Q 12-02-072JEE MainJEE Main 2024 (27 Jan, Shift 2)Easy
The electric potential at the surface of an atomic nucleus ($Z = 50$) of radius $9\times10^{-13}\ \text{cm}$ is $\alpha\times10^6\ \text{V}$. What is the value of $\alpha$? (Charge of proton $1.6\times10^{-19}\ \text{C}$)
Numerical value type. Enter your answer.
Answer: 8
Treating the nucleus as a charged sphere of radius $R = 9\times10^{-15}\ \text{m}$:
$$V = \frac{kZe}{R} = \frac{9\times10^9\times50\times1.6\times10^{-19}}{9\times10^{-15}} = 8\times10^6\ \text{V}$$
So $\alpha = 8$.
Solution by Sreeraj P, M.Sc Physics