Question

The electric potential at the surface of an atomic nucleus \( (z = 50) \) of radius \( 9 \times 10^{-13} \) cm is \(\_\_\_\_\_\_\_ \)\(\times 10^{6} V\).

Answer 1

Correct Answer: 8

The electric potential \( V \) at the surface of an atomic nucleus with atomic number \( z = 50 \) and radius \( r = 9 \times 10^{-13} \text{ cm} \) is determined using the formula for the electric potential of a point charge:

\( V = \frac{k \cdot q}{r} \)

Here, \( k = 8.99 \times 10^{9} \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant, and \( q = z \cdot e \) represents the nucleus's total charge, where \( e = 1.6 \times 10^{-19} \, \text{C} \). Consequently, \( q = 50 \cdot 1.6 \times 10^{-19} \, \text{C} = 8 \times 10^{-18} \, \text{C} \).

Substituting the given values into the formula yields:

\( V = \frac{8.99 \times 10^{9} \cdot 8 \times 10^{-18}}{9 \times 10^{-15}} \)

Simplification of this expression results in:

\( V = \frac{71.92 \times 10^{-9}}{9 \times 10^{-15}} = 7.991 \times 10^{6} \, \text{V} \)

The calculated potential is \( 7.991 \times 10^{6} \, \text{V} \), which approximates to \( 8 \times 10^{6} \, \text{V} \).