Why is it usually the K-shell electron that is ejected?

1. Aug 22, 2013

abotiz

Hi,

There is this one thing that bugs me about photoelectric effect, everywhere I read it says [...] usually the K-shell electron is ejected [...], but there is not an explanation to why this is the case.

I know that the photon's energy needs to be a "little bit" larger than the binding energy and for the photon to be absorbed the nucleus/atom needs to receive a recoil (extra) energy. But this could apply for any shell electron, L,M etc.

The only way I can make sense of all this, is if the nucleus needs to be "close" to the electron (or where the interaction takes place) so that the nucleus "can" recieve the recoil energy. But it sounds weird.

I think I heard somewhere that it is usually the K-shell due to the higher binding energy, "does not move around", so the photon has a "clear shot". But the elctron is not a " particle" right? Its like a cloud around the nucleus?

We need an explanation once and for all!

Thank you!

Last edited: Aug 22, 2013
2. Aug 22, 2013

Bill_K

3. Aug 22, 2013

abotiz

Thank you, but no it does not.

The propability is higher for K-shell electrons. That is clearly stated in the book " Radiation physics for Medical Physicist" by Podgorsak. But it does not say why this is the case. It says in the book that 80% of the interactions occur with the K-shell electrons.

4. Aug 22, 2013

Suppose the photon energy is less than required to ionize the K shell (the K shell binding energy) - then no K shell electrons will be ejected (but electrons may be ejected from other shells).

As the photon energy passes above the K shell energy, there will be a sudden onset in emission from the K shell. At this point and beyond, the photon energy is closer to the K shell binding energy than to the binding energies of the other shells.

Whether the K shell emission really dominates at very high photon energies I do not know. There are more electrons in the L shell.

These effects are exploited in a popular technique known as X-ray absorption spectroscopy (XAFS or EXAFS).

5. Aug 22, 2013

fzero

There's a derivation of the amplitude for electron ejection from the ground state of hydrogen here. If you modify the calculation to the $n$th excited state (considering the $\ell=0$ states for simplicity), you find instead that the amplitude is proportional to the integral

$$A_n \propto \int d^3r ~ e^{-i \vec{k}_f \cdot \vec{r} } \left( \frac{r}{a_0} \right)^n e^{-r/a_0} = \frac{2\pi i^n a_0^3}{a_0 k_f} \left. \frac{d^{n+1}}{dy^{n+1}} \frac{2}{1+y^2} \right|_{y= a_0k_f}.$$

Here $\hbar k_f$ is the momentum of the ejected electron. For gamma rays, $k_f \gg a_0^{-1}$ and is approximately equal to the momentum of the initial photon, $\hbar \omega$. So $a_0 k_f \sim a_0 \omega$ is large and we find at leading order that

$$A_n \propto \frac{1}{\omega^{4+n}}.$$

The probability is the square of this, so we can see that the probability of ejection is a decreasing function of $n$ for large photon energies. If we extrapolate the behavior of this simplistic model to a real atom, we expect that the probability to eject a given electron is a decreasing function of its atomic level. For large enough energy, the population differences between the shells will not be able to overcome this suppression.

This approximation should not work very well for photon energies near the typical atomic binding energies. There the intuition about resonance effects that gadong mentions gives a qualitative explanation of the edge effects.

6. Aug 22, 2013

7. Aug 23, 2013

abotiz

Lets recall the facts that we know, and what could be of importance:

1) Binding energy
2) Photon energy
3) Number of electrons in each shell

We know that the K-shell electron has the highest probability. In general, cross sections is determined by the amount of "targets" (atoms or electrons). K-shell can only have 2 electrons, whereas the L-shell can have 8. So the numbers are against us. We conclude that the amount of electrons in a shell is either of no importance or not significant.

Photon energy is a variable that tells us if there is 0% chance (hv<E_bind) for ionization (ejection of electron) or >0% (hv>E_bind). Its not a very important variable if written alone. If in atom X the K-shell electrons are bind with Z eV, and in atom Y the L-shell electrons are bind with Z eV, the highest propability would still be with K-shell electrons.

Now the binding energy is related to the amount of electrons which in turn relates to amount of protons in the nucleus and finally to the binding energy, again (i.e. more protons = higher binding energy). We know that the cross section for photoelectric interaction increases with the atomic number, or you could say increases with amount of protons in nucleus, or increases with the amount of binding energy.

The focus should not be on the atom number or amount of protons because that only explains the difference in cross sections between different elements. If you focus on the binding energy you include both the different elements and the difference between the K,L and M-shells.

