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Question

(D) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For a given tran

Vedclass · Accepted Answer

$81,588 \, cm^{-1}$

Vedclass · Answer

(D) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For a given transition $(n_1=2, n_2=4)$,the wave number is proportional to the square of the atomic number: $\bar{
u} \propto Z^2$.For Hydrogen $(H)$,$Z_1 = 1$. For Helium ion $(He^+)$,$Z_2 = 2$.Therefore,$\frac{\bar{
u}_{He^+}}{\bar{
u}_{H}} = \left( \frac{Z_2}{Z_1} ight)^2 = \left( \frac{2}{1} ight)^2 = 4$.Thus,$\bar{
u}_{He^+} = 4 	imes \bar{
u}_{H} = 4 	imes 20,397 \, cm^{-1} = 81,588 \, cm^{-1}$.

In a hydrogen atom,when an electron transitions from the $4^{th}$ orbit to the $2^{nd}$ orbit,the wave number is $20,397 \, cm^{-1}$. What will be the wave number when an electron in a $He^+$ ion transitions from the $4^{th}$ orbit to the $2^{nd}$ orbit?

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