# I Why are core electrons ejected by gammas

1. May 19, 2017

### pierce15

I read here that the reason why gammas primarily interact with core electrons in an atom is that "As the difference between the photon energy and the energy required for the process increases, the cross section of the process decrease". Is that correct? If so is there any intuition for why that would be the case? My classical intuition says that the gamma should interact mostly with electrons whose wavefunction gives the largest average distance from the nucleus.

2. May 20, 2017

### Staff: Mentor

Why? The emission is produced in the nucleus, and the innermost electrons (especially s electrons) have the largest wavefunction amplitude there, and they interact the most with the nucleus.

3. May 20, 2017

### pierce15

Correct me if I'm wrong, but e.g. in Compton scattering the gamma interacts directly with an electron. Is Compton scattering not the main process resulting in ejected core electrons?

Last edited: May 20, 2017
4. May 23, 2017

### pierce15

Just thought about it a little more, I think a more accurate classical interpretation is that the photon is more likely to interact in regions dense with electrons. However that doesn't explain why K electrons are primarily ejected instead of e.g. L...

5. May 23, 2017

### Staff: Mentor

Strange. I wrote a reply on Sunday, but somehow that didn't get saved.
At high energies, all electrons have similar probabilities to get hit, and I found a nice reference for that but I forgot where it was.
At low energies, phase space alters the probability, but you also have the photoionization process, where the nucleus absorbs the excess momentum and no photon is emitted. And that works better for electrons close to the nucleus.

6. May 23, 2017

### ZapperZ

Staff Emeritus
There is an important idea here that has not been established yet, at least, not to my satisfaction.

I'd like to see either calculation or experiment that shows the photoemission or photoionization cross-section as a function of photon frequency.

You see, intensity of photoelectrons emitted from various band or levels does not JUST depend on the difference between the photon energy and the occupied electronic energy state. If this were the case, then the photoelectrons from the Fermi energy, or the outer-most shell, will dominate! This is not what we see (see, for example, the spectra from XPS experiments on metals). So it can't be just due to "... the difference between the photon energy and the energy required for the process.... ".

Therefore, until the premise that is in the OP is established FIRST, we are discussing why the unicorn's has pink horn.

Zz.

7. May 24, 2017

### pierce15

I am having trouble seeing how having such data would clear this up. Can you explain?

Let me try to explain more clearly what I understand about this problem. Gamma rays interact with an atom according to some cross section that varies with the energy of the gamma. This may result in ionizing the atom, which I believe occurs for all 3 of the main interactions (photoelectric effect, Compton scattering, pair production). Hence the ion is typically left in an excited state, and some excited states may be more likely than others. I was of the understanding that "core electrons" are typically the ones that are ejected (though I am now unsure whether this is true), or more formally the ion is most often found in the state given by orbital theory as 1 electron missing in the state on lowest principal quantum number and angular momentum quantum number (1s). However, Mfb says that there is a reference saying the ion product states are mostly uniformly distributed.

Feel free to correct me where I am misunderstanding.

8. May 24, 2017

### ZapperZ

Staff Emeritus
Here's the problem. You are saying that A causes X. But reread the first post that you made. It isn't just that. It is also stating that B also causes X, but with a higher probability than A. So it is more of a comparison with something having shorter or longer wavelength.

You have two opposing views here. The one you quoted claim that there is a monotonic DECREASE in the x-section with shortening wavelength, while you think that there should be a monotonic INCREASE in the x-section.

My claim here is that, it is not that easy, and that this trend may not be monotonic, i.e. it may increase over a range, and then it may turn and decrease beyond that range. After all, I've seen quantum efficiency in materials do the exact same thing.

So before we start to discuss why the photoionization cross-section (we are talking about atoms here and not solids, aren't we?) behaves in such a way, don't you think it is crucial that we get the actual picture of the photoionization cross-section to begin with, and not just simply GUESS at it? It is why I'm asking for it.

