Question

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

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

Answer 1

Answer: (B)

Electric potential at surface,
V = $\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}(q=Ze)$
= $9\times 10^9\times \frac{50\times 1.6\times 10^{-19}}{9\times 10^{-15}}$
= $8\times 10^6$ Volt