Statement of Bohr's quantization condition
Calculation of shortest wavelength
Identification of part of electromagnetic spectrum
Electron revolves around the nucleus only in those orbits for which the angular momentum is some integral of $\mathit{h}∕2\mathit{\pi }$ . (where $\mathit{h}$ is Planck's constant)
Also give full credit if a student write mathematically $\mathit{m}\mathit{v}\mathit{r}=\frac{\mathit{n}\mathit{h}}{2\mathit{\pi }}$
$\frac{1}{\mathit{\lambda }}=\mathit{R}\left(\frac{1}{{\mathit{n}}_{\mathit{f}}^2}-\frac{1}{{\mathit{n}}_{\mathit{i}}^2}\right)$
For Brackett Series,
Shortest wavelength is for the transition of electrons from ${\mathit{n}}_{\mathit{i}}=\mathit{\infty }$ to ${\mathit{n}}_{\mathit{f}}=4$
$\frac{1}{\mathit{\lambda }}=\mathit{R}\left(\frac{1}{4^2}\right)=\frac{\mathit{R}}{16}$
$\mathit{\lambda }=\frac{16}{\mathit{R}}\mathrm{m}$

OR
Statement of the formula for ${\mathit{r}}_{\mathit{n}}$
Statement of the formula for ${\mathit{v}}_{\mathit{n}}$
Obtaining formula for ${\mathit{T}}_{\mathit{n}}$
Getting expression for ${\mathit{T}}_2\left(\mathit{n}=2\right)$
Radius, ${\mathit{r}}_{\mathit{n}}=\frac{{\mathit{h}}^2{\mathit{\epsilon }}_0}{\mathit{\pi }\mathit{m}{\mathit{e}}^2}{\mathit{n}}^2$
velocity, ${\mathit{v}}_{\mathit{n}}=\frac{2\mathit{\pi }{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0\mathit{h}}\frac{1}{\mathit{n}}$
Timeperiod, ${\mathit{T}}_{\mathit{n}}=\frac{2\mathit{\pi }{\mathit{r}}_{\mathit{n}}}{{\mathit{v}}_{\mathit{n}}}=\frac{4{\mathit{\epsilon }}_0^2{\mathit{h}}^3{\mathit{n}}^3}{\mathit{m}{\mathit{e}}^4}$
For first excited state of hydrogen atom $\mathit{n}=2$
${\mathit{T}}_2=\frac{32{\mathit{\epsilon }}_0^2{\mathit{h}}^3}{\mathit{m}{\mathit{e}}^4}$
On calculation we get ${\mathit{T}}_2\approx 1.22\times {10}^{-15}\mathrm{}\mathit{s}$.