For any given series of spectral lines of ato | vedclass

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Question

(C) The wavenumber is given by $\bar{
u} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For any series,$\Delta \bar{
u} = \bar{
u}_{\max}

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

(C) The wavenumber is given by $\bar{
u} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For any series,$\Delta \bar{
u} = \bar{
u}_{\max} - \bar{
u}_{\min} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_1+1^2} ight] - R_H \left[ \frac{1}{n_1^2} - \frac{1}{\infty^2} ight] = R_H \left[ \frac{1}{\infty^2} - \frac{1}{(n_1+1)^2} ight] = \frac{R_H}{(n_1+1)^2}$.For the Lyman series,$n_1 = 1$,so $\Delta \bar{
u}_{	ext{Lyman}} = \frac{R_H}{(1+1)^2} = \frac{R_H}{4}$.For the Balmer series,$n_1 = 2$,so $\Delta \bar{
u}_{	ext{Balmer}} = \frac{R_H}{(2+1)^2} = \frac{R_H}{9}$.Therefore,the ratio is $\frac{\Delta \bar{
u}_{	ext{Lyman}}}{\Delta \bar{
u}_{	ext{Balmer}}} = \frac{R_H/4}{R_H/9} = \frac{9}{4}$.

For any given series of spectral lines of atomic hydrogen,let $\Delta \bar{\nu} = \Delta \bar{\nu}_{\max} - \Delta \bar{\nu}_{\min}$ be the difference in maximum and minimum wavenumbers in $cm^{-1}$. The ratio $\Delta \bar{\nu}_{\text{Lyman}} / \Delta \bar{\nu}_{\text{Balmer}}$ is

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