A gold nucleus ( Z = 79 ) has a radius of approximately 7.0 × 10^ -15 m . Assuming its positive charge is uniformly distributed throughout the nuclear volume, determine the strength of the electric field at the surface

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Quantrex Academy · Accepted Answer

Correct answer: C. 2.32 × 10^ 21 N C^ -1

The charge of the gold nucleus is given by

q = Ze

, where

Z = 79

and

e = 1.6 \times 10^{-19} \text{ C}

.

q = 79 \times 1.6 \times 10^{-19} \text{ C}

The electric field at the surface of a uniformly charged sphere of radius

R

is given by:

E = \dfrac{1}{4\pi\epsilon_0} \dfrac{q}{R^2}

Substituting the given values:

E = \dfrac{9 \times 10^9 \times 79 \times 1.6 \times 10^{-19}}{(7.0 \times 10^{-15})^2}

E = \dfrac{1137.6 \times 10^{-10}}{49 \times 10^{-30}}

E = 23.216 \times 10^{20} \text{ N C}^{-1}

E \approx 2.32 \times 10^{21} \text{ N C}^{-1}

A gold nucleus ( Z = 79 ) has a radius of approximately 7.0 × 10^...

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