In hydrogen atoms,the transition from the sta | vedclass

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

Question

(B) The wavelength $\lambda$ of emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \ri

Vedclass · Accepted Answer

$n=5$ to $n=3$

Vedclass · Answer

(B) The wavelength $\lambda$ of emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Ultraviolet radiation corresponds to the Lyman series $(n_1 = 1)$.Infrared radiation corresponds to the Paschen series $(n_1 = 3)$,Brackett series $(n_1 = 4)$,or Pfund series $(n_1 = 5)$.Comparing the options:$(A)$ $n=6$ to $n=2$ is the Balmer series (Visible).$(B)$ $n=5$ to $n=3$ is the Paschen series (Infrared).$(C)$ $n=3$ to $n=5$ is an absorption process,not emission.$(D)$ $n=4$ to $n=2$ is the Balmer series (Visible).Therefore,the transition $n=5$ to $n=3$ results in infrared radiation.

In hydrogen atoms,the transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

Similar Questions

The maximum number of emission lines for atomic hydrogen that you would expect to see with the naked eye if the only electronic levels involved are those shown in the figure,is

In the hydrogen spectrum,the ratio of the wavelengths of the last line of the Lyman series and that of the last line of the Balmer series is:

In hydrogen spectrum,which of the following spectral series lies in the ultraviolet region?

Which series is found in the visible light region?

The wavelength of the second Balmer line in the hydrogen spectrum is $600 \ nm$. The wavelength for its third line in the Lyman series is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module