Exam 2019 question 2b (+2d)

Hello,

I don’t understand how I calculate the radii in question 2b. And I also don’t understand how I should find the optimal ratio between Rb/Ra. I only found for FCC the optimal radius, but that isn’t right.

Could someone help me?

My answer:

Also I think there is a mis calculation in the answer of 2d. Should it not been:

Thanks in advance,
Nadia

We obtain the max filling fraction when the atom at the origin and the atom at 1/4,1/4,/1/4 touch. So you need to calculate the distance between their centers expressed in d.

to your 2nd point: you are correct

Thank you!

@Nadia how did you calculate (or what was your line of reasoning) while constructing the optimal value for r_A, I’m struggeling with that. Thanks in advance!

There are 2 question is 2b.

  1. If Ra = Rb, then I used the shortest distance between center of atom A and B should be equal to there radius. 2Ra = sqrt((d/4)^2+(d/4)^2+(d/4)^2). This is with the basis: A(0,0,0) and B(d/4,d/4,d/4). (Do not use the axis of the picture, they didn’t used it in the answers and this is easier).

  2. If Ra not equal to Rb and find the max filling factor. The A atoms form a FCC lattice, so the highest filling factor for a FCC lattice is Ra = d/(2sqrt(2)) (see notes). Now we know Ra.
    The distance between center of Ra and Rb is sqrt((d/4)^2+(d/4)^2+(d/4)^2). So we can say that sqrt((d/4)^2+(d/4)^2+(d/4)^2) = Ra + Rb. Then we can calculate Rb.

I hope this is good enough, otherwise I can write it down.

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