Calculate the wave number for the shortest wavelength transition in Brackett series for atomic hydrogen.

The correct option is C $6.85\times {10}^5{\text{m}}^{-1}$
For Brackett series, all transitions take place from higher levels to $n=4$ level.
Wavenumber, $\stackrel{-}{v}=\frac{1}{\lambda }=R_H[\frac{1}{4^2}-\frac{1}{n^2}]$

For minimum or the shortest wavelength, $n=\infty$
Hence, $\stackrel{-}{v}=\frac{1}{{\lambda }_{min}}=R_H[\frac{1}{4^2}-\frac{1}{{\infty }^2}]$

$=\frac{1.09677\times {10}^7{\text{m}}^{-1}}{16}$

$\Rightarrow \stackrel{-}{v}=6.85\times {10}^5{\text{m}}^{-1}$