The radius of the nucleus of silver (atomic n | vedclass

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Question

(A) The electric potential $V$ at the surface of a nucleus is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze}{r}$.Here,$Z = 47$,$e

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$1.99 	imes 10^6 \ V$

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(A) The electric potential $V$ at the surface of a nucleus is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze}{r}$.Here,$Z = 47$,$e = 1.6 	imes 10^{-19} \ C$,$r = 3.4 	imes 10^{-14} \ m$,and $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \ N \ m^2/C^2$.Substituting the values:$V = (9 	imes 10^9) 	imes \frac{47 	imes 1.6 	imes 10^{-19}}{3.4 	imes 10^{-14}}$$V = \frac{9 	imes 47 	imes 1.6}{3.4} 	imes 10^{9 - 19 + 14}$$V = \frac{676.8}{3.4} 	imes 10^4$$V \approx 199.05 	imes 10^4 \ V = 1.99 	imes 10^6 \ V$.

The radius of the nucleus of silver (atomic number $Z = 47$) is $3.4 \times 10^{-14} \ m$. Calculate the electric potential on the surface of the nucleus. (Given: $e = 1.6 \times 10^{-19} \ C$,$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ N \ m^2/C^2$)

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