The necessary centripetal force for the rotation of electron is supplied by the electrostatic force between the electron and nucleus.
$\frac{\mathit{m}{\mathit{v}}^2}{\mathit{r}}=\left(\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\right)\left(\frac{{\mathit{e}}^2}{{\mathit{r}}^2}\right)\left[\mathrm{putting}\mathrm{Z}=1\right]$
Or, $\mathit{m}{\mathit{v}}^2=\frac{{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0\mathit{r}}..........\left(\mathit{i}\right)$
From Bohr's theory,
$\mathit{m}\mathit{v}\mathit{r}=\frac{\mathit{n}\mathit{h}}{2\mathit{\pi }}$
$\therefore \mathit{v}=\frac{\mathit{n}\mathit{h}}{2\mathit{\pi }\mathit{m}\mathit{r}}$
Putting in equation (i)
$\mathit{m}{\left(\frac{\mathit{n}\mathit{h}}{2\mathit{\pi }\mathit{m}\mathit{r}}\right)}^2=\frac{{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0\mathit{r}}$
Or, $\mathit{r}=\frac{{\mathit{\epsilon }}_0{\mathit{n}}^2{\mathit{h}}^2}{\mathit{\pi }\mathit{m}{\mathit{e}}^2}$
In general,
${\mathit{r}}_{\mathit{n}}=\frac{{\mathit{\epsilon }}_0{\mathit{n}}^2{\mathit{h}}^2}{\mathit{\pi }\mathit{m}{\mathit{e}}^2}$
$\therefore {\mathit{r}}_{\mathit{n}}\propto {\mathit{n}}^2$
(b) ${\mathit{E}}_{\mathit{n}}\propto -\frac{1}{{\mathit{n}}^2}$
Energy is minimum at $\mathit{n}=1$
As $\mathit{n}$ increases, the enrgy becomes less negative, which means energy increases.