In a hydrogen atom,a transition from n = 4 to | vedclass

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

Question

(B) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \fr

Vedclass · Accepted Answer

$5 \rightarrow 4$

Vedclass · Answer

(B) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Infrared radiation corresponds to the Paschen series $(n_f = 3)$,Brackett series $(n_f = 4)$,or Pfund series $(n_f = 5)$.The transition from $n = 4$ to $n = 3$ belongs to the Paschen series,which lies in the infrared region.Among the given options,the transition $5 ightarrow 4$ belongs to the Brackett series,which also lies in the infrared region.Therefore,the correct transition is $5 ightarrow 4$.

In a hydrogen atom,a transition from $n = 4$ to $n = 3$ results in the emission of an infrared photon. Which of the following transitions would also result in the emission of an infrared photon?

Similar Questions

What is the hydrogen spectral series? Explain with the help of a diagram.

Number of visible lines in Balmer's series

At room temperature, gaseous hydrogen is bombarded with a beam of electrons of $13.6 \ eV$ energy. The series to which the emitted spectral line belongs is:

The shortest wavelength in the Balmer series of the hydrogen atom spectrum is approximately equal to (use $R_{H} = 1.097 \times 10^7 \ \text{m}^{-1}$) (in $\text{Å}$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module