The mass of a `._(3)^(7) Li` nucleus is `0.042 u` less than the sum of the masses of all its nucleons. The binding energy per nucleon of `._(3)^(7)Li` nucleus is nearly

Options

A. 46 MeV
B. `5.6` MeV
C. `3.9` MeV
D. 23 MeV

Answer

B

Explanation

For `"" _(3) ^(7) ` li nucleus.
f Mass defect,` Delta M=0.42u `
`because 1u = 931.5 MeV//c ^(2)`
`therefore Delta M = 0.042 x 931.5 MeV//C ^(2) = 39.`
`MeV //c ^(2)`
Binding energy,` E _(b)= Delta MC ^(2)`
`= (39.1 (MeV )/(c ^(2)))c ^(2)`
`= 39 . 1 MeV`
B.E per nucleon =
`(E _(b))/( A ) = (39.1 MeV )/(A) ~~ 5.6 MeV`