Question

The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9\times {10}^{-15}\text{m}$ is

• A. $80\text{V}$
• B. $8\times {10}^6\text{V}$
• C. $9\text{V}$
• D. $9\times {10}^6\text{V}$

Answer 1

The correct option is B $8\times {10}^6\text{V}$

Given that,
Atomic number of the atom, $Z=50$
Radius of the atom, $R=9\times {10}^{-15}$

The atomic nucleus of an atom having $50$ as the atomic number will have $50$ protons.

So charge of the nucleus, $q=Ze$

where $e=1.6\times {10}^{-19}\text{C}$

Now the electric potential $\left(V\right)$ at the surface of the atom will be
$V=\frac{kq}{R}$
$\Rightarrow V=\frac{9\times {10}^9\times 50\times 1.6\times {10}^{-19}}{9\times {10}^{-15}}$
$\Rightarrow V=8\times {10}^6\text{V}$

Hence, option (b) is correct.