The series limit of the Lyman series is given | vedclass

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Question

(A) The wavenumber $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.F

Vedclass · Accepted Answer

$1.097 	imes 10^7 \, m^{-1}$

Vedclass · Answer

(A) The wavenumber $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the series limit of the Lyman series,the electron transitions from $n_2 = \infty$ to $n_1 = 1$.Substituting these values into the formula:$\bar{
u} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R(1 - 0) = R$.The Rydberg constant $R$ is approximately $1.097 	imes 10^7 \, m^{-1}$.Therefore,the wavenumber is $1.097 	imes 10^7 \, m^{-1}$.

The series limit of the Lyman series is given by a wavenumber of ......

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