Verify that equation (4-19) follows from (4-16) and (4-18).

According to Bohr model, the expression for the velocity of an electron, rotating around a proton, in terms of charge and permittivity of free space is given by,

${\mathit{v}}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}\mathbf{mr}}\mathbf{}\mathbf{}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

According to the Bohr model, only those orbits are available for the revolution of electrons, for which the angular momentum is an integral multiple of$\mathit{\hslash }$. Where,

and $\mathit{h}$is Planck’s constant $(h=6.626\times {10}^{-34}\raisebox{1ex}{${\mathrm{kgm}}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)$.

Thus, for an electron of mass m moving in an orbit of radius r and moving with velocity v, the angular momentum L is given as-

$$\begin{gathered} \mathbf{L}\mathbf{=}\mathbf{mvr} \\ \mathbf{L}\mathbf{=}\mathbf{n}\mathit{\hslash } \\ \mathbf{v}\mathbf{=}\frac{\mathbf{n}\mathbf{\hslash }}{\mathbf{mr}}\mathbf{}\mathbf{}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Consider an electron, having charge $e$, rotating around a proton in a circular orbit of radius $r$, for which the electron of mass $m$has angular momentum$L$and is moving with velocity$v$.

Using equations (1) and (2), the following relation can be obtained-

$v^2=\frac{e^2}{4\pi {\epsilon }_{0}mr}$

$\left(\frac{n^2{\hslash }^2}{m^2{r}^2}\right)=\frac{e^2}{4\pi {\epsilon }_{0}mr}$

$$\begin{gathered} n^2{\hslash }^2=\frac{mr{e}^2}{4\pi {\epsilon }_0} \\ r=\left(n^2{\hslash }^2\right)\frac{4\pi {\epsilon }_0}{m{e}^2} \end{gathered}$$

Therefore, the equation $r\mathit{=}\frac{\mathit{\left(}\mathit{4}\mathit{\pi }{\mathit{\epsilon }}_{\mathit{0}}\mathit{\right)}{\hslash }^{\mathit{2}}}{m{e}^{\mathit{2}}}{n}^{\mathit{2}}$is proved.