If the wave number of radiation emitted for t | vedclass

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Question

(A) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For hydrogen $(Z=1)$, the t

Vedclass · Accepted Answer

$\frac{7x}{144}$

Vedclass · Answer

(A) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For hydrogen $(Z=1)$, the transition from $n_2$ to $n_1=2$ is $\bar{
u}_1 = R_H (\frac{1}{2^2} - \frac{1}{n_2^2}) = R_H (\frac{1}{4} - \frac{1}{n_2^2}) = \frac{5x}{36}$.Given $\frac{1}{4} - \frac{1}{n_2^2} = \frac{5}{36 R_H}$. Assuming $R_H$ is a constant that cancels out or is normalized, we find $n_2$ by setting $\frac{1}{4} - \frac{1}{n_2^2} = \frac{5}{36} \implies \frac{1}{n_2^2} = \frac{1}{4} - \frac{5}{36} = \frac{9-5}{36} = \frac{4}{36} = \frac{1}{9}$. Thus, $n_2 = 3$.The transition from $n_2=3$ to $n_3=4$ involves absorption of radiation with wave number $\bar{
u}_2 = R_H (\frac{1}{3^2} - \frac{1}{4^2}) = R_H (\frac{1}{9} - \frac{1}{16}) = R_H (\frac{16-9}{144}) = \frac{7}{144} R_H$.Since $\bar{
u}_1 = \frac{5}{36} R_H = \frac{5x}{36}$, then $R_H = x$. Substituting $R_H = x$, we get $\bar{
u}_2 = \frac{7x}{144}$.

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