Using Bohr?s postulates, obtain the expression for total energy of the electron in the
n
th orbit of hydrogen atom.

 TOTAL ENERGY OF ELECTRON IN nth  ORBIT IS THE SUM OF KINETIC ENERGY( DUE TO MOTION OF ELECTRON IN THE ORBIT) AND ELECTROSTATIC POTENTIAL ENERGY ( DUE INTERACTION BETWEEN POSITIVELY CHARGE NUCLEUS AND NEGATIVELY CHARGE ELECTRON)

velocity of electron in nth orbit
                                           $\frac{Ze}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}} = \frac{nh}{2\mathrm{\pi }}\times v$
                                         $v = \frac{Z}{n}\frac{e^2}{2{\epsilon }_{o}h}$

radius of nth orbit
                                             $mvr = \frac{nh}{2\mathrm{\pi }}$
                                             $r = \frac{n^2}{Z}\frac{{\epsilon }_o{h}^2}{\mathrm{\pi me}} u\sin g v = \frac{Z}{n}\frac{e^2}{2{\epsilon }_{o}h}$

 TOTAL ENERGY OF ELECTRON IN nth ORBIT
         
                                                K.E. = $\frac{1}{2}m{v}^2$ = $\frac{Z{e}^2}{8{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{r}}$    [ using the relation $mvr = \frac{nh}{2\mathrm{\pi }}$ $\Rightarrow v = \frac{nh}{2\mathrm{\pi mr}}$]
                                                P.E. = $\frac{-Z{e}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{r}} = 2K.E.$

TOTAL ENERGY OF ELECTRON NUCLEUS SYSTEM = .K.E +  P.E
                                                                                            = K.E - 2K.E. = - K.E

HENCE THE TOTAL ENERGY OF ELECTRON IN nth ORBIT = $\frac{-Z{e}^2}{8{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{r}}$