The excellent book on the Introduction to FIB by Gianuzzi et. al. states the following:
"FIBs are most often used to create features of high aspect ratio (i.e., deep narrow trenches). Sputtered material and backsputtered ions may therefore deposit on surfaces that are in close proximity to the active milling site (e.g., the sidewalls of a deep narrow trench)."
However, backsputtered ions are ions that exit out of the material. I presume these ions exit from the back of the sample. Thus, how could these ions possibly contribute to redeposition of the milled structure?
I tried to visually represent my question with an image, which is attached.
Backsputtered does not mean exit from the bottom of the sample. It means they go backwards. Backsputtered ions refer to the sputtering ions that are coming from the incident beam, since you have two systems: the milling beam and the material.
Here you can find slides that will help with visualizing the situation.
https://nanohub.org/resources/24597/download/2016.03.11-ECE695Q-L24.pdf
If you mill solid bulk material by FIB then lots of gallium ions will be back-scattered as as well as many atoms from the target material will be sputtered out, see eg. the SRIM/TRIM code by Ziegler. In an ideal world these ions and atoms would all be pumped away by the pumps of the vacuum system but in reality only few will, and the rest will all get stuck on any cold surface near-by, most likely the surface of your sample.
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