An atom of atomic number Z = 50 is having nuclear radius = 9 × 10^–13 cm. Potential at the surface of the nucleus is

The correct option is 8 × 10⁶ V.

To find the electric potential at the surface of the nucleus, we treat the nucleus as a spherical charge distribution. According to electrostatics, the potential $V$ at the surface of a sphere of radius $R$ containing a total charge $q$ is given by the formula:
$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$
where:
• $\frac{1}{4\pi {\epsilon }_0}\approx 9\times {10}^9\text{N·m}{}^2\text{/C}{}^2$ is the electrostatic constant in SI units.
• $q$ is the total charge of the nucleus.
• $R$ is the radius of the nucleus.

First, let us calculate the total charge $q$ of the nucleus. The nucleus contains $Z$ protons, where $Z$ is the atomic number. Given:
• Atomic number, $Z=50$
• Charge of a single proton, $e\approx 1.6\times {10}^{-19}\text{C}$
Therefore, the nuclear charge $q$ is:
$q=Z\times e=50\times 1.6\times {10}^{-19}\text{C}=8\times {10}^{-18}\text{C}$

Next, we convert the nuclear radius $R$ from centimeters to meters to align with SI units:
• Given radius, $R=9\times {10}^{-13}\text{cm}$
• Since $1\text{cm}={10}^{-2}\text{m}$, we have:
$R=9\times {10}^{-13}\times {10}^{-2}\text{m}=9\times {10}^{-15}\text{m}$

Now, substitute the values of the electrostatic constant, the total charge $q$, and the radius $R$ into the potential formula:
$V=(9\times {10}^9)\times \frac{8\times {10}^{-18}}{9\times {10}^{-15}}$

Simplify the expression by canceling out the factor of 9:
$V=8\times \frac{{10}^9\times {10}^{-18}}{{10}^{-15}}$
$V=8\times {10}^{9-18-(-15)}$
$V=8\times {10}^{-9+15}$
$V=8\times {10}^6\text{V}$

Thus, the electric potential at the surface of the nucleus is 8 × 10⁶ V.