The electric potential at the surface of an atomic nucleus ($$Z = 50$$) of radius $$9 \times 10^{-13}$$ cm is $$\alpha \times 10^6$$ V. What is the value of $$\alpha$$?
(Charge of proton $$1.6 \times 10^{-19}$$ C)

Electric potential at the surface of a nucleus:

$$V = \frac{1}{4\pi\epsilon_0}\frac{Ze}{R}$$

$$Z = 50$$, $$R = 9 \times 10^{-13}$$ cm $$= 9 \times 10^{-15}$$ m.

$$V = 9 \times 10^9 \times \frac{50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}$$

$$= 9 \times 10^9 \times \frac{80 \times 10^{-19}}{9 \times 10^{-15}}$$

$$= 10^9 \times \frac{80 \times 10^{-19}}{10^{-15}}$$

$$= 10^9 \times 80 \times 10^{-4} = 80 \times 10^5 = 8 \times 10^6$$ V

So $$\alpha = 8$$.

The answer is $$\boxed{8}$$.