Question

12. If the binding energy per nucleon in $L{{i}^{7}}$ and $H{{e}^{4}}$ nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction $L{{i}^{7}}+p\xrightarrow{{}}2\,{{\,}_{2}}H{{e}^{4}}$ is:

• A. 17.3 MeV
• B. 8.4 MeV
• C. 2.4 MeV
• D. 19.6 MeV

Answer 1

Answer: (A)

17.3 MeV

Answer 2

Binding energy of $L{{i}^{7}}=39.20\,MeV$ Binding energy of $H{{e}^{4}}=28.24\,MeV$ Therefore binding energy of $2H{{e}^{4}}=56.48\,MeV$ Now energy of reaction is $=56.48-39.20=17.28\text{ }MeV$ $=17.3\text{ }MeV$