The electrostatic energy of Z protons uniformly distributed throu… - Vidyadip

MCQ

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

$E=\frac{3}{5} \frac{Z(Z-1) e^2}{4 \pi \varepsilon_0 R}$

The measured masses of the neutron, ${ }_1^1 \mathrm{H},{ }_7^{15} \mathrm{~N}$ and ${ }_8^{15} \mathrm{O}$ are $1.008665 \mathrm{u}, 1.007825 \mathrm{u}$, $15.000109 \mathrm{u}$ and $15.003065 \mathrm{u}$, respectively. Given that the radii of both the ${ }_7^{15} \mathrm{~N}$ and ${ }_8^{15} \mathrm{O}$ nuclei are same, $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2$ ( $c$ is the speed of light) and $e^2 /\left(4 \pi \varepsilon_0\right)=1.44 \mathrm{MeV} \mathrm{fm}$. Assuming that the difference between the binding energies of ${ }_7^{15} \mathrm{~N}$ and ${ }_8^{15} \mathrm{O}$ is purely due to the electrostatic energy, the radius of either of the nuclei is

$\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right.$ )

A
$2.85 \mathrm{fm}$
B
$3.03 \mathrm{fm}$
✓
$3.42 \mathrm{fm}$
D
$3.80 \mathrm{fm}$

✓

Answer

Correct option: C.

$3.42 \mathrm{fm}$

c
$E_0=\frac{3}{5} \times \frac{8 \times 7}{R} \times \frac{e^2}{4 \pi \varepsilon_0}=\frac{3}{5} \times \frac{8 \times 7}{R} \times 1.44 MeV$

$E_N=\frac{3}{5} \times \frac{7 \times 6}{R} \times \frac{e^2}{4 \pi \varepsilon_0}=\frac{3}{5} \times \frac{7 \times 6}{R} \times 1.44 MeV$

$\text { so }\left|E_0-E_N\right|=\frac{3}{5} \times \frac{1.44}{R} \times 7(2)$

Now mass defect of N atom

$8 \times 1.008665+7 \times 1.007825-15.000109$

$=0.1239864 u$

So binding energy $=0.1239864 \times 931.5 MeV$ and mass defect of O atom

$7 \times 1.008665+8 \times 1.007825-15.003065$

$=12019044 u$

So binding energy $=0.12019044 \times 931.5 MeV$

So $\left|B_0-B_N\right|=0.0037960 \times 931.5 MeV$

from $(i)$ and $(ii)$ we get

$R=3.42 fm \text {. }$

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