The frequency of a certain line of the Lyman | vedclass

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Question

(A) The frequency of a spectral line in the $H$ atom is given by $
u = R_H c (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. For a Lyman series line,$n_1 = 1$.

Vedclass · Accepted Answer

$n_2 = 3$ to $n_1 = 1$

Vedclass · Answer

(A) The frequency of a spectral line in the $H$ atom is given by $
u = R_H c (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. For a Lyman series line,$n_1 = 1$. Let the transition be $n_2 = n$. Then $
u = R_H c (1 - \frac{1}{n^2})$. Condition $(i)$ states: $
u(n 	o 1) = 
u(n' 	o 1) + 
u(n'' 	o 2)$. Substituting the Rydberg formula: $(1 - \frac{1}{n^2}) = (1 - \frac{1}{n'^2}) + (\frac{1}{4} - \frac{1}{n''^2})$. For $n=3$,$
u(3 	o 1) = R_H c (1 - \frac{1}{9}) = \frac{8}{9} R_H c$. Checking the conditions for $n=3$: $(i)$ $
u(3 	o 1) = 
u(2 	o 1) + 
u(3 	o 2) = R_H c (1 - \frac{1}{4}) + R_H c (\frac{1}{4} - \frac{1}{9}) = R_H c (1 - \frac{1}{9}) = \frac{8}{9} R_H c$. This matches. Thus,the transition corresponds to $n_2 = 3$ to $n_1 = 1$.

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