Question

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

$[R_H=1.097\times {10}^5{cm}^{-1}]$

• A. $27425c{m}^{-1}$
• B. $23456c{m}^{-1}$
• C. $34265c{m}^{-1}$
• D. $18453c{m}^{-1}$

Answer 1

Answer: (A)

$27425c{m}^{-1}$

The shortest wavelength transition inn the balmer series corresponds to the transiton $n_1=2,n_2=\infty$.

For hydrogen spectrum,

Wave number, $\overline{\nu }$ or $\frac{1}{\lambda }=R_H[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]$

Putting the values and we get,

$\overline{\nu }=27425c{m}^{-1}$