The frequency of one of the lines in the Pasc | vedclass

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Question

(C) For the Paschen series,the lower energy level is $n_1 = 3$.The frequency $
u$ is given by the Rydberg formula: $
u = R_H 	imes c 	imes \left(

Vedclass · Accepted Answer

$4$

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(C) For the Paschen series,the lower energy level is $n_1 = 3$.The frequency $
u$ is given by the Rydberg formula: $
u = R_H 	imes c 	imes \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Given $
u = 2.340 	imes 10^{11} \ Hz$,$R_H = 1.097 	imes 10^7 \ m^{-1}$,and $c = 3 	imes 10^8 \ m/s$.$2.340 	imes 10^{11} = (1.097 	imes 10^7) 	imes (3 	imes 10^8) 	imes \left( \frac{1}{3^2} - \frac{1}{n_2^2} ight)$.$2.340 	imes 10^{11} = 3.291 	imes 10^{15} 	imes \left( \frac{1}{9} - \frac{1}{n_2^2} ight)$.$\frac{2.340 	imes 10^{11}}{3.291 	imes 10^{15}} = \frac{1}{9} - \frac{1}{n_2^2}$.$7.11 	imes 10^{-5} = 0.1111 - \frac{1}{n_2^2}$.$\frac{1}{n_2^2} = 0.1111 - 0.0000711 \approx 0.1110$.$n_2^2 \approx \frac{1}{0.1110} \approx 9.009$.$n_2 \approx 3$. However,the Paschen series requires $n_2 > n_1$. Given the options,if $n_2 = 4$,$
u = 3.291 	imes 10^{15} 	imes (1/9 - 1/16) = 1.599 	imes 10^{14} \ Hz$. The provided frequency value in the question appears to be inconsistent with standard hydrogen transitions. Based on the options provided for a transition in the Paschen series,$n_2 = 4$ is the first line.

The frequency of one of the lines in the Paschen series of a hydrogen atom is $2.340 \times 10^{11} \ Hz$. The quantum number $n_2$ which produces this transition is

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