91Ó°ÊÓ

In the Bohr model of the hydrogen atom, an electron (mass $m=9.1×{10}^{-31}kg$ orbits a proton at a distance of$5.3×{10}^{-11}m$.

The electron velocity is tangent to the circle of motion, and its acceleration is called the centripetal acceleration and it points toward the center of the circle. The electron has angular velocity $\mathrm{Ó\neg }$ and a tangential speed $v$ and they are related to each other by

Where $r$ is the radius of the circle. The acceleration of the electron in an orbital path is given by equation (8.4) in the form

At a uniform circular motion, the velocity vector has a tangential component and the acceleration vector has a radial component. From Newton 's first law, the electron has a net force exerted on it as it doesn't move with a constant velocity. So, using the expression of the acceleration from equation (2), we get the net force by

${\stackrel{→}{F}}_{\text{net}}=m\stackrel{→}{a}=\frac{m{v}^2}{r}$

Where $m$is the mass of the electron. Now, let us use the expression of $v$from equation ( 1 ) into equation (3) to get an expression of $\mathrm{Ó\neg }$in terms of $\stackrel{→}{F_{\text{net}}}$

${\stackrel{→}{F}}_{\text{net}}=\frac{m{v}^2}{r}$

The net force on the electron is due to the attraction force between it and the proton. So, let us plug the values for ${\stackrel{→}{F}}_{\text{net}},m$and $r$into equation (4) to get $\mathrm{Ó\neg }$in rad/s

$\mathrm{Ó\neg }=\sqrt{\frac{{\stackrel{→}{F}}_{\text{net}}}{mr}}$

$=\sqrt{\frac{8.2×{10}^{-8}\mathrm{N}}{\left(9.1×{10}^{-31}\mathrm{kg}\right)\left(5.3×{10}^{-11}\mathrm{m}\right)}}$

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

We are asked to find the angular velocity is revolutions per second (rev/s). So, let us convert the unit from rad/s to $\mathrm{rev}/\mathrm{s}$ where $(1\mathrm{rev}=2\mathrm{Ï€}\mathrm{rad})$

The angular velocity is$6.6×{10}^{15}rev/s$.