# ionising atom(s) with gamma rays

if a gamma ray hits an electron and transfers energy, does it hit that electron (ionising the atom), transfer all its energy and stop or does it pass through multiple electrons, transferring a portion of its energy each time?

i.e.) if a gamma ray has enough energy to ionise multiple atoms, will it, or will it just give all its energy to the one electron?

also if is has the energy to ionise multiple atoms, about how many electrons can it remove (for a range of ray energies)? (i know it is a bit general, but lets say for the first ionisation of potassium)

also, why is the dose equivalent so low? is it because it can only ionise 1 atom by +1, or because its ionizations (and hence energy) are dispersed throughout many atoms (making the meaning less energy transferred per ionisation)?

It's tempting to think of photoionisation as the photon coming in like a billiard ball and knocking out an electron. However this is a very misleading representation of the process.

A gamma ray is poorly modeled as a photon or photon(s) because the energy in it is delocalised. If you wanted to use a photon description you'd have to treat the ray as a superposition of photons in different positions. No doubt it can be done, but it's far easier and physically more appropriate to treat the gamma radiation as a wave.

Also, we think of photoionisation as ejection of an electron from a specific atomic orbital. However the electron correlations in atoms mix up the orbitals so strictly speaking an atom that contains more that one electron does not have distinct atomic orbitals. Instead we have a wavefunction that describes all the electrons and isn't separable.

So before the interaction we have a wavefunction describing the gamma radiation, and a wavefunction describing the atom. As the two interact they become entangled and we can not longer describe them as separate objects. Instead we have some bigger wavefunction describing the combined system. After the interaction we have some final state, e.g. an ion and free electron, that again are described by wavefunctions. The probability of any particular final state is calculated using Fermi's golden rule.

This may seem an awful lot of waffle, but the point is that the gamma ray doesn't interact with a single electron but with the whole system. So it's quite reasonable to expect final states that include a doubly charged ion and two free electrons. And indeed this process does occur and is called double ionisation.

However the energy exchange between the gamma ray and the perturbed system always occurs in units of $h\nu$ i.e. an integral number of photons. The obvious case is simple ionisation where a single free electron is generated with kinetic energy equal to $h\nu$ minus the binding energy. However it's possible to have two photon ionisation where the KE minus binding energy is equal to $2h\nu$.

• is that to say that in an atom, electrons are just a wave of probability and lose their individual identity? – ziggy May 19 '15 at 11:25
• @ziggy: An atom always contains a fixed number of electrons, e.g. a carbon atom always contains 6 electrons, however you cannot assign a specific energy to an electron. If you use the atomic orbital approximation then any particular electron is in a superposition of being in the 1s, 2s and 2p orbitals. In principle all electrons are identical i.e. having identical probabilities for which orbital they are in. I'm not sure that the phrase lose their individual identity is really appropriate - I'd have to sit down and think about it. – John Rennie May 19 '15 at 11:36
• ok thanks, also, is it possible for a gamma ray to ionise two different atoms, passing on a portion of its energy to each atom? – ziggy May 19 '15 at 11:40
• @ziggy: a gamma ray can ionise a molecule that contains two atoms, because the two atoms are interacting and described by a single wavefunction. However if you have two isolated and non-interacting atoms, e.g. two atoms widely separated in space, then a gamma ray cannot ionise both in a single exchange of $h\nu$ of energy. It cannot give $\tfrac{1}{2}h\nu$ to one atom then later $\tfrac{1}{2}h\nu$ to an unconnected atom. – John Rennie May 19 '15 at 11:44
• @JohnRennie - Compton scattering is another path allowing ionization without consuming all of the photon's initial energy. And the resulting photon could go on to Compton scatter yet again - I never really calculated out the cross sections for that happening. – Jon Custer May 19 '15 at 13:32