Hi!
In the exam of 2019 I do not quite understand in 2a why the primitive unit cell vectors do not include the B atom. Is it because the B atom has a different environment and is thus a different lattice point? But why then should you discard the B atom?
Thanks in advance for the clarification 
Cheers
So first of all, let’s see where the Atoms are located.
The A atoms are located at the corners and at the face centers, this corresponds to a FCC lattice of atoms A, which yields the primitive cells:
(a/2, a/2, 0)
(0, a/2, a/2)
(a/2, a/2, 0)
With these 3 vectors we can map out EVERY A atom of the FCC lattice starting from point (0,0,0).
However, we can’t map out the B atoms.
I honestly can’t make out where the B atoms are, but the idea is, that you write down the position of all the B atoms.
That is your basis, now you need to try to reduce it as much as possible, you need to do this by asking yourself the question “which B-atoms can I reach from a linear combination of primitive lattice vectors + another B-atom point” → Eliminate those B-atoms from your basis as they are redundant and shouldn’t be part of your basis.
I believe this leads to the answer that you only need 1 B-atom as a basis, and that you can reach the other B-atoms from that basis, which is why the answer is the way it is.
first of all thank you for your answer.
Do you mean that from every B atom, you can reach another B atom by using the primitive lattice vectors you used for the A atoms?
That should indeed be the case, if you look at the homework of lecture 9 or 10, there is an exercise with graphene, look at the basis there, they omitted an atom from the basis there as well because it was reachable from another atom in the basis + a x^, (ax^ was a primitive lattice vector).
Alright I will look into that! Thanks 
Hi,
I’ll add a bit on this answering directly to the question.
Your suggestion is true, you cannot have a lattice that includes the B atoms and A atoms and fulfills the properties of a lattice. You can also think that if all types of atoms are included in the lattice, what tells you which lattice point is of which kind?
That’s where basis come into play. First we choose a kind of atom (A in this case) and we use it as the “background”. Then with the basis we can say, atoms of type A will sit in positions (0,0,0) from the lattice points and atoms of type B in positions (smt, smt, smt) from the lattice points.
What you choose to make your lattice is up to you, but there are usually some options simpler than others.
Hope this clarifies a bit more.
Hi! Thank you for your answer. Just to make sure, why can’t I have a lattice that includes both atoms? And what if the B atoms are also A atoms, do you then have the same answer?
Two reasons:
Because then you would have a different environment. Picture the NaCl of minitest 3. Na sees Cl the closest. Cl sees Nas the closest. The positioning is the same but the atoms are not.
If we accepted that we could have different atoms in the lattice. Then what tells you what kind of atom goes into each lattice point? Ot would only get confusing.
And yes if atoms A and B are the same then you have another solid and you can include both of them in the same lattice.
With the lattice vectors proposed in the solution and basis A(0, 0, 0), I can’ t see how B(1/4, 1/4, 1/4) reaches a B atom. It seems to me that (1/4, 1/4, 1/4) just points to empty space when you start at (0, 0, 0).
This is dependend on what you choose your (0,0,0) point to be.
But I would agree that if you take the crossing of the axis as this point it would not.
If you however take the bottom left point to be (0,0,0) it would be correct. For this to work, you do however also have to ignore the arrows on the axis.
There’s another discussion on this in the forum. You can check it out. Indeed the origin is taken different than in the picture.