The nucleus of the deuterium atom, called the deuteron, consists of a proton and a neutron. Calculate the deutron's binding energy, given atomic mass, i.e., the mass of a deuterium nucleus plus an electron is measured to be `2.014 102 u`.

Options

A. `0.002388 MeV`
B. `2.014102 MeV`
C. `2.16490 MeV`
D. `2.224 MeV`

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) +M_(n)-M(D)]`
`M(D) =`mass of deutron `=2.016490 u-2.014102 u`
`=0.002388 u`
As `I u` corresponds to `931.494 MeV` energy, therefore, mass defect corresponds to energy,
`E_(b) =0.002388 xx 931.5=2.224MeV`.