Electrostatic potential

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Question

The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9.0 \times 10^{- 13} c m$ is___ $\times 10^{4} V$ .

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Solution

$V = k \frac{Q}{R} = 9 \times 10^{9} \times \frac{50 \times 1.6 \times 10^{- 19}}{9 \times 10^{- 15}} = 800 \times 10^{4} V$

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