Question

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

• A. $80V$
• B. $8\times {10}^{6}V$
• C. $9V$
• D. $9\times {10}^{5}V$

Answer 1

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

$\text{Step 1: Total charge on the nucleus}$

$q=Ze$

Where $Z$ is the atomic number of nucleus

$\therefore q=50\times 1.6\times {10}^{-19}C=8\times {10}^{-18}C$

$\text{Step 2: Potential at the surface of the nucleus}$

There is symmetrical distribution of charge around the nucleus so we can assume the shape of a nucleus to be a sphere and the Potential at the surface of a nucleus is given as:

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

$=\frac{9\times {10}^9\times 8\times {10}^{-18}}{9\times {10}^{-15}}$$=8\times {10}^{6}Volts$

Hence, Option (B) is correct.