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So I'm a bit confused about this question

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This question asks for the miller indices for the "families of planes". Is there a single set of Miller indices for each cubic unit cell which I can use to present all of the planes for that unit cell?

For a) I have: $(1, 0, 0)$, $(-1, 0, 0)$

For b) I have: $(0, 1, 0)$, $(0, -1, 0)$, $(0, 3, 0)$, $(0, -3, 0)$

For c) I have: $(3, 2, 0)$, $(-3, -2, 0)$ and I have no idea how to find the others for this one.

Also I noticed that the planes for each cubic unit cell has the same direction. I know that enclosing miller indices in square brackets represents a direction but isn't this just a vector, not a representation of a family of planes?

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  • $\begingroup$ for (c) isnt it supposed to be {230} instead of {320}? $\endgroup$ Sep 5 at 13:16

2 Answers 2

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Assume a 3D lattice and denote its reciprocal lattice basis vectors as $\vec{b}_{1,2,3}$. The symbol $\left(h,k,l\right)$ stands for all the planes orthogonal to the vector $h\vec{b}_{1}+k\vec{b}_{2}+l\vec{b}_{3}$ (also written $\left[h,k,l\right]$ as you stated), so in fact there is no difference between $\left(1,0,0\right)$ and $\left(-1,0,0\right)$ for instance.

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  • $\begingroup$ Ohh thank you for clarifying that for me. So it would be sufficient for me to write (1,0,0), (0,1,0,) and (3, 2, 0) for a), b) and c), respectively? What confused me is the number of planes, I thought I had to express them individually $\endgroup$
    – Coconut
    Nov 30, 2017 at 14:30
  • $\begingroup$ You are correct for a) and c). For b) one usually writes $\left(0,3,0\right)$ because you represent planes that cross the unit cell $x$-axis at $\frac{1}{3}$, $\frac{2}{3}$ and etc.. $\endgroup$
    – eranreches
    Nov 30, 2017 at 14:34
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You almost answered the question yourself. One uses curly parentheses for family of planes.

  • For (a) you have {100}
  • For (b) you have {030}
  • For (c) you have {320}

Realise that there is a redundancy when it comes to defining family of planes. One can use more than one notation.

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