The photon emitted when an electron drops from a 5d orbital to a 2p orbital has a wavelength of 434.084 nm.
Rydberg's formula lists the spectral energy variations of electron transitions in relation to their wavelengths. When single electron-containing atoms in their gaseous state are heated, they absorb energy and initiate electron transitions between their varying energy levels, resulting in the formation of spectral lines. Gaining energy allows an electron to move from less energetic orbits to more energetic orbits.
Given, electrons shifts from 5d to 2p orbital.
The sub-shells d and p given are irrelevant to the because the sub-shells within an energy level are degenerate in the hydrogen atom (of the same energy).
The wavelength of the photon can be calculated by Rydberg expression,
[tex]\frac{1}{\lambda }=R[\frac{1}{n^2_1}-\frac{1}{n^2_2}][/tex]
Where,
R= Rydberg's constant = [tex]1.097*10^{-2}\ nm^{-1}[/tex]
[tex]n_1[/tex] and [tex]n_2[/tex] are the sub-shells.
[tex]n_1[/tex] = 2
[tex]n_2[/tex] = 5
[tex]\frac{1}{\lambda }=1.097*10^{-2}[\frac{1}{2^2}-\frac{1}{5^2}]\\\\\frac{1}{\lambda }=1.097*10^{-2}[\frac{1}{4}-\frac{1}{25}]\\\\\frac{1}{\lambda }=1.097*10^{-2}[\frac{21}{100}]\\\\\frac{1}{\lambda} = 0.0023037\\\\\lambda=\frac{1}{0.0023037}\\\\ \lambda=434.084\ nm[/tex]
Hence, the wavelength of the photon is 434.084 nm.
To learn more about Rydberg's expression refer here
https://brainly.com/question/13185515
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Your question is incomplete, here is the complete question.
For hydrogen atom, what is the wavelength of the photon emitted when an electron drops from a 5d orbital to a 2p orbital? the rydberg constant is 1.097 x 10-2 nm-1.
