The wavelength of the Lyman series is given b | vedclass

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

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

What is emission line spectrum?

The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition of an electron from the:

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$,the emitted radiation has a frequency ($R=$ Rydberg constant,$C=$ velocity of light).

In a hydrogen atom,when an electron transitions from the fourth orbit to the ground state,find the wavelength of the emitted photon in $nm$.

$v_{1}$ is the frequency of the series limit of the Lyman series,$v_{2}$ is the frequency of the first line of the Lyman series,and $v_{3}$ is the frequency of the series limit of the Balmer series. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module