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When imaging with SEM (Scanning Electron Microscopy) an incident electron ejects an electron from the surface of the sample, which is then detected in order to create an image. In that case I guess that the process is dominantly electromagnetic (right?), so can I write such an equation to describe the process:

$$e^- + X \overset{\gamma*}{\rightarrow} e^- + X^+ + e^-$$

where $\gamma*$ describes a virtual photon, $X$ the target atom and $X^+$ the target atom ionized after the interaction?

My real question is actually about an ion (say $He^+$) hitting a target atom and sputtering a secondary electron. Is it still the electron of the $He^+$ interacting electromagnetically with the target electron, or something else that I cannot think of? And if yes, can I then describe the process as follows:

$$He^++X \overset{\gamma*}{\rightarrow} He^+ + X^+ + e^-$$

? Would there be a chance that the incident ion loses its electron, like:

$$He^+ + X \overset{\gamma*}{\rightarrow} He^{++} + X^+ + 2e^-$$

?

Thanks a lot in advance for your answers.

Pxx
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  • An incident energetic electron or ion entering a solid will interact with electrons and nuclei in the solid through Coulomb forces, transferring energy through scattering processes described by Rutherford more than 100 years ago. Some of those scattering processes will transfer enough energy to an electron in the solid to enable it to escape the solid. No photons needed. – Jon Custer Mar 02 '18 at 19:55
  • When I said virtual photon I meant electromagnetic interaction which is the same as Coulomb interaction but from a particle physics point of view I guess. However I think that the process I describe is inelastic while Rutherford scattering is generally considered elastic, if I am not mistaken? In that case maybe Mott scattering would be more adequate? – Pxx Mar 02 '18 at 19:58
  • Why do you think that transfer of energy from one particle to another is not elastic? Rutherford scattering is indeed elastic. – Jon Custer Mar 02 '18 at 20:00
  • I guess the main reason is because my book about helium ion microscopy says "During inelastic interactions energy is transferred to the electrons in the sample and results in the emission of secondary electrons and electromagnetic radiation." XD – Pxx Mar 02 '18 at 20:03
  • I guess that the guy considers the emitted photon as part of the inelastic process? – Pxx Mar 02 '18 at 20:03
  • Yes, once you have an excited electron in the band structure, it will start thermalizing. Some of that process may be phonon mediated, which you might interpret as inelastic. Further, hitting something with energetic electrons can result in various x-ray producing processes through excitation of core electrons. – Jon Custer Mar 02 '18 at 20:33
  • So would you then suggest that this process is well modeled by Mott scattering? Or do you still think that Rutherford is more relevant? – Pxx Mar 02 '18 at 20:36
  • Perhaps start with https://journals.aps.org/pr/pdf/10.1103/PhysRev.93.981 and see where that leads you. There is not a simple answer. – Jon Custer Mar 02 '18 at 20:40
  • if you are thinking still of imaging this shows the approximations https://warwick.ac.uk/fac/sci/physics/current/postgraduate/regs/mpagswarwick/ex5/techniques/structural/ionscattering/ . – anna v Mar 03 '18 at 06:17

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One has to give boundary conditions,energy and momentum of beam. For low energies the behavior can be well modeled , even though the underlying interaction is quantum mechanical and for accuracy, it will always be described by virtual photon exchanges.

Look at the photoelectric effect, a real photon hits a neutral atom

photel

The black electron line represents the "orbital" of the electron before interacting with the incoming photon. If the photon frequency is higher than the binding energy of the electron, it will ionize the Z nucleus.

In the case of surfaces there will be a collective diagram between conduction band electrons, the ones with the weakest binding energy to the lattice, which will be kicked out if the conduction band binding energy is supplied by the photon.

The diagram can be extended to a virtual photon exchange between the He+ and the last electron of the nucleus, or the conduction band of a surface.

This is simple for single nuclei. For nuclei in a lattice the specific interactions will depend on the way electrons are distributed , in bands, the conduction band being the less bound state. Again a specific model has to be invoked, but the classical models in the first link are adequate to describe the statistical behavior of the scattering.

anna v
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