The electric potential at the surface of an atomic nucleus ( Z = 50...

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

$V = \dfrac{kQ}{R} = \dfrac{k \cdot \left(Ze\right)}{R} . . . \left(1\right)$

From equation (1), it follows that

$\begin{cases}V & = & \dfrac{9 \times {10}^{9} \times 50 \times 1 . 6 \times {10}^{- 19}}{9 \times {10}^{- 13} \times {10}^{- 2}}V \\ & = & 8 \times {10}^{6}V\end{cases}$

Hence, $\alpha = 8$.