Derive an expression for potential and kinetic energy of an electron in any orbit of a hydrogen atom according to Bohr’s atomic model. How does, P.E change with increasing n.

According to Bohr electrostatic force of attraction, F_e between the revolving electrons and nucleus provides the necessary centripetal force F_c to keep them in their orbits .
Thus in orbit of an hydrogen atom.
${\mathrm{F}}_{\mathrm{c}}={\mathrm{F}}_{\mathrm{e}}\Rightarrow \frac{{\mathrm{mv}}^2}{\mathrm{r}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^2}$
The relation between the orbit radius and the electron velocity is $\mathrm{r}=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\left({{\mathrm{mv}}^2}^{ }\right)}$
The kinetic energy (K) of the electron in hydrogen atom is
${\mathrm{K}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{n}}^2=\frac{{\mathrm{e}}^2}{8{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}$
The potential energy of electron in the n^th orbit is 
${\mathrm{U}}_{\mathrm{n}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{\left(\mathrm{e}\right)\times (-\mathrm{e})}{{\mathrm{r}}_{\mathrm{n}}}=-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}$
with increasing n, radius of orbit increases. Hence potential energy of electron increases