If R is the Rydberg constant for hydrogen,the | vedclass

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

Question

(B) For the Lyman series,the wave number $\bar{
u}$ is given by the Rydberg formula:$\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac

Vedclass · Accepted Answer

$\frac{3R}{4}$

Vedclass · Answer

(B) For the Lyman series,the wave number $\bar{
u}$ is given by the Rydberg formula:$\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$where $n_1 = 1$ and $n_2 = 2, 3, 4, \dots$For the first line of the Lyman series,we take $n_1 = 1$ and $n_2 = 2$.Substituting these values into the formula:$\bar{
u} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight)$$\bar{
u} = R \left( 1 - \frac{1}{4} ight)$$\bar{
u} = R \left( \frac{3}{4} ight) = \frac{3R}{4}$

If $R$ is the Rydberg constant for hydrogen,the wave number of the first line in the Lyman series will be

Similar Questions

What is emission line spectrum?

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 $\text{ eV}$)

Number of visible lines in Balmer's series

The wavelength of radiation emitted is ${\lambda _0}$ when an electron jumps from the third to the second orbit of a hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom,the wavelength of radiation emitted will be

Which of the following spectral series in a hydrogen atom gives a spectral line of $4860 \mathring A$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module