Question

Using Bohr's postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition energy levels .

Answer 1

According Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge $+e$. Then the centripetal force is provided by Coulomb force of gravitational attraction. So,
$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$
$m{v}^2=\frac{k{e}^2}{r}..................\left(1\right)$
Where, $m=$ mass of electron
$r=$ radius of electron orbit
$v=$ velocity of electron
Again,
$mvr=\frac{nh}{2\pi }$
$v=\frac{nh}{2\pi mr}$

Substituting $v$ in (1) we get,
$m{\left(\frac{nh}{2\pi mr}\right)}^2=\frac{k{e}^2}{r}$

$\frac{n^2{h}^2}{4{\pi }^{2}mr{e}^2}=r...........\left(2\right)$

Using equation (2) we get:
$E_k=\frac{k{e}^{2}4{\pi }^{2}km{e}^2}{2n^2{h}^2}$

$=\frac{2{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$

(ii) Potential energy
$E_p=\frac{-k\left(e\right)\times \left(e\right)}{r}$

$=\frac{k{e}^2}{r}$
$=-k{e}^2\times \frac{4{\pi }^{2}km{e}^2}{n^2{h}^2}$
$=-\frac{4{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$

Hence, total energy of the element in the $n^{th}$ orbit
$E=E_P+E_K$

$=-\frac{4{\pi }^{2}km{e}^4}{n^2{h}^2}+\frac{2{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$

$=-\frac{2{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}=-\frac{13.6}{n^2}cV$

In $H$-atom, when an electron jumps from the orbit $n_i$ to orbit $n_f$ the wavelength of the emitted radiation is given by:
$\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
Where, $R\to$ Rydberg's constant $=1.09678\times {10}^{-10}{m}^{-1}$

For Balmer series, $n_f=2and{n}_i=3,4,5,....\left(i\right)$
$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n_i^2})$
Where, $n_i=3,4,5,....$
These spectral lines lie in the visible region.