QJEEAdv16Phb8ba: JEE Advanced 2016 Physics - Class 12 JEEAdv16Phb8ba Solution: The electrostatic energy of Z protons uniformly distributed…

Question

Verified Solution • 2025–26 Session moderate

The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by

E = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi {\varepsilon _0}R}}

The measured masses of the neutron,

_1^1H

,

_7^{15}N

and

_8^{15}O

are 1.008665u, 1.007825u, 15.000109u and 15.003065u, respectively. Given that the radii of both the

_7^{15}N

and

_8^{15}O

nuclei are same, 1 u = 931.5 MeV/c² (c is the speed of light) and e²/(4

\pi

{{\varepsilon _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 electrostatic energy, the radius of either of the nuclei is (1 fm = 10^{$-$ 15} m)

Accepted Answer

The binding energies of $_7^{15}N$ and $_8^{15}O$ are given by

$B{E_N} = \Delta {m_N}{c^2} = (8{m_n} + 7{m_p} - {M_N}){c^2}$

$= (8 \times 1.008665 + 7 \times 1.007825 - 15.000109 \times 931.5$

= 115.49 MeV,

$B{E_O} = \Delta {m_O}{c^2} = (7{m_n} + 8{m_p} - {M_O}){c^2}$

$= (7 \times 1.008665 + 8 \times 1.007825 - 15.003065 \times 931.5$

= 111.95 MeV.

The difference in binding energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta BE = B{E_N} - B{E_O} = 115.49 - 111.95$

= 3.54 MeV. ........ (1)

The electrostatic energies of $_7^{15}N$ and $_8^{15}O$ are given by

${E_N} = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi { \in _0}R}} = {3 \over 5}{{7(7 - 1)1.44} \over R}$

$= {{36.288} \over R}$ MeV-fm

${E_O} = {3 \over 5}{{8(8 - 1)1.44} \over R} = {{48.384} \over R}$ MeV-fm.

The difference in electrostatic energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta E = {E_O} - {E_N} = (12.096/R)$ ...... (2)

Since, $\Delta E = \Delta BE$ , equations (1) and (2) give $R = 12.096/3.54 = 3.42$ fm.

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