State the basic postulates of Bohr's theory of atomic spectra. Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom?

Postulates of bohr's theory :
→ The electrons revolve around the nucleus of the atom in circular orbits.
→ The angular momentum of an electron in particular orbits is integral multiples of $\mathrm{h}/2\mathrm{\pi }$
$\therefore \mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi }}$ 
→ where 'h' is plank's constant 'n' is principal quantum number.
→ The electron revolves in definite orbits its energy remains constant. So these are known as stationary orbits.
→ If the electron absorbs the energy, then it jumps from lower energy level to heigher energy level. Similarly if the electrons jumps from higher energy level to lower energy level, it loses energy in the form of radiations.
$\mathrm{\Delta E}={\mathrm{E}}_2-{\mathrm{E}}_1=h\nu$
Bohr's theory of Hydrogen atom :
→ Consider an electron revolving around the nucleus in the n^th stationary orbit as shown in figure

Radii of Bohr's stationary orbits :
→ As the centripetal force is provided by the electrostatic force of attraction between the nucleus and electrons.
$\frac{{\mathrm{mv}}_{\mathrm{n}}^2}{{\mathrm{r}}_{\mathrm{n}}}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{\left(\mathrm{Ze}\right)\mathrm{e}}{{\mathrm{r}}_{\mathrm{n}}^2}$
→ where r_n= radius of the n^th orbit
${\mathrm{mv}}_{\mathrm{n}}^2=\frac{{\mathrm{Ze}}^2}{4\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}_{\mathrm{n}}}\to \left(1\right)$
→ V_n= velocity of the electron in the n^th orbit.
→ According to Bohr, ${\mathrm{mv}}_{\mathrm{n}}{\mathrm{r}}_{\mathrm{n}}=\mathrm{n}\frac{\mathrm{h}}{2\mathrm{\pi }}\to \left(2\right)$
where n = 1, 2, 3 -----------
→ squaring equation (2) and dividing by equation (1), we get
$\therefore {\mathrm{r}}_{\mathrm{n}}=\left[\frac{{\mathrm{ϵ}}_0{\mathrm{h}}^2}{{\mathrm{\pi me}}^2}\right]\frac{{\mathrm{n}}^2}{\mathrm{z}}\to \left(3\right)$
Total energy of electron in stationary orbits :
→ The kinetic energy of electron in the n^th orbit is
${\mathrm{K}}_{\mathrm{n}}=1/2{\mathrm{mv}}_{\mathrm{n}}^2=\frac{{\mathrm{ze}}^2}{8\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}_{\mathrm{n}}}$
→ The potential energy of electron in the n^th orbit is
${\mathrm{U}}_{\mathrm{n}}=\frac{-{\mathrm{ze}}^2}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{n}}}$
→ Total energy of electron in the n^th orbit is
$\begin{array}{l}{\mathrm{E}}_{\mathrm{n}}={\mathrm{K}}_{\mathrm{n}}+{\mathrm{U}}_{\mathrm{n}}=\frac{{\mathrm{ze}}^2}{8\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}_{\mathrm{n}}}+\left[-\frac{{\mathrm{ze}}^2}{4\mathrm{\pi }{\in }_0{\mathrm{r}}_{\mathrm{n}}}\right]\\ \therefore {\mathrm{E}}_{\mathrm{n}}=\frac{-{\mathrm{ze}}^2}{8\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}_{\mathrm{n}}}\\ \therefore {\mathrm{E}}_{\mathrm{n}}=-\frac{13.6{\mathrm{z}}^2}{{\mathrm{n}}^2}\mathrm{ev}\end{array}$
as n increases the total energy of the electron increases