I am told that, when a heavy charged particle passes through a substance, some (small number of the) collisions occur with electrons of relatively large energy. I am then told that the maximum energy of these secondary electrons is $4\dfrac{v}{M}E$, where $M$ and $E$ are the mass and energy of the incident particle, respectively. The example of the case of a proton with energy $E = 10 \ \text{MeV}$ is given, stating that secondary electrons of different energy may be produced, the maximum energy being equal to $20 \ \text{keV}$.

I'm not quite sure that I understand what this means, and I'm also not sure how the $20 \ \text{keV}$ result was calculated. If we take $4\dfrac{v}{M}E$, and presume that $v$ is the velocity of the incident particle (the proton), then we get $\dfrac{4v}{1.6726 \times10^{-27} \ \text{kg}}\times 10 \ \text{MeV}$, but, without a value for $v$, I don't understand how this is calculated.

I would greatly appreciate it if people who are more familiar with the theory of "collisions of particles" would please take the time to clarify this.

Related: Questions about a heavy charged particle passing through a substance , https://en.wikipedia.org/wiki/Delta_ray

  • $\begingroup$ Consider simple kinematics of a particle of mass $m_{1}$ hitting a particle of mass $m_{2}$ head on. What is the maximum energy transferred? $\endgroup$
    – Jon Custer
    May 22, 2020 at 13:18
  • $\begingroup$ @JonCuster Is it this icecube.wisc.edu/~tmontaruli/801/Exercise4_801.pdf ? I'm a novice, so this is all new to me. $\endgroup$ May 22, 2020 at 13:41
  • $\begingroup$ That reference is complex. I might suggest springer.com/cda/content/document/cda_downloaddocument/… instead - simple non-relativistic kinematics. Equation 2.7 does the trick. You might benefit from a book on ion beam analysis, including the Handbook of Modern Ion Beam Analysis (Ed Tesmer and Nastasi, 1995), or the book that the link points to (Fundamentals of Nanoscale Film Analysis, Terry Alford et al., 2007). $\endgroup$
    – Jon Custer
    May 22, 2020 at 13:54
  • $\begingroup$ @JonCuster Ahh, yes, I found the same equation on page 167 of Classical Mechanics, 5th edition, by Kibble and Berkshire. $\endgroup$ May 22, 2020 at 14:17
  • 1
    $\begingroup$ To get actual stopping powers, you should get SRIM (srim.org) $\endgroup$
    – Jon Custer
    May 22, 2020 at 16:36

1 Answer 1


For the general, non-relativistic case, standard kinematics can be applied for an electron-ion interaction just as for an ion-ion interaction (also known as Rutherford scattering). Taking from Fundamentals of Nanoscale Film Analysis, the maximum energy (equation 2.7), one sees in a head-on collision is

${E_{2} \over E_{0}} = $ ${4M_{1}M_{2} \over (M_{1}+M_{2})^{2}}$

Here $E_{2}$ is the energy of the scattered electron with mass $M_{2}$, $E_{0}$ is the incident energy of the incoming ion with mass $M_{1}$ (yes, slightly confusing but in Rutherford backscattering one wants to know $E_{1}$, the scattered energy of the incident ion).

Since electrons are much lighter than even a proton, this simplifies to

${E_{2} \over E_{0}} = $ ${4M_{2} \over M_{1}}$

In the case of a proton, with a mass roughly 2000 times that of an electron, the maximum energy transferred is $4/2000$ or 0.2%, so a 10MeV proton can give an electron about 20keV.

Now, I'm not sure quite where you got the equation you start with, since the units do not work out at all. But going back to fundamental kinematics gets you to the right answer.

  • $\begingroup$ Thanks again! I really appreciate you taking the time to clarify this for me. $\endgroup$ Jul 30, 2020 at 14:21

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