The electric potential at the surface of an a | vedclass

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(B) The electric potential $V$ at the surface of a nucleus is given by the formula $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.Here,the charge of the n

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$8 	imes 10^6 \, V$

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(B) The electric potential $V$ at the surface of a nucleus is given by the formula $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.Here,the charge of the nucleus $Q = Ze = 50 	imes 1.6 	imes 10^{-19} \, C$.The radius $R = 9.0 	imes 10^{-13} \, cm = 9.0 	imes 10^{-15} \, m$.The constant $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \, N \cdot m^2/C^2$.Substituting these values:$V = (9 	imes 10^9) 	imes \frac{50 	imes 1.6 	imes 10^{-19}}{9.0 	imes 10^{-15}}$$V = 10^9 	imes 50 	imes 1.6 	imes 10^{-19} 	imes 10^{15}$$V = 80 	imes 10^5 \, V = 8 	imes 10^6 \, V$.

The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9.0 \times 10^{-15} \, m$ is

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