Question

The wavenumber for the shortest wavelength transition in the Balmer series of atomic hydrogen is:

• A. $2.725\times {10}^6{m}^{-1}$
  
  
• B. $2.725\times {10}^{-6}{m}^{-1}$
• C. $3.725\times {10}^6{m}^{-1}$
• D. $2.725\times {10}^{6}c{m}^{-1}$

Answer 1

The correct option is A $2.725\times {10}^6{m}^{-1}$



The Rydberg's equation for H-atom is:
$\frac{1}{\lambda }=\overline{v}\left(wavenumber\right)=R_H[\frac{1}{{n_1}^2}-\frac{1}{{n_1}^2}]$
For Balmer series, $n_1=2$ and $n_2=3,4,5,.......,\infty$
For shortest wavelength $\left(\lambda \right)$, $n_2$ has to be maximum i.e. infinity. Then:
(value of $R_{H}constant=1.09\times {10}^7{m}^{-1}$)
$\overline{v}=R_H[\frac{1}{4}-\frac{1}{\infty }]=\frac{R_H}{4}=\frac{1.09\times {10}^7}{4}$

= $2.725\times {10}^6{m}^{-1}$