In a hydrogen atom's ground state, the average separation between the proton and the electron is 5.3 × 10^ -11 m . When the atom transitions into its first excited state, this average separation increases to four times

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Correct answer: A. 5.1 × 10^ -9 N

The average separation between the proton and the electron in the first excited state is given as:

r = 4 \times 5.3 \times 10^{-11} \text{ m} = 21.2 \times 10^{-11} \text{ m}

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

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

Substituting the standard values

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

and

\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}{(21.2 \times 10^{-11})^2}

F = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{449.44 \times 10^{-22}}

F = \dfrac{23.04 \times 10^{-29}}{449.44 \times 10^{-22}}

F \approx 5.1 \times 10^{-9} \text{ N}

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

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