Specify the quantum number for the limit wave | vedclass

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

Question

(A) The wavelength $\lambda$ of a spectral line in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} -

Vedclass · Accepted Answer

$n_1 = 1, n_2 = \infty$

Vedclass · Answer

(A) The wavelength $\lambda$ of a spectral line in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1$ is the lower energy level and $n_2$ is the higher energy level.For the limit wavelength (also known as the series limit),the wavelength is at its minimum,which corresponds to the maximum possible energy transition.This occurs when the electron transitions from an infinite energy level $(n_2 = \infty)$ to the specific lower energy level $(n_1)$ of the series.Therefore,for the limit wavelength,the quantum numbers are $n_1$ (fixed for the series) and $n_2 = \infty$.

Specify the quantum number for the limit wavelength of the spectral series of a hydrogen atom.

Similar Questions

The first wavelength of the Paschen series in hydrogen is $18,800 \, \mathring A$. What is the minimum wavelength of the Paschen series in $\mathring A$?

The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first member of the Balmer and Paschen series in a hydrogen atom,respectively,then the ratio of the respective frequencies,$f_1 / f_2$,is:

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

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module