Question

A 13.1 eV photon is emitted from a hydrogen atom. What is the Balmer formula n value corresponding to this emission? What is the Balmer formula m value corresponding to this emission?

I thought I could use E = (13.6 eV)/(n^2) to find n and plug it into the Balmer formula [lambda = (91.1nm)/((1/m^2)-1/n^2))], but it did not work. Any help? Thanks!

Answer 1

`n=5` and `m=1`
I guess here the problem is one of rounding up.

Explanation:

The energy of an electron in the `n`th Bohr level is `-(13.6\ "eV")/n^2`

So, the energy released as a photon when an electron transits from the `n`th level to the `m`th level is

`Delta E = (-(13.6\ "eV")/n^2)-(-(13.6\ "eV")/m^2) = (13.6\ "eV")(1/m^2-1/n^2)`

Now this energy released is obviously smaller than `(13.6\ "eV")/m^2`, so in our case we must have `m=1`.

Using

`13.1\ "eV" = (13.6\ "eV")(1/1^2-1/n^2)`

we get

`n^2=27.2`

The trouble is, there is no integer `n` for which the square will be 27.2#.

Using the closest integer possible gives `n=5`.

The discrepancy can be easily explained, however. Using `n=5` and `m=1` gives us

`Delta E = 13.056\ "eV"`

which, when rounded off to three significant figures is `13.1\ "eV"`