The frequency corresponding to the transition | vedclass

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Question

(B) The wavenumber $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the transition

Vedclass · Accepted Answer

$24.66 	imes 10^{14} \ Hz$

Vedclass · Answer

(B) The wavenumber $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the transition $n_2 = 2$ to $n_1 = 1$,$\bar{
u} = 109677 \ cm^{-1} 	imes (1 - 1/4) = 109677 	imes 0.75 = 82257.75 \ cm^{-1}$.Converting to meters: $\bar{
u} = 8225775 \ m^{-1}$.The frequency $
u$ is calculated as $
u = c 	imes \bar{
u}$.$
u = (3 	imes 10^8 \ m/s) 	imes (8225775 \ m^{-1}) = 2.4677 	imes 10^{15} \ Hz \approx 24.66 	imes 10^{14} \ Hz$.

The frequency corresponding to the transition $n = 2$ to $n = 1$ in a hydrogen atom is:

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