First slide

Pontial due to

Question

The radius of nucleus of silver (atomic number = 47) is 3.4 x 10^-14m. The electric potential on the surface of nucleus is (e = 1.6 x 10^-19C)

Easy

A

1.99 x 10⁶ volt
B

2.9 x 10⁶ volt
C

4.99 x 10⁶ volt
D

0.99 x 10⁶ volt

Solution

$V = \frac{1}{4 {πε}_{0}} \cdot \frac{(Ze)}{r} = 9 \times 10^{9} \times \frac{47 \times 1.6 \times 10^{- 19}}{3.4 \times 10^{- 14}} = 1.99 \times 10^{6} V$

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