Question

Using Bohr's postulates derive the expression for the frequency of radiation submitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number $n_i$) to the lower state, $\left(n_f\right)$
when electron in hydrogen atom jumps from energy state $n_i=4to{n}_f=3,2,1$. Identify the special series to which the emission lines holding 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:
$E_k=\frac{1}{2}m{v}^2=\frac{k{e}^2}{r}$

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

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

Potential energy:
$E_p=\frac{-k\left(e\right)\times \left(e\right)}{r}=\frac{-k{e}^2}{r}$

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

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

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 uindergoes transition from higher energy state to the lower energy state $\left(n_f\right)$ is.
$hv=E_m-E_{nf}$

$hv=-\frac{2{\pi }^2{k}^{2}m{e}^4}{h^2}\times \frac{1}{n_i^2}-[\frac{-2{\pi }^2{k}^{2}m{e}^4}{h^2}\times \frac{1}{n_i^2}]$

$hv=\frac{2{\pi }^2{k}^{2}m{e}^4}{h^2}\times [\frac{1}{n_i^2}-\frac{1}{n_i^2}]$

$v=\frac{2{\pi }^2{k}^{2}m{e}^4}{h^3}\times [\frac{1}{n_i^2}-\frac{1}{n_i^2}]$

$v=\frac{c2{\pi }^2{k}^{2}m{e}^4}{c{h}^3}\times [\frac{1}{n_i^2}-\frac{1}{n_i^2}]$

$\frac{2{\pi }^2{k}^{2}m{e}^4}{c{h}^3}=R=$Rydberg constant

$R=1.097\times {10}^{7}m-1$

Thus, $v=Rc\times [\frac{1}{n_i^2}-\frac{1}{n_i^2}]$

Now higher state, $n_i=4$
Lower state, $n_f=3,2,1$

For the transition:
$n_i=4to{n}_f=3,\to PaschenSeries$
$n_i=4to{n}_f=2,\to BalmerSeries$
$n_i=4to{n}_f=1,\to LymanSeries$