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Take a particle on BCC lattice and from that we have to find number of particles at a distance of $\displaystyle \sqrt{11} \cdot \frac{a}{2}$ (where a is side length of cube). For this, the answer that I saw somewhere gave 24 particles. However, I am not getting as many as that. Could someone please explain how 24 is coming?

Karsten
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Anne
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  • $$11 = 3^2 + 1^2 + 1^2$$ – Karsten Aug 29 '23 at 19:06
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    (3,1,1), (3,1,-1), (3,-1,1), (3,-1,-1), (-3,1,1), (-3,1,-1), (-3,-1,1), (-3,-1,-1) is 8, and permuting the 3 from the x to the y and z axes is 3, and 8x3 = 24. Not unreasonable. – Karsten Aug 29 '23 at 19:07
  • Is there a practical application? – Karsten Aug 29 '23 at 19:10
  • @Karsten But, wouldn't some of them get repeated? – Anne Aug 29 '23 at 19:48
  • @Karsten I don't think there is, but it came up in a discussion, so I was curious. – Anne Aug 29 '23 at 19:49
  • @Karsten However, could you explain what you did there, since the coordinates you mentioned correspond to distance root(11) not root(11)/2, or perhaps I am missing something? – Anne Aug 29 '23 at 19:53

1 Answers1

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To get to a distance of $\displaystyle \sqrt{11} \cdot \frac{a}{2}$, you have to go half a step in one direction, half a step in another direction, and one and a half steps in a third. In the scheme below, I am using a/2 as the yard stick to have to write less:

(3,1,1), (3,1,-1), (3,-1,1), (3,-1,-1), (-3,1,1), (-3,1,-1), (-3,-1,1), (-3,-1,-1)

(1,3,1), (1,3,-1), (-1,3,1), (-1,3,-1), (1,-3,1), (1,-3,-1), (-1,-3,1), (-1,-3,-1)

(1,1,3), (1,-1,3), (-1,1,3), (-1,-1,3), (1,1,-3), (1,-1,-3), (-1,1,-3), (-1,-1,-3)

You can check that each of these is unique. With a little bit of fiddling, you can also check that you end up on a lattice point (there is a 1/2, 1/2, 1/2 translation, and you can add unit cell translations on top of that).

Karsten
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