Question

The binding energy of deuteron ${(}_1^{2}H)$ is 1.15 MeV per nucleon and an alpha particle ${(}_2^{4}He)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction
${}_1^{2}H{+}_1^{2}H{\to }_2^{4}He+Q$
the energy Q released is

• A. 1 MeV
• B. 11.9 MeV
• C. 23. MeV
• D. 931 MeV

Answer 1

The correct option is C

23. MeV

Binding energy of ${}_1^{2}H=1.15\times$ number of nucelons = 1.15 $\times$ 2 = 2.3 MeV. Total binding energy of reactants = 2.3 + 2.3 = 4.6 MeV. Binding energy of ${}_2^{4}He=7.1\times$ number of nucleons = 7.1 $\times$ 4 = 28.4 MeV. Therefore, Q = 28.4 - 4.6 = 23.8 MeV.
Hence the correct choice is (c)