I tried to slove this question but im not sure if am right,please help me out..

{a} Calculate the average binding enrgy per nucleon in the deutrium nucleus
{b} The energy that binds an orbiting electron to the hydrogen nucleus is 13.4 eV. Calculate the ratio of the binding energy per nucleon to the binding energy per electron in deutrium. Which particle is held more tightly, the electron or the neutron?

Ans:
{a} Deutrium has 1 neutron and 1 proton
1876.120 MeV/c^2 {mass of deutrium} - 0.511 MeV/c^2{mass of electron} =
1875.609 MeV/c^2 {mass of nucleus}

938.27 MeV/c^2 {mass of proton} + 939.57 MeV/c^2 {mass of neutron} =
1877.84 MeV/c^2

Mass difference = 1877.84 MeV/c^2 - 1875.609 MeV/c^2 {mass of nucleus} =
2.231 MeV/c^2

2.231 / 3 nucleons = 1.1155 MeV/c^2
= 1.12 MeV/c^2

{b} 1.6605*10^-27kg ------ 931.5 MeV/c^2
??? ------ 1.1155 MeV/c^2

E = mc^2 ----- {1.1155 / 931.5} * {1.6605*10^-27} * {2.998 * 10^8}^2 =
= 0.0178726 * 10^-11 J
1eV ------- 1.6 * 10^-19 J
13.4eV ------- ??? = 13.4 * {1.6 * 10^-19} = 21.44 * 10^-19 J

Ratio --- 0.0178726 * 10^-11 J / 21.44 * 10^-19 J = 8.336 * 10^8

Therefore,the raitio of the binding energy per nucleon to the binding energy per electron in deutrium is 1 : 8.34 and the neutron is held more tightly than the electron.

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