Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.72x10⁻¹¹m around a stationary proton. What is the speed of the electron in its orbit? Express your answer with the appropriate units.

Assume that both the proton and the electron are point charges. By Coulomb's Law, at a distance of [tex]r[/tex] from each other, the magnitude of the electrostatic force between the two would be:

[tex]\displaystyle F = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex],

Where:

[tex]k \approx 8.99 \times 10^{9} \; {\rm N\cdot m^{2} \cdot C^{-2}}[/tex] is the Coulomb constant.
[tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the magnitude of electrostatic charges on the two point charges. For the proton and the electron, [tex]q_{1} = q_{2} = e = 1.602 \times 10^{-19}\; {\rm C}[/tex].
[tex]r = 5.72 \times 10^{-11}\; {\rm m}[/tex] as given in the question.

Assuming that all other forces on this electron are negligible. The magnitude of the resultant force on this electron would be equal to that of this electrostatic attraction from the proton:

[tex]\displaystyle F_\text{net} = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex].

This question assumes that the electron is in a centripetal motion around the proton at a radius of [tex]r = 5.72 \times 10^{-11}\; {\rm m}[/tex] and an unknown orbital speed of [tex]v[/tex] (needs to be found.)

Assume that the speed of the electron is only a small fraction of the speed of light, such that the effects of relativity on the motion of the electron is negligible. Resultant force on this electron should ensure that:

[tex]\displaystyle F_\text{net} = \frac{m\, v^{2}}{r}[/tex],

Where [tex]m[/tex] is the mass of the electron. Under the assumptions, [tex]m \approx 9.11 \times 10^{-31}\; {\rm kg}[/tex] (the mass of a stationary electron) should be a reasonably good estimate.

Equate the two expressions for the resultant force [tex]F_\text{net}[/tex] and solve for orbital speed [tex]v[/tex]:

[tex]\displaystyle \frac{k\, q_{1}\, q_{2}}{r^{2}} = F_\text{net} = \frac{m\, v^{2}}{r}[/tex].

[tex]\begin{aligned} v &= \sqrt{\frac{k\, q_{1}\, q_{2}}{m\, r}} \\ &\approx \sqrt{\frac{(8.99 \times 10^{9})\, (1.602 \times 10^{-19})\, (1.602 \times 10^{-19})}{(9.11 \times 10^{-31})\, (5.72 \times 10^{-11})}}\; {\rm m\cdot s^{-1}} \\ &\approx 2.10 \times 10^{6}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

(Approximately [tex]0.007\, c[/tex], which is in line with the assumption.)

In other words, the orbital speed of this electron would be approximately [tex]2.10 \times 10^{6}\; {\rm m\cdot s^{-1}}[/tex] under the assumptions.