# Emission in hydrogen atom - difference between electrons and photons

A gas composed of hydrogen atoms in the ground state is bombarded with electrons of energy 12.5eV.

a) What wavelengths emitted can we expect to observe?
b) If we replaced the incident electrons with incident photons of the same energy, what would happen?

I know the answers to both parts, I just don't understand part b. In part a), the electrons in the ground state will have enough energy to go up to the 3rd energy level: (-13.6eV) + 12.5eV = -1.1eV. When the electron de-excites it will have 3 possible transitions: 3->1 3->2 and 2->1. Then you just calculate the wavelength of these emitted photons.

For part b, the answer claims that there will be no photon emissions because the photon energy must be exactly equal to the difference between the two levels and it won't be absorbed.

My question is, why is it that incident electrons of 12.5eV can give energy to a -13.6eV ground state electron, but photons cannot?

Hi :)

As you said, 12.5 eV is enough to reach up to n=3. But 12.5 eV is _not_ the position of a state.
The difference between photons and electrons is that electrons can give to the atom _part_ of their 12.5 eV energy. A photon has to give 12.5 eV or nothing.

In this case, a 12.5 eV electron can
– give 10.2 eV to excite an atom in n=2 and continue flying through the gas with its remaining 2.3 eV of kinetic energy
– give 12.08 eV to excite an atom in n=3 and continue flying with its 0.42 eV of kinetic energy.

A photon cannot give its 12.5 eV energy because there's no state at this energy.

:)