# Modern Physics. Find the correct option(s)?

Apr 6, 2017

I got option
(C) $3.42 f m$

#### Explanation:

Writing equation for constituents of given nuclei
"_7^15$\text{N} = 7 p + 8 n - B {E}_{N}$ .......(1)
"_8^15$\text{O} = 8 p + 7 n - B {E}_{O}$ .......(2)
where $p$ is proton, $n$ neutron and $B E$ is binding energy of nucleus. When constituents of nucleus are brought together binding energy is released.

Subtracting (1) from (2)
$\text{O"-"N} = 8 p + 7 n - B {E}_{O} - \left(7 p + 8 n - B {E}_{N}\right)$
$\implies \text{O"-"N} = p - n - \Delta B E$

Inserting given values we get
$15.003065 - 15.000109 = 1.007825 - 1.008665 - \Delta B E$
$\implies 0.002956 = - 0.00084 - \Delta B E$
$\implies - \Delta B E = 0.003796 = 3.535974 M e v$ ......(3)

Given electrostatic energy of nucleus is
$E = \frac{3}{5} \frac{Z \left(Z - 1\right) {e}^{2}}{4 \pi {\epsilon}_{0} R}$

Calculating respective electrostatic energies we get
${E}_{\text{O}} = \frac{3}{5} \frac{8 \left(8 - 1\right) {e}^{2}}{4 \pi {\epsilon}_{0} R}$ ......(4)
${E}_{\text{N}} = \frac{3}{5} \frac{7 \left(7 - 1\right) {e}^{2}}{4 \pi {\epsilon}_{0} R}$ .......(5)

As difference of binding energy is purely due to electrostatic energy, therefore, we have
$\Delta B E = {E}_{\text{N"-E_"O}}$

Inserting values from equations (3), (4) and (5) and absorbing $- v e$ sign we get
$3.535974 = \frac{3}{5} \frac{8 \left(8 - 1\right) {e}^{2}}{4 \pi {\epsilon}_{0} R} - \frac{3}{5} \frac{7 \left(7 - 1\right) {e}^{2}}{4 \pi {\epsilon}_{0} R}$
$\implies 3.535974 = \frac{3}{5} \times \left[8 \left(8 - 1\right) - 7 \left(7 - 1\right)\right] \times \left(\frac{{e}^{2}}{4 \pi {\epsilon}_{0} R}\right)$
$\implies 3.535974 = \frac{3}{5} \times 14 \times 1.44 \times \frac{1}{R}$
$\implies R = \frac{3}{5} \times 14 \times 1.44 \times \frac{1}{3.535974}$
$\implies R = 3.42 f m$