Question

How do you calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 6 to the n = 3 principal energy level? Recall that for hydrogen `E_n` =`-2.18x10^-18` `J(1/n^2)`?

Answer 1

You can do it like this:

Explanation:

The energy of the electron is given by:

`sf(E_n=-(2.18xx10^(-18))/(n^2)" ""J")`

Where `sf(n)` is the principle quantum number.

We find the difference between the electrons energy in the `sf(n=6)` and `sf(n=3)` levels and then find the frequency of the electromagnetic radiation which this corresponds to.

The difference in energy is given by:

`sf(DeltaE=-[(2.18xx10^(18))/(6^2)]-[-(2.18xx10^(-18))/(3^2)])`

`sf(DeltaE=2.18xx10^(18)[1/9-1/36])`

`sf(DeltaE=0.1816xx10^(-18)" ""J")`

To find the frequency of this emission line use the Planck Expression:

`sf(E=hf)`

`:.` `sf(f=E/h)`

`:.f` `sf(=(0.1816xx10^(-18))/(6.626xx10^(-34))=2.74xx10^(14)" ""s"^(-1))`

This transition is part of The Paschen Series of lines which occurs in the infra - red region of the electromagnetic spectrum.

This corresponds to a wavelength of 1094 nm.