Question

In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius $0.053\mathrm{nm}$around a stationary proton. How many revolutions per second does the electron make?

Short answer

Revolutions per second for electron is$6.6\times {10}^{15}\mathrm{rev}/\mathrm{s}$.

Step-by-step solution

Formula for force

The two particles exert forces on each other, with the force decreasing as the distance between them grows and increasing as the size of the charges increases.

The force that exists between a proton and an electron is known as the proton-electron force.

Coulomb 's law force is,

${\mathrm{F}}_{\mathrm{C}}=\frac{\mathrm{K}\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}=\mathrm{K}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^2}$

The electrostatic constant is $9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$$9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$.

Because the electron revolves around the nucleus, it has a centripetal force, which is given by

${\mathrm{F}}_{\mathrm{e}}={\mathrm{m\omega }}^2\mathrm{r}$

Calculation for revolution per second

Angular velocity,

${\mathrm{F}}_{\mathrm{C}}={\mathrm{F}}_{\mathrm{e}}$

$\mathrm{K}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^2}={\mathrm{m\omega }}^2\mathrm{r}$

$\mathrm{\omega }=\sqrt{\frac{{\mathrm{Ke}}^2}{{\mathrm{mr}}^3}}$

$=\sqrt{\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right){\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{\left(9.1\times {10}^{-31}\mathrm{kg}\right){\left(0.053\times {10}^{-9}\mathrm{m}\right)}^3}}$

$=4.12\times {10}^{16}\mathrm{rad}/\mathrm{s}$

We discovered the angular velocity in $\mathrm{rad}/\sec$, so we'll convert it to $\mathrm{rev}/\mathrm{s}$.

localid="1649151421673" $\mathrm{\omega }=\left(4.12\times {10}^{16}\mathrm{rad}/\mathrm{s}\right)\left(\frac{\mathrm{rev}}{2\mathrm{\pi rad}}\right)$

$=6.6\times {10}^{15}\mathrm{rev}/\mathrm{s}$