Question

The radius of a gold nucleus $(Z=79)$ is about $7\times {10}^{-15}\text{m}$. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the flux through the imaginary sphere of double the radius as of gold nucleus.

• A. $1.4\times {10}^{-6}{\text{N-m}}^2/\text{C}$
• B. $2.4\times {10}^{-6}{\text{N-m}}^2/\text{C}$
• C. $3.4\times {10}^{-6}{\text{N-m}}^2/\text{C}$
• D. $4.4\times {10}^{-6}{\text{N-m}}^2/\text{C}$

Answer 1

The correct option is A $1.4\times {10}^{-6}{\text{N-m}}^2/\text{C}$

Given:
Radius of gold nucleus, $r=7\times {10}^{-15}\text{m}$

Atomic number of gold, $Z=79$

So, number of positive charged proton in the nucleus, $n=79$

Therefore, total charge , $q=ne=+79\times 1.6\times {10}^{-19}\text{C}$


This charge is inside the imaginary sphere.
So, electric flux $\varphi$ through the imaginary sphere

$\varphi =\frac{q}{{ϵ}_0}$

$\Rightarrow \varphi =\frac{79\times 1.6\times {10}^{-19}}{9\times {10}^{-12}}=1.4\times {10}^{-6}{\text{N-m}}^2/\text{C}$

So, option $\left(a\right)$ is correct.