In a hydrogen atom's ground state, the average separation between the proton and the electron is 5.3 × 10^ -11 m . Determine the Coulomb force between them at this separation.

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Correct answer: B. 8.2 × 10^ -8 N

The Coulomb force between the proton and the electron is given by Coulomb's law:

F = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q_1 q_2}{r^2}

Substituting the given values:

q_1 = e = 1.6 \times 10^{-19} \text{ C}

q_2 = e = 1.6 \times 10^{-19} \text{ C}

r = 5.3 \times 10^{-11} \text{ m}

\dfrac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2

F = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2}

F = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{28.09 \times 10^{-22}}

F = \dfrac{23.04 \times 10^{-29}}{28.09 \times 10^{-22}}

F \approx 0.82 \times 10^{-7} \text{ N}

F = 8.2 \times 10^{-8} \text{ N}

In a hydrogen atom's ground state, the average separation between...

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