Calculate the binding energy of the deuteron, which consistss of a proton and a neutron, given that the atomic mass of the deuteron is 2.014102 amu,. Take mass of proton `(m_(p))=1.007825` amu, mass of a neutron `(m_(n))=1.008665` amu and 1amu=931.5 MeV

Options

A. 0.002388 MeV
B. 2.014102 Me V
C. 2.16490 Me V
D. 2.224 Me V

Answer

D

Explanation

Atomic mass M (H) of hydrogen and nuclear mass `(M_(n)`) are M(H) = 1.007825 u and `M_(n) = 1.008665`u
Mass defect ,
`Delta m = [M (H) + Mn - M (D)]`
M(D) = mass of deuteron = 2.016490 u - 2.014102 u = 0.002388 u
As 1 u corresponds to 931. 494 MeV energy , therefore , mass defect corresponds to energy , `E_(b) = 0.002388 xx 931.5 = 2.224` Me V