The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by $E = \frac{3}{5} \frac{Z (Z - 1) e^{2}}{4 π ε_{0} R}$ The measured masses of the neutron, $_{1}^{1} H$ , $_{7}^{15} N$ and $_{8}^{15} O$ are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the $_{7}^{15} N$ and $_{8}^{15} O$ nuclei are same, $1 u = 931.5 M e V / c^{2}$ (c is the speed of light) and $e^{2} / (4 π ε_{0}) = 1.44 M e V$ fm. Assuming that the difference between the binding energies of $_{7}^{15} N$ and $_{8}^{15} O$ is purely due to the electrostatic energy, the radius of either of the nuclei is $(1 f_{m} = 10^{- 15})$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

3.80 fm

3.03 fm

2.85 fm

3.42 fm

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} B i n d i n g e n e r g y o f n i t o g e n a t o m = [8 \times 1.008665 + 7 \times 1.007825 - 15.000109] 931.5 \\ B i n d i n g e n e r g y o f o x y g e n a t o m = [7 \times 1.008665 + 8 \times 1.007825 - 15.003065] 931.5 \\ ∴ D i f f e r e n c e i n B E = 0.0037960 \times 931.5 = 3.54 M e V \\ N o w, D i f f e r e n c e i n e l e c t r o s t a t i c e n e r g y = E_{O} - E_{N} = \frac{3}{5} (\frac{8 \times 7}{R} - \frac{7 \times 6}{R}) 1.44 M e V f m = \frac{12.096}{R} M e V f m \\ B o t h d i f f e r e n c e s a r e e q u a l, \frac{12.096}{R} M e V f m = 3.54 M e V \\ \Rightarrow R = 3.42 f m \end{array}$