# Electrons bombard gaseous hydrogen - photon energies emitted?

## 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 ($\frac{1}{n^{2}_{f}}$ - $\frac{1}{n^{2}_{i}}$)

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

ΔE = -13.6 ($\frac{1}{n^{2}_{f}}$ - 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$\rightarrow$3 = 0.66 ev
4$\rightarrow$2 = 2.55 ev
4$\rightarrow$1 = 12.75 ev

3$\rightarrow$2 = 1.89 ev
3$\rightarrow$1 = 12.09 ev

2$\rightarrow$1 = 10.2 ev

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maajdl
Gold Member
I'm unsure how to word it.
13 eV is enough to reach the n=3 level which has an energy of 12.09 eV above ground state.
13 eV is enough to reach the n=4 level which has an energy of 12.75 eV above ground state.
13 eV is not enough to reach the n=5 level which has an energy of 13.056 eV above ground state.

it asks what photon energies will be emitted so would I have to write them all down?
Yes.
As you saw, there are not so many lines.
You might present them in a level diagram, or in a table, if you like to present things ...

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Thanks maajdl