Is elastic scattering of electrons by a solid possible?

[Philip Koeck]

In electron microscopy of thin solid specimens elastic scattering is treated as the main process responsible for formation of (phase contrast) images and diffraction patterns.
However, if an electron changes direction it should lose energy by producing a breaking radiation photon.
How can it be elastically scattered then?

 

[hutchphd]

The short answer is Quantum Mechanics. Check out this article on Low Energy Electron Diffraction

 

[gentzen]

However, if an electron changes direction it should lose energy by producing a breaking radiation photon.
How can it be elastically scattered then?

This may only be a question of terminology. (Not talking about bremsstrahlung vs "breaking radiation" vs "braking radiation" here.) The interactions of an electron with the "shielded" ions of the crystal are called elastic scattering, because the ions are so much heavier than the electron, such that the electron nearly doesn't change its energy in those interactions, especially compared to its change in direction.

As long as the change in energy of the electron is insignificant for the resulting diffraction pattern (or if the portion of electrons with a significant change in energy is insignificant), talk of elastic scattering seems justified.

You may worry that producing a bremsstrahlung photon could act like a measurement and "localize" the electron. But the created photons are quite delocalized, therefore their creation doesn't localize the electron either.

 

[hutchphd]

I do not think it is terminology. Classical mechanics cannot begin to really to explain the behavior of electrons in crystals (the fact that the nearly free electron gas models works at all is because of quantum effects) The fact that there is a large elastic component scattered from the surface from the surface is inherently a quantum phenomenon.
It is also true that electrons in "orbit" around atoms do not spiral into the nucleus.

 

[Philip Koeck]

I do not think it is terminology. Classical mechanics cannot begin to really to explain the behavior of electrons in crystals (the fact that the nearly free electron gas models works at all is because of quantum effects) The fact that there is a large elastic component scattered from the surface from the surface is inherently a quantum phenomenon.
It is also true that electrons in "orbit" around atoms do not spiral into the nucleus.

So completely elastic scattering really contradicts classical electrodynamics just like stable orbitals do. In both cases one would classically expect energy loss due to braking radiation (got the spelling right this time).
In a way elastic scattering is a limiting case of a bound electron, isn't it?

 

[Lord Jestocost]

In electron microscopy of thin solid specimens elastic scattering is treated as the main process responsible for formation of (phase contrast) images and diffraction patterns.
However, if an electron changes direction it should lose energy by producing a breaking radiation photon.
How can it be elastically scattered then?

With regard to elastic scattering, I only remember a remark by Peter Sigmund:

“In elastic scattering the identity of the collision partners and their total kinetic energy is conserved during interaction.”

 

[gentzen]

So completely elastic scattering really contradicts classical electrodynamics just like stable orbitals do.

But this "elastic scattering" is something different than what is meant by that word in electron microscopy. A scatter event with a free electron is an inelastic event in that context, even if the event would be completely elastic in the literal sense. The point is that the electron "itself" loses a significant amount of energy.

In a way elastic scattering is a limiting case of a bound electron, isn't it?

No. To see why, look at the classical two body problem. The bound case corresponds to an ellipse here, whose limiting case is a parabola. But the scattering case corresponds to a hyperbola.

 

[Lord Jestocost]

With regard to elastic scattering, I only remember a remark by Peter Sigmund:

“In elastic scattering the identity of the collision partners and their total kinetic energy is conserved during interaction.”

To my mind, this definition is prerequisite in order to cover electron diffraction at atom lattices by means of quantum mechanics where now the "lattice" is the collision partner.

 

[Philip Koeck]

But this "elastic scattering" is something different than what is meant by that word in electron microscopy. A scatter event with a free electron is an inelastic event in that context, even if the event would be completely elastic in the literal sense. The point is that the electron "itself" loses a significant amount of energy.

If the electrons in a transmission electron microscope (100 to 300 keV) are scattered by the nuclei of a thin specimen then the transfer of energy to the nuclei should be essentially zero, is that right?
In that case elastic scattering actually means that the TEM-electron loses no energy, or?

 

[gentzen]

If the electrons in a transmission electron microscope (100 to 300 keV) are scattered by the nuclei of a thin specimen

The shielding effect of the electrons is actually included. So it is less scattering by the nuclei, but more scattering by the potential landscape resulting from the mean distribution of negative charge from electrons and positive charge from nuclei.

then the transfer of energy to the nuclei should be essentially zero, is that right?

Yes, that is right.

In that case elastic scattering actually means that the TEM-electron loses no energy, or?

Yes. The energy loss events are modeled separately. In the simplest case, one uses a continuous slowdown model based on the mean stopping power (which can be measured, at least in principle).

More sophisticated models include stochastic events where the electron loses a significant amount of energy in a single event, especially the creation of (fast) secondary electrons. (Not sure whether this is important for TEM. It certainly is important for SEM, where the secondary electrons often provide the best available signal.)

 

[Philip Koeck]

The shielding effect of the electrons is actually included. So it is less scattering by the nuclei, but more scattering by the potential landscape resulting from the mean distribution of negative charge from electrons and positive charge from nuclei.

I guess it's treated as a stationary electrostatic potential landscape to a very good approximation.

Yes. The energy loss events are modeled separately. In the simplest case, one uses a continuous slowdown model based on the mean stopping power (which can be measured, at least in principle).

More sophisticated models include stochastic events where the electron loses a significant amount of energy in a single event, especially the creation of (fast) secondary electrons. (Not sure whether this is important for TEM. It certainly is important for SEM, where the secondary electrons often provide the best available signal.)

At least in TEM of metals energy losses due to excitation of plasmons are the most important.
Electron energy loss spectra (EELS) always start with a broad and high plasmon peak (more like a bump).
But now we're talking about inelastic scattering.

 

[gentzen]

I guess it's treated as a stationary electrostatic potential landscape to a very good approximation.

Indeed, the computations for elastic scattering are "always" done (as far as I know) with a stationary electrostatic potential landscape.

I didn't notice before that the use of a potential landscape has the conservation of energy (of the electron itself) already built-in, in a certain sense. Even so one could still "directly" include energy transfer in various ways if desired (for example by transforming into the center of mass reference frame before the interaction and back afterwards, or by including damping factors in the propagation of the electron), the more typical approach would be to introduce scattering with quasiparticles like phonons to account for the energy transfer.

Typical computer simulations will treat different scattering processes like phonon scattering, "elastic scattering", plasmon scattering, ... as independent. But what I didn't notice before, and why I decided to write this, is that if such scattering processes are used to compensate for not "directly" including energy transfers that could have been included "rigorously" into "elastic scattering" (like using the center of mass reference frame), then the randomness of those quasiparticle processes would have to be correlated to the randomness of the "elastic scattering" process to obtain "similarly rigorous" results. So it is a nice example of how nonlocal randomness would sneak into a quasiparticle picture, if used "too liberally".