The electric potential at the surface of an atomic nucleus (Z = 50) of radius \( 9.0\times10^{-15}\,\mathrm{m} \) is (Take \( \frac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \) and \( e=1.6\times10^{-19}\,\mathrm{C} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 8\times10^{6}\,\mathrm{V} \)
The nucleus carries the charge of its \( Z \) protons, \( q=Ze \). At its surface and outside it, the charged sphere gives the same potential as a point charge at its centre. So the potential at the surface is \( V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R} \).
The charge of the nucleus is:
\( \begin{aligned} q &= Ze \\ &= 50\times1.6\times10^{-19}\,\mathrm{C} \\ &= 8\times10^{-18}\,\mathrm{C} \end{aligned} \)
The potential at the surface is:
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R} \\ &= \frac{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}\times8\times10^{-18}\,\mathrm{C}}{9.0\times10^{-15}\,\mathrm{m}} \\ &= 8\times10^{6}\,\mathrm{V} \end{aligned} \)