In terms of Rydberg's constant R, the wave nu | vedclass

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

Question

(C) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.

Vedclass · Accepted Answer

$\frac{5R}{36}$

Vedclass · Answer

(C) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,the transition occurs to the $n_1 = 2$ energy level.The first line of the Balmer series corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.Substituting these values into the formula:$\bar{
u} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$$\bar{
u} = R \left( \frac{1}{4} - \frac{1}{9} ight)$$\bar{
u} = R \left( \frac{9 - 4}{36} ight) = \frac{5R}{36}$.

In terms of Rydberg's constant $R,$ the wave number of the first Balmer line is

Similar Questions

The ratio of minimum wavelengths of Lyman and Balmer series will be

The wavelength of the first spectral line in the Balmer series of a hydrogen atom is $6561 \mathring A$. The wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom is

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?

$A$ hydrogen atom is excited from the ground state to a state with principal quantum number $n = 4$. The number of spectral lines emitted in the emission spectrum is:

In a hydrogen-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \ nm$ is emitted. What will be the wavelength (in $nm$) of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module