Question

Using Bohr's postulates, derive the expression for the frequency of radiation emitted when an electron in a hydrogen atom undergoes transition from a higher energy state (quantum number $n_i$) to a lower state, $\left(n_f\right)$.
When an electron in a hydrogen atom jumps from energy state $n_i=4$ to $n_f=3,2,1,$ identify the spectral series to which the emission lines belong.

Answer 1

In the hydrogen atom,
Radius of electron orbit,
$r=\frac{n^2{h}^2}{4{\pi }^{2}km{e}^2}$ ..... (i)
Kinetic energy of electron,
$E_k=\frac{1}{2}m{v}^2=\frac{k{e}^2}{2r}$
Using equation (i), we get
$E_k=\frac{k{e}^2}{2}\frac{4{\pi }^{2}km{e}^2}{n^2{h}^2}=\frac{2{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$
Potential energy,
$E_p=\frac{-k\left(e\right)\times \left(e\right)}{r}=\frac{-k{e}^2}{r}$
Using equation (i), we get
$E_p=-k{e}^2\times \frac{4{\pi }^{2}kkm{e}^2}{n^2{h}^2}=-\frac{4{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$
Total energy of electron,
$E=\frac{2{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}-\frac{4{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$
$=-\frac{2{\pi }^2{k}^{2}m{e}^4}{n^2{h}^2}$
$=-\frac{2{\pi }^2{k}^{2}m{e}^4}{h^2}\times \left(\frac{1}{n^2}\right)$
Now, according to Bohr's frequency condition when electron in hydrogen atom undergoes transition from higher energy state to the lower energy state $\left(n_f\right)$ is,
$hv=E_{n_i}-E_{n_f}$
or \left(\frac{-2 \pi^2k^2me^4}{h2} \times \frac{1}{n^2_f} \right)\)
or $hv=\frac{2{\pi }^2{k}^{2}m{e}^4}{h^2}\times (\frac{1}{n_f^2}-\frac{1}{n_i^2})$
$v=\frac{2{\pi }^2{k}^{2}m{e}^4}{h^3}\times (\frac{1}{n_f^2}-\frac{1}{n_i^2})$
$v=\frac{c2{\pi }^2{k}^{2}m{e}^4}{c{h}^3}\times (\frac{1}{n_f^2}-\frac{1}{n_i^2})$
$\frac{2{\pi }^2{k}^{2}m{e}^4}{c{h}^3}$ = R (Rydberg constant)
$R=1.097\times {10}^7{m}^{-1}$
Thus,
$v=Rc\times (\frac{1}{n_f^2}-\frac{1}{n_i^2})$
Now, higher state,
$n_i=4$
Lower state,
$n_f=3,2,1$
For the transition
$n_i=4to{n}_f=3\to$ Paschen series
$n_i=4to{n}_f=2\to$ Balmer series
$n_i=4to{n}_f=1\to$ Lyman series