If the binding energy per nucleon in $Li^7$ and $He^4$ nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction $Li^7+p \rightarrow 2{ }_2 He^4$ is
Answer & explanation
Correct answer: option 4
B.E. of $Li^7=39.20$ MeV and $He^4$ = 28.24 MeV
Hence binding energy of $2 He^4$ = 56.48 MeV
Energy of reaction = 56.48 − 39.20 = 17.28 MeV