Ratio of the wavelengths of the first line of | vedclass

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

Question

(C) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the firs

Vedclass · Accepted Answer

$5:27$

Vedclass · Answer

(C) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$,which implies $\lambda_L = \frac{4}{3R}$.For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$,which implies $\lambda_B = \frac{36}{5R}$.The ratio is $\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.

Ratio of the wavelengths of the first line of the Lyman series and the first line of the Balmer series is

Similar Questions

The light emitted in the transition $n = 3$ to $n = 2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light. Find the maximum work function that a metal can have so that $H_{\alpha}$-light can emit photoelectrons from it. (in $\text{ eV}$)

Name the different series obtained in the hydrogen spectrum and give the general formula for finding the wave number.

If the series limit of the Lyman series for a Hydrogen atom is equal to the series limit of the Balmer series for a hydrogen-like atom,then the atomic number of this hydrogen-like atom will be:

What is the maximum wavelength of the $H_{\alpha}$ line in the Balmer series of the hydrogen atom (in $nm$)?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module