The electric potential at the surface of an atomic nucleus (Z=50) of radius 9.0 x 10^-15 m is : 

8 x 10⁶ V

 The correct answer is Option 2: 8 x 10⁶ V

• Atomic Number ($Z$): 50

• Charge of a proton ($e$): $1.6 \times 10^{-19}\text{ C}$

• Total Charge ($Q$): $Z \times e = 50 \times 1.6 \times 10^{-19}\text{ C} = 80 \times 10^{-19}\text{ C}$

• Radius ($r$): $9.0 \times 10^{-15}\text{ m}$

• Coulomb's Constant ($k$): $9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$