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Question: If per nucleon binding energy for \({}_{3}^{7}Li\) and \({}_{2}^{4}He\) are 5.60 MeV and 7.06 MeV, ...

If per nucleon binding energy for 37Li{}_{3}^{7}Li and 24He{}_{2}^{4}He

are 5.60 MeV and 7.06 MeV, then the energy released in the nuclear process 37Li+11p22He4{}_{3}^{7}Li +_{1}^{1}p \rightarrow 2_{2}He^{4}is

A

19.6 MeV

B

2.4 MeV

C

8.4 MeV

D

17.3 MeV

Answer

17.3 MeV

Explanation

Solution

E =