Zz.

9. May 24, 2017

### pierce15

Either I am misunderstanding you or that's not what I meant. I am talking about everything here with a fixed photon wavelength. For a fixed wavelength, what state is the ion product most likely to be in? The original quote which is rather imprecisely worded says that for a fixed wavelength, higher energy electrons are more likely to be ejected (assuming you can treat them as independent particles), i.e. K electrons. What's increasing is "the energy of the process" (i.e. the difference between the photon energy and the ionization energy of that electron with the photon energy fixed). The photoionization cross section may not be monotonic as a function of energy of the photon, but the original claim is that the photoionization cross section is always higher for K electrons than for other electrons.

10. May 24, 2017

### hilbert2

Let's say we model an electron orbiting a nucleus with a very simple, crude model of a "box potential", where the electron is confined inside a cube with side length that is similar to the expectation value of the electron-nucleus distance. If you have a small cube, with side length similar to the orbit radius of inner electrons of a heavy atom, the spacings between energy eigenvalues are larger than in the case where the side lengths are similar to outer orbital radii of a heavy atom. As a gamma ray photon has high energy, it's more likely to interact with the more confined electron.

Actually this can be seen with simple dimensional analysis, as in natural units energy has dimensions of reciprocal length and therefore you can expect large energy photons to be better absorbed by something confined in small space than by something with lots of room to move in.

11. May 24, 2017

### pierce15

This is exactly my question: why should a photon interact more with something that is more confined? Do you have an explanation more rigorous than dimensional analysis as to why this should be the case?

12. May 24, 2017

### hilbert2

Because the energy levels, and also the energy differences between levels, of an electron in a cubic box are proportional to $\frac{\hbar^2 \pi^2}{8mL^2}$, where $L$ is the length of the sides. Therefore decreasing the side length by 1/2 will increase the energy spacings by 4-fold. A photon is most likely to cause a transition where the energy difference is similar to the photon energy. A simplified classical analogue to explain this is that when an electron is bouncing back and forth in a small box, the frequency of its motion is higher than in a large box and therefore its more likely to interact with a photon of high frequency/energy (which is what gamma photons are).

13. May 24, 2017

### Staff: Mentor

We are talking about photon energies sufficient to bring the electron to the continuum. How would the density of bound energy levels in the atom matter?

14. May 24, 2017

### pierce15

This is what I am trying to figure out.

Are you suggesting this is a resonance effect, i.e. the probability increases when the "frequency" of the electron is similar to the frequency of the gamma?

I have an idea for how this could be quantified in the case of low-energy gammas, i.e. ones where the Compton scattering and pair production are dominated by the photoelectric effect. You can use Fermi's golden rule to get an estimate of the photoelectric cross section of atomic hydrogen by applying an EM plane wave perturbation to the usual Hamiltonian, where the initial state of the atom is $| 100 \rangle$ and using a free electron as an approximation of the final state. Perhaps the same method can be applied to a larger atom by assuming that the initial state is the ground state and the final state is the direct product of a nontrivial ionized state with a free electron.

15. May 25, 2017

### hilbert2

But the ionization energy is even larger than the spacing between bound state energies, requiring an even higher energy photon. I know this is a very heuristic argument, as the individual electrons don't of course have their own wavefunctions in a many-particle system.

16. May 25, 2017

### hilbert2

Yes, this can be demonstrated with a charged particle in harmonic oscillator potential, with hamiltonian $H = \frac{\hat{p}^2}{2m}+\frac{1}{2}k\hat{x}^2$ and by adding a time-dependent perturbation $\hat{V}'(t)=C\hat{x}\sin{\omega t}$ that models the time-varying electric field of EM radiation. With time-dependent perturbation theory you can show that at least in the linear regime (relatively small C), the ability of the perturbation to excite the oscillator to higher energy levels is greatest when the frequency $\sqrt{k/m}$ is in resonance with the field frequency $\omega$.