Question

Find the binding energy of the nucleus of lighium isotope `._(3)Li^(7)` and hence find the binding energy per nucleon in it.

Answer 1

Correct Answer - `B.E=[3M_(1H^(1))+4m_(0n^(1))-M_(3Li^(7))]931 MeV`
`=39.22 MeV, (B.E)/(A)=(39.22)/(7)=5.6 MeV`
`B.E=[3M_(1H^(1))+4m_(0n^(1))-M_(3Li^(7))]931 MeV=39.22MeV, (B.E.)/(A)=(39.22)/(7)=5.6 MeV`