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Question

(A) The wave number $\bar{
u}$ of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula: $\bar{
u} = R_H \left( \frac{1}

Vedclass · Accepted Answer

$n=3$ to $n=1$

Vedclass · Answer

(A) The wave number $\bar{
u}$ of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula: $\bar{
u} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Given that $\bar{
u} = \frac{8}{9} R_H$,we substitute this into the formula:$\frac{8}{9} R_H = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Dividing both sides by $R_H$,we get $\frac{8}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$.For the emission spectrum,$n_1 $\frac{8}{9} = 1 - \frac{1}{n_2^2} \implies \frac{1}{n_2^2} = 1 - \frac{8}{9} = \frac{1}{9}$.Thus,$n_2^2 = 9$,which means $n_2 = 3$.Therefore,the electron jumps from $n=3$ to $n=1$.

The wave number of the spectral line in the emission spectrum of hydrogen will be equal to $\frac{8}{9}$ times the Rydberg's constant if the electron jumps from

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