Wave number of a spectral line for a given tr | vedclass

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Question

(D) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_

Vedclass · Accepted Answer

$\frac{9y}{4} \ cm^{-1}$

Vedclass · Answer

(D) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For $He^{+}$ $(Z=2)$,the wave number is $y = R_H (2)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) = 4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) ......(i)$.For $Li^{2+}$ $(Z=3)$,the wave number is $\bar{
u}' = R_H (3)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) = 9 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) ......(ii)$.Dividing equation $(ii)$ by equation $(i)$:$\frac{\bar{
u}'}{y} = \frac{9 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)}{4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)} = \frac{9}{4}$.Therefore,$\bar{
u}' = \frac{9y}{4} \ cm^{-1}$.

Wave number of a spectral line for a given transition is $y \ cm^{-1}$ for $He^{+}$,then its value for $Li^{2+}$ for the same transition is

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