What is the wave number of the lowest energy | vedclass

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Question

(D) The Rydberg formula for the wave number of a spectral line is given by: $\bar{
u} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the Paschen seri

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$\bar{
u} = R_H (\frac{7}{144}) 	ext{ cm}^{-1}$

Vedclass · Answer

(D) The Rydberg formula for the wave number of a spectral line is given by: $\bar{
u} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the Paschen series, the transition occurs to the energy level $n_1 = 3$.The lowest energy transition corresponds to the transition from the immediate next energy level, which is $n_2 = 4$.Substituting these values into the formula:$\bar{
u} = R_H (\frac{1}{3^2} - \frac{1}{4^2})$$\bar{
u} = R_H (\frac{1}{9} - \frac{1}{16})$$\bar{
u} = R_H (\frac{16 - 9}{144})$$\bar{
u} = R_H (\frac{7}{144}) 	ext{ cm}^{-1}$.

What is the wave number of the lowest energy transition associated with the Paschen series?

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