What is the wavenumber of the photon emitted | vedclass

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(A) The wavenumber $(\bar{
u})$ for a transition in a hydrogen atom is given by the Rydberg formula: $\bar{
u} = R_{H} 	imes Z^2 	imes \left( \fra

Vedclass · Accepted Answer

$23032$

Vedclass · Answer

(A) The wavenumber $(\bar{
u})$ for a transition in a hydrogen atom is given by the Rydberg formula: $\bar{
u} = R_{H} 	imes Z^2 	imes \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Here,$R_{H} = 109677 \ cm^{-1}$,$Z = 1$ (for hydrogen atom),$n_1 = 2$,and $n_2 = 5$.Substituting the values: $\bar{
u} = 109677 	imes 1^2 	imes \left( \frac{1}{2^2} - \frac{1}{5^2} ight)$.$\bar{
u} = 109677 	imes \left( \frac{1}{4} - \frac{1}{25} ight)$.$\bar{
u} = 109677 	imes \left( \frac{25 - 4}{100} ight) = 109677 	imes \frac{21}{100}$.$\bar{
u} = 109677 	imes 0.21 = 23032.17 \ cm^{-1}$.Rounding to the nearest integer,we get $23032 \ cm^{-1}$.

What is the wavenumber of the photon emitted during transition from the orbit $n=5$ to that of $n=2$ in hydrogen atom (in $cm^{-1}$)?
$\left[R_{H}=109677 \ cm^{-1}\right]$

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What is the wavenumber of the photon emitted during transition from the orbit $n=5$ to that of $n=2$ in hydrogen atom (in $cm^{-1}$)?$\left[R_{H}=109677 \ cm^{-1}\right]$