The binding energy per nucleon of `._(5)B^(10)` is 8 MeV and that of `._(5)B^(11)` is 7.5 MeV. The energy required to remove a neutron from `._(5)B^(11)` is (mass of electron and proton are `9.11 xx 10^(-21) kg` and `1.67 xx 10^(-27) kg` respectively.

Options

A. 2.5 M eV
B. 8.0 M eV
C. 0.5 M eV
D. 5.7 M eV

Answer

A

Explanation

`E_(1)`= Binding energy of `""_(5)B^(11) = 7.5 xx 11` MeV= 82.5 MeV
`E_(2)=` Binding energy of `""_(5)B^(10) = 8 xx 10= 80` MeV
Energy given `= E_(1)- E_(2)= 82.5 - 80= 2.5MeV`