Question

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

Answer 1

$R_H=$Rydberg constant=${109737cm}^{-1}$

The transition from $n=2$ to $n=\infty$ will have shortest wavelength

wave number=$\overrightarrow{v}=\frac{1}{\lambda }=R_H(\frac{1}{4}-\frac{1}{\infty })=\frac{R_H}{4}=27434.25{cm}^{-1}\left(approx\right)$