Discussion Forum

The electrostatic energy of Zprotons uniformly distributed throughout a spherical nucleus of radius R is given by $E=\frac{3}{5}\frac{Z(Z-1)e^{2}}{4\pi\epsilon_{0}R}$.The measured masses of the neutron, $_{1}^{1}H,_{7}^{15}N$and $_{8}^{15}O$are  1.008665u. 1.007825u, 15.000109u and 15.003065 u. respectively. Given that the radii of both the $_{7}^{15}N$ and$_{8}^{15} O$ nuclei are same,1 u =931.5MeV/c² ,c is the speed of light and $e^{2}/(4\pi \epsilon_{0})=1.44$ 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^-15m)

Answer:

Explanation:

Electrostatic energy = Binding energy of N - Binding energy of O

=[(7M_H+8M_n-M_N)]-[(8M_H+7M_n-M_O)]xc²

=[-M_H+M_n+M_o-M_N]c²

=[- 1.007825 + 1.008665 + 15.003065- 15.000109]x 931.5 = + 3.5359 MeV

$\triangle E=\frac{3}{5}\times \frac{1.44\times 8\times7}{R}-\frac{3}{5}\times \frac{1.44\times 7\times 6}{R}=3.5359 MeV$

$R=\frac{3\times 1.44\times 14}{5\times 3.5359}=3.42 fm$