Question

The binding energy per nucleon of deuteron ${(}_1^{2}H)$ and helium nucleus ${(}_2^{4}He)$ is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

• A. 13.9 MeV
• B. 26.9 MeV
• C. 23.6 MeV
• D. 19.2 MeV

Answer 1

The correct option is C 23.6 MeV

${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^4+Q$
Total binding energy of helium nucleus = $4\times 7=28MeV$
Total binding energy of each deutron = $2\times 1.1=2.2MeV$
Hence energy released = $28-2\times 2.2=23.6MeV$