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## Homework Statement

A beam of 13.0 eV electrons is used to bombard gaseous hydrogen in ground state.

What photon energies will be emitted?

## Homework Equations

ΔE = -13.6 ([itex]\frac{1}{n^{2}_{f}}[/itex] - [itex]\frac{1}{n^{2}_{i}}[/itex])

However hydrogen is in the ground state therefore n_i= 1:

ΔE = -13.6 ([itex]\frac{1}{n^{2}_{f}}[/itex] - 1)

## The Attempt at a Solution

Rearranging the above equation I got n_f = 4.76.

Am I correct in saying "the photon can be emitted from a maximum energy state of n=4"?

I'm unsure how to word it.

WRT the initial question, it asks what photon

*energies*will be emitted so would I have to write them all down?

ie

4[itex]\rightarrow[/itex]3 = 0.66 ev

4[itex]\rightarrow[/itex]2 = 2.55 ev

4[itex]\rightarrow[/itex]1 = 12.75 ev

3[itex]\rightarrow[/itex]2 = 1.89 ev

3[itex]\rightarrow[/itex]1 = 12.09 ev

2[itex]\rightarrow[/itex]1 = 10.2 ev