Answer

Easy one to start with, we saw in the video that we can define the information depth as 3x the IMFP. So, in this case, our information depth is 7.26 nm!

Answer

A higher IMFP suggests we can see deeper into our material, we learned that IMFP may be affected by many factors including the atomic mass, so it is no surprise to see that TiO2, with it’s high mass, results in the most restricted path length of travelling photoelectrons. Higher path lengths result in a higher signal from substrate layers (in this case, Ag). This means that, in increasing order of signal intensity, we will see Ag/TiO2, Ag/unknown, Ag/SiO2 and Ag/graphite.

Answer

Again here, we’re going to work to our approximation that information depth = 3λ. Clearly the thicker the silica overlayer, the more photoelectrons we will lose to scattering processes, so let’s go through 1 by 1.

2nm and 6nm: The smallest ID will be that from Mg kα, which ≈ 8.01 nm, so clearly we can use any source to observe the substrate for these samples.

9nm: This is now thicker than we should be able to observe using Mg kα radiation, so we need to try others. The ID from Al kα is ≈ 9.27nm, so we should just be able to observe it, though it will likely be a weak signal.

13nm: This is clearly above the capabilities of Al kα radiation, so now we will move to Ag Lα (ID = 16.8 nm) to observe the substrate.

30nm: Finally we look at our thickest film. Cr kα radiation results in an ID ≈ 28.02, so we can see that even this will not be enough to see the substrate in this system. We must use the Ga kα source, with a whopping 44.49 nm ID!