Question

Gold nucleus radius R is represented by the symbol Au atomic number 79 and mass number 197 taking e as the elementary charge and E° as the permittivity of free space, what is the electric field strength at the surface of an isolated gold nucleus?

Answer 1

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• $\red{\boxed{E = \frac{1} {4\pi {E}_{0}} \frac{79e} {{R}^{2}}}}$

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Here,

❥︎Atomic Number,$Z = 79$

❥︎Elementary Charge = $e$

❥︎Permitivity of Free Space = $E₀$

❥︎Radius of Gold Nucleus = $R$

Then, total charge of the Gold nucleus

❥︎$Q =$ (No. of Proton) x (Elementary Charge)

❥︎$Q =$(Atomic Number) x $e$

❥︎$Q = Ze = 79e\:\:C$

Electric Field Strength by a point charge Q at distance R is given by

❥︎$E = \frac{1} {4\pi {E}_{0}} \frac{Q}{{R}^{2}}$

❥︎$E = \frac{1} {4\pi {E}_{0}} \frac{Ze} {{R}^{2}}$

❥︎$E = \frac{1} {4\pi {E}_{0}} \frac{79e} {{R}^{2}}$