The series limit of the Lyman series is given | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 ......

Similar Questions

The ratio of wavelengths of the second line in the Balmer series and the first line in the Lyman series of a hydrogen atom is:

Line spectra are due to

If the electron in a hydrogen atom transits from the second orbit to the first orbit,in what region would the emitted radiation be?

If radiation of all wavelengths from ultraviolet to infrared is passed through hydrogen gas at room temperature,absorption lines will be observed in the:

Vedclass Test Series

Exam Paper Generator

Online Exam Module