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

To find the electric potential \( V \) at the surface of an atomic nucleus with atomic number \( z = 50 \) and radius \( r = 9 \times 10^{-13} \text{ cm} \), we use the formula for the electric potential due to a point charge: 

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

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

Substituting the known values into the formula:

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

Simplifying the expression:

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

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

Answer 2

The electric potential at the surface of a nucleus is given by:

\(V = \frac{kQ}{R} = \frac{kZe}{R},\)

where:
- \(k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\),
- \(Z = 50\),
- \(e = 1.6 \times 10^{-19} \, \text{C}\),
- \(R = 9 \times 10^{-13} \, \text{cm} = 9 \times 10^{-15} \, \text{m}\).

Substituting the values:

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

Simplify:

\(V = 8 \times 10^6 \, \text{V}.\)
The Correct answer is: 8