If Delta lambda_L is the difference between t | vedclass

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

Question

(D) For the Lyman series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$ where $n = 2, 3, 4, \dots$.

Vedclass · Accepted Answer

$9.6$

Vedclass · Answer

(D) For the Lyman series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} ight)$ where $n = 2, 3, 4, \dots$. Shortest wavelength $(\lambda_{L, min})$ occurs at $n = \infty$: $\frac{1}{\lambda_{L, min}} = R \implies \lambda_{L, min} = \frac{1}{R}$. Longest wavelength $(\lambda_{L, max})$ occurs at $n = 2$: $\frac{1}{\lambda_{L, max}} = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies \lambda_{L, max} = \frac{4}{3R}$. $\Delta \lambda_L = \lambda_{L, max} - \lambda_{L, min} = \frac{4}{3R} - \frac{1}{R} = \frac{1}{3R}$. For the Balmer series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} ight)$ where $n = 3, 4, 5, \dots$. Shortest wavelength $(\lambda_{B, min})$ occurs at $n = \infty$: $\frac{1}{\lambda_{B, min}} = \frac{R}{4} \implies \lambda_{B, min} = \frac{4}{R}$. Longest wavelength $(\lambda_{B, max})$ occurs at $n = 3$: $\frac{1}{\lambda_{B, max}} = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36} \implies \lambda_{B, max} = \frac{36}{5R}$. $\Delta \lambda_B = \lambda_{B, max} - \lambda_{B, min} = \frac{36}{5R} - \frac{4}{R} = \frac{36 - 20}{5R} = \frac{16}{5R}$. Therefore,$\frac{\Delta \lambda_B}{\Delta \lambda_L} = \frac{16/5R}{1/3R} = \frac{16}{5} 	imes 3 = \frac{48}{5} = 9.6$.

If $\Delta \lambda_L$ is the difference between the shortest and longest wavelengths of the Lyman series and $\Delta \lambda_B$ is the difference between the shortest and longest wavelengths of the Balmer series,then $\frac{\Delta \lambda_B}{\Delta \lambda_L} = $

Similar Questions

Whenever a hydrogen atom emits a photon in the Balmer series,

The ratio of the longest to shortest wavelengths in the Brackett series of hydrogen spectra is

The series limit of the Lyman series is given by a wavenumber of ......

The shortest wavelength in the Lyman series is $91.2 \, nm$. The largest wavelength of the series is.....$nm$

In the hydrogen spectrum,the ratio of the wavelengths for Lyman-alpha radiation to Balmer-alpha radiation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module