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`.

Calculate the binding energy per nucleon of ._(20)Ca^(40) nucleus. Mass of (._(20)Ca^(40)) = 39.962591 am u .

Find the binding energy per nucleon of (Ca^40_20) nucleus given m^40(_20) Ca = 39.962589u , m_p = 1.00783u , m_n=1.00876u , TAKE 1amu = 931MeV/C^2

Calculate the binding energy per nucleon of "_20Ca^40 nucleus. Given mass of "_20Ca^40 nucleus = 39.962589 a.m.u., mass of proton = 1.007825 a.m.u., mass of neutron = 1.008665 a.m.u. and 1 a.m.u. = 931.5 MeV.

Calculate the binding energy per nucleon of ._26Fe^(56) nucleus. Given that mass of ._26Fe^(56)=55.934939u , mass of proton =1.007825u and mass of neutron =1.008665 u and 1u=931MeV .

Calculate the binding energy per nucleon of ._(20)^(40)Ca . Given that mass of ._(20)^(40)Ca nucleus = 39.962589 u , mass of proton = 1.007825 u . Mass of Neutron = 1.008665 u and 1 u is equivalent to 931 MeV .