A ground state hydrogen atom absorbs a photon of light having a wavelength of 94.92 nm. It then gives off a photon having a wavelength of 433.9 nm. What is the final state of the hydrogen atom?

Question

Please explain each step..

Answer 1

Hi Jessica D.,

So you have the Bohr atom, with electron energy levels described by -k/n^2 , where k is a constant, and n is the principal quantum number of the particular level the electron is residing in (usually, ground state, unless perturbed).

Now, you may assume that the electron is being transferred among three numbered energy levels in this problem - if it were being ionized completely, you would need to use the additional datum that E1 = -13.6 eV. But since you weren't clued in to this, the electron must have been in well-specified energy levels.

The wavelengths of the absorbed and emitted light correspond to differences in the energies of those energy levels. So if the ground state has a reference energy value of 1/1^2 (drop the minus sign throughout for convenience), the n₁ level has energy 1/n₁^2, and the n₂ level has energy 1/n₂^2 . OK so far?

So just express the two energy changes in terms of ratios: 433.9/94.92 (note that the energies are reciprocal to the wavelengths, so the ratio is inverted relative to the respective energy level differences!) = 4.57122 = (1-1/n₁^2)/(1/n₁^2 - 1/n₂^2 ) . The rest is just plugging away at algebra, form the common denominators, and then drop all the ( n₁^2 * n₂^2 ) common denominators. That gives: n₂^2 *(n₁^2 -1) / (n₂^2 - n₁^2 ) = 4.57122 .

If you throw this expression up onto a 2-D spreadsheet for n1 and n2 and look for generated values, you find that n₁ = 5 and n₂ = 2 results in a very close match (4.57142). But you should definitely do it yourself, to make sure that you can set it up.

I'm not Bohring you too much, am I?

-- Cheers, -- Mr. d.