The electrostatic energy of `Z` protons uniformly distributed throughout a spherical nucleus of radius `R` is given by
`E = (3 Z(Z- 1)e^(2))/5( 4 pi e _(0)R)`
The measured masses of the neutron `_(1)^(1) H, _(7)^(15) N and , _(8)^(16)O are 1.008665 u, 1.007825 u , 15.000109 u and 15.003065 u, ` respectively Given that the ratio of both the `_(7)^(12) N` and `_(8)^(15) O` nucleus are same , 1 u = = 931.5 Me V`c^(2) ` (c is the speed of light ) and `e^(2)//(4 pi e_(0)) = 1.44 MeV` fm Assuming that the difference between the binding energies of `_7^(15) N and `_(8)^(15) O ` is purely due to the electric energy , The radius of the nucleus of the nuclei is

Options

A. 2.85 fm
B. 3.03 fm
C. 3.42 fm
D. 3.80 fm

Answer

C

Explanation

Binding energy of `""_(7)N^(15)` can be written as
`(B.E)_(N) =(8m_(n) + 7m_(p) -M_(N))c^(2)`
`=(8 xx 1.008665 + 7 xx 1.007825 - 15.000109) xx 931.5` MeV
=115.49 MeV
Similarly binding energy of `""_(8)O^(15)` can be written as
`(BE)_(O)=(7m_(n)+8m_(p) -M_(O))c^(2)=111.95` MeV
Difference of binding energies:
`Delta(B.E.)=(B.E)_(N)-(BE)_(O)=115.49-111.95`
`rArr Delta(BE)=3.54 MeV`.........(i)
Electrostatic energies can be written as follows:
`E_(N) =3/5.(Z(Z-1)e^(2))/(4piepsilon_(0)R) =3/5(7(7-1) xx 1.44)/R = (36.288)/R MeV`-fm
`E_(0)=3/5(Z(Z-1)e^(2))/(4piepsilon_(0)R) =3/5(8(8-1)xx 1.44)/R =(48.384)/R` MeV-fm
=3.42 fm