Question: If the binding energy per nucleon in $Li^{7}$and $He^{4}$ nuclei are respectively 5.60 MeV and 7...

Question

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 $Li^{7} + p \rightarrow 2\mspace{6mu}_{2}He^{4}$ is.

Answer 1

Solution

B.E. of $Li^{7} = 39.20\mspace{6mu} MeV$ and $He^{4} = 28.24\mspace{6mu} MeV$

Hence binding energy of $2He^{4} = 56.48\mspace{6mu} MeV$

Energy of reaction $= 56.48 - 39.20 = 17.28\mspace{6mu} MeV$ .

Question: If the binding energy per nucleon in \(Li^{7}\)and \(He^{4}\) nuclei are respectively 5.60 MeV and 7...

Solution