The electric potential at the surface of an atomic nucleus ( Z = 50 ) of radius 9 × 10 - 13 cm is α × 10 6 V…

The electric potential at the surface of an atomic nucleus can be written as

$$\begin{gathered} V=\frac{kQ}{R} \\ =\frac{k·\left(Ze\right)}{R} ...\left(1\right) \end{gathered}$$

From equation (1), it follows that

$\begin{array}{rcl}V& =& \frac{9\times {10}^9\times 50\times 1.6\times {10}^{-19}}{9\times {10}^{-13}\times {10}^{-2}} \mathrm{V}\\ & =& 8\times {10}^6 \mathrm{V}\end{array}$

Hence, $\alpha =8$.