The wavelength of the Lyman series is given b | vedclass

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(A) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2

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$\frac{4}{3 	imes 10967} 	ext{ cm}$

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(A) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $R_H$ is the Rydberg constant.For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$. The longest wavelength (first line) corresponds to $n_2 = 2$.Substituting these values: $\frac{1}{\lambda} = R_H \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R_H \left[ 1 - \frac{1}{4} ight] = R_H \left( \frac{3}{4} ight)$.Therefore,$\lambda = \frac{4}{3 R_H}$.Given $R_H \approx 10967 	ext{ cm}^{-1}$,we get $\lambda = \frac{4}{3 	imes 10967} 	ext{ cm}$.

The wavelength of the Lyman series is given by which of the following expressions?

Similar Questions

The frequencies for the series limit of the Balmer and Paschen series are $f_1$ and $f_3$,respectively. If the frequency of the first line of the Balmer series is $f_2$,then the relation between $f_1, f_2$,and $f_3$ is:

The number of spectral lines in a hydrogen atom is . . . . . . .

At room temperature, gaseous hydrogen is bombarded with a beam of electrons of $13.6 \ eV$ energy. The series to which the emitted spectral line belongs is:

Line spectra are due to

Assertion : In Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason : Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom.

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