Bremsstrahlung emissions of an electron dependent on electron energy?

Question

I'm reading a book entitled Nuclear Electronics by P.M. Nicholson. It's giving me quite a bit of trouble as I go through (an hour or two in, and I'm only on page 3) partly because I don't have a ton of background knowledge on the subject, though I do know a bit (please keep this in mind when answering).

In chapter 1, "An Outline of Detection Methods", section 1.1.3 "Electrons and positrons", the book says

For energies up to 10 MeV, electrons lose their energy to the detection medium mainly by excitation and ionization of the electrons of the medium, as in the case of heavily charged particles.

So far, so good. Then it says

For higher energy electrons the loss of energy as bremsstrahlung becomes increasingly important and the intensity of this varies as $Z^2$, where $Z$ is the atomic number of the medium. Thus, for example, 9 MeV electrons in lead lose as much energy due to bremsstrahlung as due to ionization.

After skimming the Wikipedia article on bremsstrahlung radiation, I have this understanding: generally, electrons decelerate as they come near a nucleus and emit a photon to uphold the law of conservation of energy, losing energy in the process. This in and of itself makes sense.

However, I didn't really find anything in the article (or understand anything, one of the two) that mentioned this was dependent on the particle energy. I mean, the electron slows down because of the electromagnetic field, loses some energy in the form of a photon, and goes on its merry way. I'm not sure how the electron's energy plays into this.

Secondly, I'm not sure I understand the book's example - firstly with how the energy plays into it (especially with what exactly it means with the 10 MeV threshold in the first part of the quote) and secondly what it means by "the intensity of this varies" and how $Z^2$ shows that.

If this is too broad, I can split it up into two questions, but I thought the questions were intertwined enough they could be asked together. Please provide an answer on the more intuitive side - I don't need to do tons of math with this, I just want to understand it. Any help would be much appreciated!

Answer 1

In simple terms the electron can lose energy by

• inelastic collisions with the atomic electrons of the material which will cause excitation or ionisation of the atom and the parameter which characterises this process is called the collisional (or collision or electronic) stopping power.

• inelastic collisions with the nuclei of the material which will cause the electrons to accelerate resulting in the emission of electromagnetic radiation called bremsstrahlung and the parameter which characterises this process is called the radiative stopping power.

There are other processes but they have a smaller effect on electrons and for details of these have a look at @dmckee 's reference although you may find other articles less comprehensive but easier to understand.

The stopping power $\dfrac 1 \rho \dfrac{dT}{dx}$ is the rate of change of kinetic energy of the electron $T$ with distance $x$ divided by the density of the material $\rho$.

This NIST website shows you how these two processes compare for different materials and for a range of electron energies.

From the graphs you will note that the collisional stopping power does not depend very much on the target material.
The collisional stopping power is in fact proportional to $\dfrac {N_{\rm A}Z} {A}$, the electron density, a slowly varying function with changing $Z$.

The critical energy, which is about $10 \rm MeV$ (you quoted $9\, \rm MeV$) for lead, is the energy of the electrons when the collisional stopping power is equal to the radiative stopping power.
The blue dashed line indicates this condition.

The radiative stopping power is proportional to $\dfrac {N_{\rm A} Z^2}{A}$ and this is where your $Z^2$ factor comes from.

The ratio of the stopping powers $\dfrac{\rm radiative}{\rm collisional} \approx \dfrac{TZ}{n}$ where $T$ is the kinetic energy of the electron, $Z$ the atomic number of the material and $n$ a constant which seems to vary between $700$ and $800$ MeV in the literature.
This can be used to estimate to estimate the critical energy.

Using the values of critical density found from the NIST data, Al = $50$ MeV, Cu = $24$ MeV and Pb = $10$ MeV, the values of $n$ turn out to be in rough agreement with the formula, Al = $650$ MeV, Cu = $700$ MeV and Pb = $820$ MeV

I think that @dmckee's comment answer the energy dependence question.

Answer 2

Having it dependent on particle energy turns out to actually make a decent amount of sense. Imagine two cars, one going very fast, and one going at a more sedentary pace. If the first car all of a sudden needs to slow down, this will take more energy than the second car needing to suddenly slow down. Another way to look at it: when playing dodgeball in middle school, you always avoid the throws by the class jock. Why? Because that ball's much faster speed when thrown will lead to a much harder hit (a higher expense of energy) when it suddenly comes to a stop when it hits you as opposed to the balls thrown by normal humans.

The second part of this question ends up splitting into two parts, the first on the $Z^2$ bit and the second on the 10 MeV maximum mentioned in the first quote. When the book says "the intensity of this varies as $Z^2$ [bolding my own]" it really means that the intensity, which we'll call $I$, is a function dependent on $Z$ that can be defined as $I(Z)=kZ^2$ where $k$ is a constant. What $I(Z)$ represents is the part of the energy lost by the electron that is lost due to bremsstrahlung.

Finally, the part about the 10 MeV maximum. There's a point at which energy loss due to bremsstrahlung becomes equally important or more important than energy loss due to ionization. This happens at a certain energy of the electron which varies from medium to medium. The 10 MeV maximum appears to be the maximum for lead.

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Note: this was written based on the path I took to figure it out. It was heavily based on an hbar conversation (be warned - there are significant breaks in the conversation at the beginning, and it's a long conversation) - many thanks to Loong and Emilio Pisanty for their help, as well as to dmckee for his comments and pointers.