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

The minimum wavelength of Lyman series lines is $P$. What is the maximum wavelength of these lines?

The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is expressed as $E_n = \frac{-13.6}{n^2} \, eV$. The shortest and longest wavelengths of the Lyman series will be:

The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the

In the spectrum of hydrogen,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

In a hydrogen atom,a transition from $n = 4$ to $n = 3$ results in the emission of an infrared photon. Which of the following transitions would also result in the emission of an infrared photon?

Vedclass Test Series

Exam Paper Generator

Online Exam Module