Calculate the binding energy per nucleon of `._20Ca^(40)` nucleus. Given
Channel: Doubtnut
Date: 02/08/2020

Calculate the binding energy per nucleon of `._20Ca^(40)` nucleus. Given  m `._20Ca^(40)=39.962589u` , `M_(p) =1.007825u` and `M_(n)=1.008665 u` and  Take `1u=931MeV`.