The electric potential at the surface of an a | vedclass

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(B) Given: Atomic number $Z = 50$, Radius $r = 9 	imes 10^{-15} \,m$, and elementary charge $e = 1.6 	imes 10^{-19} \,C$.The electric potential $V$

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$8 	imes 10^6 \,V$

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(B) Given: Atomic number $Z = 50$, Radius $r = 9 	imes 10^{-15} \,m$, and elementary charge $e = 1.6 	imes 10^{-19} \,C$.The electric potential $V$ at the surface of a charged sphere (nucleus) is given by the formula:$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}$Since the total charge $q = Z e$, we substitute the values:$V = (9 	imes 10^9) 	imes \frac{50 	imes 1.6 	imes 10^{-19}}{9 	imes 10^{-15}}$$V = 9 	imes 10^9 	imes \frac{80 	imes 10^{-19}}{9 	imes 10^{-15}}$$V = 10^9 	imes 80 	imes 10^{-19} 	imes 10^{15}$$V = 80 	imes 10^5 = 8 	imes 10^6 \,V$Thus, the electric potential is $8 	imes 10^6 \,V$.

The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9 \times 10^{-15} \,m$ is

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