Universal curve of inelastic mean free path - why large at low energy? I have seen the "universal" curve of inelastic mean free path (MFP) of electrons in many places (both experimental data, and just sketches of the curve) e.g. 

https://en.wikipedia.org/wiki/Inelastic_mean_free_path


*

*Why does the MFP rise at low energy? I think this is because there are no scattering mechanisms at low energy - but don't have any good details or resources to back up that claim. 

*If the MFP is very large at low energy, then why do TEM's require large energies for imaging? Perhaps it is the technical trouble of being able to detect  ~1-10 eV electrons above the noise level (?)

 A: A bit late to answer your question...

*

*Yes, well it is more that there are more phenomena at the minimum of the curve to inelastically scatter the electrons, and the amount of phenomena decreases at very low energies. See also here for more information, and here for a paper in the 80s.


*TEMs operate at high electron energies mainly because of the obtainable resolution. Resolution is mainly limited by spherical aberration of the lenses, and this gets better at higher energies. The lower the wavelength, the more energy per electron and the higher the obtainable resolution.
It is possible to image with low energy electrons, however they need to be accelerated back to sufficient energy for the electron optics to be able to image them properly. In our LEEM the electron optics are operated at 15 kV, but just before the sample, the electrons are deaccelerated to 0-100 Volts and reflect back elastically. The electrons are accelerated back to the electron optics and then an image of the surface is made.
A: 

*

*Why does the MFP rise at low energy? I think this is because there are no scattering mechanisms at low energy - but don't have any good details or resources to back up that claim.


First, it is important to realize that the plot is illustrating the inelastic mean free path for electronic scattering. It doesn't take other types of scattering into account, such as the scattering of electrons from phonons.
Because we are only considering electronic scattering, the most important energy scale in a solid is characterized roughly by the "plasma frequency," which for many solids is on the order of tens of eV. Of course, in a real solid there is no single "plasma frequency," but nevertheless the electronic absorption does usually have a strong peak near some tens of eV.
Once the energy of the scattering electron is below the plasma frequency it is hard to make electronic excitations. This is why the mean free path goes up at low energies.



*If the MFP is very large at low energy, then why do TEM's require large energies for imaging? Perhaps it is the technical trouble of being able to detect  ~1-10 eV electrons above the noise level (?)


Electrons are charged and as such they scatter a lot when you shoot them into a solid material. Because they scatter so much they are hard to shoot through a thick sample. They are also pretty hard to even shoot through a thin sample. If you want to use electrons as probe particle then you hope that they don't alter the sample too much and you hope that they pass though in a way that is amenable to theoretical description. In order to get enough electrons through the sample to collect spectra in a TEM you will usually pump up the beam energy. At typical TEM beam energies you can even possibly use the dipole approximation for scattering and try and understand the spectra in a similar way as inelastic x-ray spectra (photons don't interact as much as electrons and in this sense are "better" probe particles).
Even if there is no (or very little) inelastic scattering, an electron is still going to scatter. Unless we shoot the electrons in at high energy it will be hard to know where the scattered electrons are. Basically we want to try and do transmission spectra, but this means we have to collect the transmitted electrons. If we shoot them in at really low energy we will just never see them again. Either because they didn't make it through the sample or because they elastically scattered away.
