1
$\begingroup$

The miller indices of the middle plane is $(2,0,0)$ and it's easy to see why it's so but I've read that one should reduce the Miller indices down to lower integers so the miller Indices of this middle plane should be $(1,0,0)$. If that's so why does the author use $(2,0,0)$ to designate the plane? enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

This is because the $(1 0 0)$ planes do not represent a true set of lattice planes. A true lattice plane is one whose successive stacking will generate the whole lattice.

Consider the two set of planes denoted by the Miller indices $(100)$ and $(200)$. The $(100)$ planes have a d spacing $d_{100}=a$ and you may think of them as the opposite faces of the cube. Now, if you calculate the d spacing for the $(200)$ planes, you get $d_{200}=\frac{a}{2}$ and you may think of these as the faces of the cube and another plane midway, justifiying the d spacing as $\frac{a}{2}$. However, the $(100)$ planes do not include all the lattice points in the crystal since you're missing out on the body centre points. Hence the $(200)$ set represents the true lattice planes.

Edit: I realise I've misunderstood your question. The $(200)$ and the $(100)$ planes are not equaivalent due to the discussion above.

$\endgroup$
5
  • $\begingroup$ So when do I reduce the Miller indices down to lower integers and when is it incorrect? $\endgroup$
    – Kashmiri
    Nov 23, 2020 at 17:17
  • $\begingroup$ It would depend on which plane you're interested in. The figure in your question shows the $(200)$ planes shaded. If it had the opposite faces shaded, then it would be the $(100)$ planes. $\endgroup$
    – AlphaBaal
    Nov 23, 2020 at 17:24
  • $\begingroup$ Could you please elaborate on that? This is thoroughly confusing to me. $\endgroup$
    – Kashmiri
    Nov 25, 2020 at 3:55
  • $\begingroup$ You might want to have a look here physics.stackexchange.com/questions/596123/… $\endgroup$
    – Kashmiri
    Nov 25, 2020 at 4:37
  • $\begingroup$ Perhaps you can try looking at the application of Miller indices to X ray diffraction. In XRD, each peak is labelled by Miller indices and you'll find no (100) peak for a BCC structure. The book 'The Oxford Solid State Basics' by Steven Simon is a great book to learn this. $\endgroup$
    – AlphaBaal
    Nov 25, 2020 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.