A hydrogen atom comprises one proton and one electron. We can assume that the electron revolves in a circle of radius 0.53 angstrom ( 1 angstrom = 10^ -10 m , abbreviated as Å) with the proton situated at the centre.

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Quantrex Academy · Accepted Answer

Correct answer: C. 2.18 × 10^6 m/s

The centripetal force for the circular motion of the electron is provided by the electrostatic force of attraction between the proton and the electron.

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

v = \sqrt{\dfrac{e^2}{4\pi\epsilon_0 mr}}

Substituting the standard values

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

,

m = 9.1 \times 10^{-31} \text{ kg}

,

r = 0.53 \times 10^{-10} \text{ m}

, and

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

:

v = \sqrt{\dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9.1 \times 10^{-31} \times 0.53 \times 10^{-10}}}

v = \sqrt{\dfrac{23.04 \times 10^{-29}}{4.823 \times 10^{-41}}}

v = \sqrt{4.77 \times 10^{12}}

v = 2.18 \times 10^6 \text{ m/s}

Alternatively, using Bohr's quantization condition for the ground state (

n=1

):

mvr = \dfrac{h}{2\pi}

v = \dfrac{6.63 \times 10^{-34}}{2 \times 3.14 \times 9.1 \times 10^{-31} \times 0.53 \times 10^{-10}} = 2.18 \times 10^6 \text{ m/s}

A hydrogen atom comprises one proton and one electron. We can...

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