Question

The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9.0\times 10^{-15}m$ is:

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

Answer 1

Answer: (A)

$V=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q}{R}=\frac{1}{4\pi {\epsilon }_0}\frac{(Ze)}{R}$
$=9\times 10^9\times \frac{50\times 1.6\times 10^{-19}}{9.0\times 10^{-15}}$
$=8\times 10^{6}V$