21-4.48=23.73MeV\approx 23.8MeV$