Obtain the equations for the frequency of radiation and the wave number when an electron makes a transition from a higher energy state to a lower energy state in a hydrogen atom.

(N/A) According to Bohr's third postulate,when an electron makes a transition from a higher energy state with quantum number $n_{i}$ to a lower energy state with quantum number $n_{f}$ $(n_{i} > n_{f})$,a photon is emitted with energy equal to the difference in energy between the two states.
The energy of an electron in the $n_{i}$ state is given by:
$E_{n_{i}} = -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{i}^{2}}$
The energy of an electron in the $n_{f}$ state is given by:
$E_{n_{f}} = -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{f}^{2}}$
The energy of the emitted photon is $h \nu_{if} = E_{n_{i}} - E_{n_{f}}$.
Substituting the expressions:
$h \nu_{if} = -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{i}^{2}} - \left( -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{f}^{2}} \right)$
$h \nu_{if} = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2}} \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$
Thus,the frequency of the emitted radiation is:
$\nu_{if} = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3}} \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$
Since the wave number $\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}$,we divide the frequency by the speed of light $c$:
$\bar{\nu} = \frac{1}{\lambda_{if}} = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c} \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$
Here,$R = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c}$ is the Rydberg constant,with a theoretical value of approximately $1.097 \times 10^{7} \ m^{-1}$.