The binding energy per nucleon for `._(3)Li^(7)` will be, if the mass of `._(3)Li^(7)` is 7.0163 a.m.u.

Calculate the binding energy per nucleon of "_3Li^7 nucleus. Given mass of "_3Li^7 nucleus = 7.01599 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 the binding energy per nucleon for a _6C^(12) nucleus. Nuclear mass of _6C^(12) = 12.000000 a.m.u.,mass of hydrogen nucleus =1.007825 a.m.u. and mass of neutron = 1.008665 a.m.u.

The binding energy per nucleon from the following data, mass of Li^(7) =7.01653amu mass of proton =1.00759amu mass of neutron = 1.00898 amu is

Assertion: Lower value of binding energy per nucleon indicates greater stability. Reason : Binding energy per nucleon of ._(3)Li^(7) is greater than that of ._(2)He^(4) . Hence ._(3)Li^(7) is less stable than ._(2)He^(4) .

The binding energy per nucleon of ""_(7)N^(14) nucleus is: (Mass of ""_(7)N^(14) = 14.00307 u ) mass of proton = 1.007825 u mass of neutron = 1.008665 u