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# The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by $E = \dfrac{3}{5}\dfrac{{Z(Z - 1){e^2}}}{{4\pi {\varepsilon _o}R}}$. The measured masses of the neutron, $_1^1H,{}_7^{15}Nand{}_8^{15}O$ are 1.008665u, 1.007825u,15.000109u and 15.003065u, respectively. Given that the radii of both the ${}_7^{15}Nand{}_8^{15}O$nuclei are same, $1u = 931.5MeV/{c^2}$ (c is speed of light) and $\dfrac{{{e^2}}}{{4\pi {\varepsilon _o}}} = 1.44MeVfm.$ Assuming that the difference between the binding energies of ${}_7^{15}Nand{}_8^{15}O$ is purely due to the electrostatic energy. The radius of either of the nuclei is ($1fm = {10^{ - 15}}m$):A. 2.85fmB. 3.03fmC. 3.42fmD. 3.80fm

Last updated date: 20th Jun 2024
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Hint: mass of proton, neutron and nuclei of ${}_7^{15}Nand{}_8^{15}O$ using mass energy equivalence proposed by einstein which is $E = m{c^2}$ find the binding energy of ${}_7^{15}Nand{}_8^{15}O$ by calculating energy of each individual neutrons and protons and then subtracting it from the total energy of those nuclei once you know the binding energy find the binding energy using the given expression in the question and then compare to get the radius of the spherical nucleus.

Let's calculate the radius of ${}_7^{15}N$. It has 7 protons and 8 neutrons. Now total mass of 7 protons and 8 neutrons will be,
${m_{total}} = 7{m_{proton}} + 8{m_{neutron}}$
$\Rightarrow {m_{total}} = 7(1.007825) + 8(1.008665)\\ \Rightarrow {m_{total}} = 15.124095u$
But given mass of ${}_7^{15}N$is $15.000109u$
So difference in mass will be,
$\Delta m = {m_{total}} - m({}_7^{15}N)$
$\Rightarrow \Delta m = 15.124095u - 15.000109u \\ \Rightarrow \Delta m = 0.123386u$
Now using mass energy equivalence we will find the binding energy. Therefore, binding energy,
$E = \Delta m{c^2}\\ \Rightarrow E= 0.123386 \times 931.5 MeV \\ \Rightarrow E= 114.93 MeV$
Now using the formula of binding energy given in the question. We have,
$E = \dfrac{3}{5}\dfrac{{Z(Z - 1){e^2}}}{{4\pi {\varepsilon _o}R}}\\ \Rightarrow E= \dfrac{3}{5}\dfrac{{7(7 - 1)}}{R}1.44MeV$
As $\dfrac{{{e^2}}}{{4\pi {\varepsilon _o}}} = 1.44MeVfm.$
$\Rightarrow E = \dfrac{3}{5}\dfrac{{42}}{R}1.44MeV \\ \Rightarrow E = \dfrac{{36.288}}{R}MeV$
Comparing both these energies we have,
$R = \dfrac{E}{{36.288}}fm \\ \therefore R= \dfrac{{114.93}}{{36.288}}fm \approx 3.03fm$

Hence Option B is correct.

Note: This question looks scary but if you read the question carefully and apply the basics of modern physics specially mass energy equivalence it is lot easier also here you don’t have to lot of calculations as values of some of the constant are given in the question.one more thing its asking radius of either of the nuclei you don’t have to calculate the both because total number of nucleons in both the nuclei are same so radius of both the nuclei will almost be same.