Question

Calculate the wavenumber for the longest wavelength transition in balmer series of atomic hydrogen.

• A. $1.52\times {10}^{-2}{\text{cm}}^{-1}$
• B. $1.52\times {10}^{-4}{\text{cm}}^{-1}$
• C. $1.52\times {10}^4{\text{cm}}^{-1}$
• D. $1.52\times {10}^{-2}{\text{cm}}^{-1}$

Answer 1

The correct option is C $1.52\times {10}^4{\text{cm}}^{-1}$

For longest wavelength transition in balmer series,
$n_1=2$
$n_2=3$
According to Rydberg equation
${\text{Wavenumber = R}}_h\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$=109677\times (\frac{1}{2^2}-\frac{1}{3^2})$
$=1.52\times {10}^4{\text{cm}}^{-1}$