Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.03 nm. It then gives off a photon having a wavelength of 93 nm. What is the final state of the hydrogen atom?

n_f= ?

Answer 1

You don't express the wavelengths to the same precision, but they are close enough to indicate that the electron is back in the ground state.

A wavelength of 93.06 nm doesn't actually correspond to any changes in energy levels.

1 --> 6 == 93.845 nm
1 --> 7 == 93.139 nm
1 --> 8 == 92.687 nm

Perhaps, it's back to the drawing-board. Actually, this was a very poor question that you were given. The numbers are not given with sufficient precision to give an answer.

If I had been constructing such a question I would have asked something like this...
A ground state electron absorbs a photon with a wavelength of 97.32 nm. It emits a photon with a wavelength of 486.6 nm. In what energy level is the electron?

And when you do the math you will find that the electron moved from n=1 to n=4 as it absorbed UV light with a wavelength of 97.32 nm. It then went from n=4 to n=2 emitting a photon in the visible spectrum with a wavelength of 486.6 nm. This makes the second energy level the final state of the electron.

In your original question, the energy levels around 93 nm are simply too close for the precision indicated in the question. In other words, you need more decimal places. And you need a better original wavelength, one that actually matches a transition energy.