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. `30.3 fm`
C. `3.42 fm`
D. `3.80 fm`

Answer

C

Explanation

`E_(0)=(3)/(5)xx(8xx7)/(R )xx(e^(2))/(4 pi epsilon_(0))=(3)/(5)xx(8xx7)/(R)xx1.44 MeV`
`E_(N)=(3)/(5)xx(7xx6)/(R)xx(e^(2))/(4 pi epsilon_(0))=(3)/(5)xx(7xx6)/(R)xx1.44 MeV`
so `|E_(0)-E_(N)|=(3)/(5)xx(1.44)/(R)xx7(2)`
Now mass defect of N atom
`8xx1.008665+7xx1.007825-15.000109`
`=0.1239864 u`
So binding energy `=0.1239864xx931.5 MeV` and mass defect of O atom
` 7xx1.008665+8xx1.007825-15.003065`
`=12019044 u`
So binding energy `= 0.12019044xx931.5 MeV`
So `|B_(0)-B_(N)|=0.0037960xx931.5 MeV`
from (i) and (ii) we get
`R=3.42 fm`.