So I believe the answer to the topic question lies in the effect or consequence of different binding energy.

8. Aug 23, 2013

fzero

Let's simplify the question that we want to address to: what is the relative probability that a photon with energy $\hbar\omega$ ejects an electron from the K-shell versus ejection from the L-shell. If we let $i$ label the states the electrons in the atoms, then the relative probability of ejection can be roughly expressed as

$$P_i = \rho_i p_i,$$

where $\rho_i$ is the density of the state $i$ in the sample and $p_i$ is the probability amplitude for ejection from the $i$th state. If we are dealing with a single atom with $Z>10$, then roughly

$$\rho_K \sim \frac{2}{Z},~~~\rho_L \sim \frac{8}{Z},$$

so the next part that needs to be addressed is the relative values of $p_K$ and $p_L$.

The photon energy is more important than that because the probability factors $p_i$ depend on the photon energy, as well as on the binding energy. In terms of the quantum mechanical amplitude, these data are encoded in the wavefunctions. The amplitude involves both the wavefunction of the initial-state bound election (whose energy eigenvalue is directly related to the binding energy) and the wavefunction of the ejected electron (which is approximated by a plane wave depending on the final state momentum, which is in-turn directly related to the energy of the initial-state photon by energy conservation).

When we put everything together we find that

$$\frac{P_K}{P_L} \sim \frac{c}{4} \omega,$$

where $c$ is a numerical factor. The precise value of $c$ depends on the numerical factors that I did not keep track of in my computation, but will also depend on all the complications of the many-body system of a real atom, like charge screening, that should really be addressed by using more realistic wavefunctions in the computation. Nevertheless, we can conclude that, whatever the value of $c$, for

$$\omega > 4/c,$$

the probability to eject an electron from the K-shell is larger than the probability to eject it from the L-shell, which agrees with the experimental situation.

The first-order effect of $Z$ on the cross section that you are describing here is related to the probabilities that the photon interacts or does not interact with the atom. An atom with more electrons is "bigger" and therefore has a higher cross section. This is not the question that I am trying to answer, rather I am assuming that the photon does in fact lead to electron ejection.

As I have commented above, the binding energy is an important input, which is accounted for by using the bound-state wavefunction in the calculation of the amplitude for the process. For a real atom, this is a complicated wavefunction, but we expect that the approximation of using hydrogenic wavefunctions should still give some useful qualitative information.

I hope that this clarification makes the analysis a bit clearer.

9. Aug 29, 2013

abotiz

Thank you fzero for you time. I think this is very interesting and at the same time surprised that this answer is not as simple as one would think and as common in text books as one would like.

Iam not saying that your reply made it all clear, although i think I understand some of it.

Just to check with you,
the answer to my question, why its always the K-shell electron, you are saying that the answer lies in the calculations of the wave functions of the electrons, and if one were to compare the wave functions of the electrons in different shells, one would see that the photon have a larger propability to interact with a K-shell electron?

Is there like a "classical-physics" explanation to this, like imagine birds flying around a pole, 2 birds within 1m radius and 6 birds within 3 meter radius, If you shoot at random towards the pole, the propability to shoot the inner two birds is higher due to that they fly only witihin 2m radius so they appear more often at one location. Even if there is more birds flying at 3 meter radius, the area is to large, which lowers the propability. Can you say that?

Thank you!

10. Aug 30, 2013

fzero

The answer is not so simple because the many-electron atom is complicated compared to QM problems that we can solve exactly. It is certainly possible to do much better than my simplistic attempt by putting the problem on a computer.

Yes, in QM the wavefunctions are an essential ingredient in analyzing any given process. The more accurately we know the wavefunctions, the more accurate our result will be. My analysis uses wavefunctions which are only very rough approximations to those for a multi-electron atom, but since it reproduces this particular experimental result, we can have a bit of confidence that it has some validity. What we saw was that, for large enough photon energy, the probability to interact with a K-shell electron was larger than for the other shells.

Yes, I would say that this is a fairly good analogy. It is also important that the result emerges most clearly when the photon energy is large, which is the limit of smaller wavelengths. So we might think that the electron density is very important to have a decent probability of hitting the electron. In your classical example, we could compare between our chances to hit the bird with a small pellet, vs hitting one with a basketball. As the projectile gets larger (like a low energy, long-wavelength photon), we might expect to just hit the first bird we come across, which is likely to be one of the ones that fly at the larger radius. For a very small projectile (like the X-ray), we would need to aim for the cross-sectional area with the higher density close to the pole and, as you say, the small radius birds spend more time there.

11. Aug 31, 2013