State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong.
OR
Calculate the orbital period of the electron in the first excited state of hydrogen atom.

According to Bohr’s quantization of angular momentum, the stationary orbits are those in which angular momentum of electron
is an integral multiple of $\frac{\mathrm{h}}{2\mathrm{\pi }}\mathrm{i}.\mathrm{e}.,$
$\mathrm{mvr}=\mathrm{n}\frac{\mathrm{h}}{2\mathrm{\pi }}\text{ Where, }\mathrm{n}=1,2,3$
For Brackett series,
$\begin{array}{r}\frac{1}{{\mathrm{\lambda }}_{min}}=\mathrm{n}=\frac{1}{\sin \mathrm{C}}\\ \Rightarrow \frac{1}{{\mathrm{\lambda }}_{min}}=\frac{\mathrm{R}}{16}\end{array}$$\begin{array}{r}\mathrm{v}=\frac{{\mathrm{ze}}^2}{2{\mathrm{h\epsilon }}_0\mathrm{n}}\\ \mathrm{v}=\frac{{\mathrm{e}}^2}{2{\mathrm{h\epsilon }}_0}\end{array}$
Where,
$\begin{array}{r}\mathrm{R}=\text{ Rydberg constant }\\ =1.096\times {10}^7{\mathrm{m}}^{-1}\end{array}$
 $\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\lambda }}_{min}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{16}{1.096\times {10}^7}\mathrm{m}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=14.599\times {10}^{-7}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=14599\text{Å}\end{array}$
Brackett series is found in infrared region $\left({10}^6\text{Å to }7000\text{Å}\right)$ of electromagnetic spectrum.
OR
For hydrogen atom z=1 and n=1
$\therefore$Velocity of electron,
$\begin{array}{r}\mathrm{v}=\frac{{\mathrm{ze}}^2}{2{\mathrm{h\epsilon }}_0\mathrm{n}}\\ \mathrm{v}=\frac{{\mathrm{e}}^2}{2{\mathrm{h\epsilon }}_0}\end{array}$
Therefore, $\mathrm{v}=\frac{\mathrm{c}}{137}$, where , c is velocity of light.
Orbital period 
$\begin{array}{l}=\frac{\text{ Distance }}{\text{ Velocity }}=\frac{2\mathrm{\pi r}}{\mathrm{c}}\times 137\\ =\frac{2\times 3.14\times 5.29\times {10}^{-11}\times 137}{3\times {10}^3}\\ =1.517\times {10}^{-16}\sec \end{array}$