Using Bohr’s postulates, derive the expression for the total energy of the electron revolving in nth orbit of hydrogen atom. Find the wavelength of <img alt="H_{\alpha }" src="https://entrancecorner.codecogs.com/gif.latex?H

Using Bohr’s postulates, derive the expression for the total energy of the electron revolving in n^th orbit of hydrogen atom. Find the wavelength of $H_{\alpha }$ line, given the value of Rydberg constant, $R= 1\cdot 1\times 10^{7}m^{-1}\cdot$ .

Answers (1)

For electron orbit
$\frac{mV_{n}^{2}}{r_{n}}= \frac{1}{4\pi \varepsilon _{0}}\; \frac{e^{2}}{r_{n}^{2}}$
$V_{n}= \frac{e}{\sqrt{4\pi \varepsilon _{0}mr_{n}}}---(i)$
From Bhor's second postulate
$mVr_{n}= \frac{nh}{2\pi }$
$V_{n}= \frac{nh}{2\pi mr_{n}}---(2)$
from (1) & (2)
$\frac{nh}{2\pi mr_{n}}= \frac{e}{\sqrt{4\pi \varepsilon _{0}mr_{n}}}$
$r_{n}= \left ( \frac{h^{2}}{m} \right )\left ( \frac{h}{2\pi } \right )^{2}\left ( \frac{4\pi \varepsilon _{0}}{e^{2}} \right )---(3)$
Total energy $= \frac{-e^{2}}{8\pi \varepsilon _{0}r_{n}}$
substituting value of $r_{n}$ from (3)
$E_{n}= \frac{-me^{4}}{8n^{2}\varepsilon _{0}^{2}h^{2}}$
     $= \frac{-2\cdot 18\times 10^{-18}}{n^{2}}J$
$=\frac{-13\cdot 6}{n^{2}}\, eV$
for $H_{\alpha }$ line in Balmer series $n_{1}= 2,n_{2}= 3$
$\therefore \frac{1}{\lambda }= R\left ( \frac{1}{2^{2}}-\frac{1}{3^{2}} \right )= R\times \frac{5}{36}$
$\lambda= \frac{36}{5R}= \frac{36}{5\times 1\cdot 1\times 10^{7}}= \frac{36}{5\cdot 5}\times 10^{-7}$
                $= 6\cdot 563\times 10^{-7}$