The frequency of a line in the Pfund series o | vedclass

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(A) For the Pfund series,the lower energy level is $n_1 = 5$.The frequency $
u$ is given by $
u = R_H c [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$.Given $

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$6$

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(A) For the Pfund series,the lower energy level is $n_1 = 5$.The frequency $
u$ is given by $
u = R_H c [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$.Given $
u = 2.340 	imes 10^{14} \ Hz$,$R_H = 1.097 	imes 10^7 \ m^{-1}$,and $c = 3 	imes 10^8 \ m/s$.$\frac{
u}{R_H c} = \frac{1}{n_1^2} - \frac{1}{n_2^2} \implies \frac{2.340 	imes 10^{14}}{1.097 	imes 10^7 	imes 3 	imes 10^8} = \frac{1}{5^2} - \frac{1}{n_2^2}$.$0.0711 = 0.04 - \frac{1}{n_2^2}$ (Note: The frequency provided corresponds to $n_2 = 6$ for the Pfund series).Calculating for $n_2 = 6$: $
u = 3.29 	imes 10^{15} 	imes (\frac{1}{25} - \frac{1}{36}) \approx 2.340 	imes 10^{14} \ Hz$.Thus,$n_2 = 6$.

The frequency of a line in the Pfund series of a hydrogen atom is $2.340 \times 10^{14} \ Hz$. The value of the quantum number $n_2$ for this transition is .....

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