The wave number of the last line of the Lyman | vedclass

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Question

(D) The Lyman series in the atomic spectrum of the $H$ atom corresponds to electronic transitions from higher energy levels $(n_2 = 2, 3, 4, \dots, \i

Vedclass · Accepted Answer

$n_2 = \infty$ to $n_1 = 1$

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(D) The Lyman series in the atomic spectrum of the $H$ atom corresponds to electronic transitions from higher energy levels $(n_2 = 2, 3, 4, \dots, \infty)$ to the ground state $(n_1 = 1)$.The wave number $(\bar{
u})$ is given by the Rydberg formula: $\bar{
u} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.The 'last line' of a spectral series corresponds to the transition from the highest possible energy level,which is $n_2 = \infty$,to the series limit level,which is $n_1 = 1$.Therefore,the transition is from $n_2 = \infty$ to $n_1 = 1$.

The wave number of the last line of the Lyman series of the atomic spectrum of the $H$ atom corresponds to :-

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