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Crystal Planes, Miller Indices: Cubic Lattice

  1. Oct 4, 2005 #1
    Just when I thought I understood the concept of planes and miller indices, I got stuck on a 'test your understanding' Q in my book.

    I can't understand that there can be two or more (110) planes in a crystal lattice? I thought there can be only one such plane. The question asks me to find the distance between the nearest (110) planes in a simple cubic lattice, with the value of lattice constant given.

    I can solve the Q, but I need to know if there can be really two (110) planes in a crystal lattice, and where will they be? I discussed it with a few classmates, and they don't get it either. We all agree that there can be only one (110) plane in a cubic lattice.

    Any help will be greatly appreciated!
     
  2. jcsd
  3. Oct 4, 2005 #2

    Astronuc

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    Here is a nice simple reference.

    Now, with respect to the faces of the a simple cube, there are six combinations of Miller indices of the family {100} which describe the planes which are 'crystographically' identical, i.e. one can rotate the cube and obtain the same plane.

    Thus (100), (010), (001) and ([itex]\bar{1}[/itex]00), (0[itex]\bar{1}[/itex]0), (00[itex]\bar{1}[/itex]) are identical.

    The distance between parallel planes is simply the lattice parameter.

    Similarly, the family of planes {110} are crystographically indentical -

    (110), (011), (101), and their complements. One can orient the cube and get the same plane.

    Now what would be the distance between parallel cubes.

    Two visualize, place two cubes side-by-side. Consider the cube a unit cell which is repeated in three dimensions.
     
  4. Oct 4, 2005 #3

    Gokul43201

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    All planes parallel to some (hkl) plane are designated (hkl). So, in a crystal, there are a macroscopic number of such planes, all parallel to each other. For instance, in a cubic lattice, the (100) plane is a cube face. You can (if you wish to denounce convention) call the parallel cube faces (1/2 0 0), (1/3 0 0), etc. but the convention requires that you use the simplest whole number ratio. Hence all such planes are (100) planes.

    Alternatively, think of it this way : you choose some arbitrary lattice point as origin and the plane in the immediate neighboring unit-cell with intercepts a/h, b/k, c/l is designated (hkl). Now, there's no reason that you can not displace your origin by (pa,qb, rc) {p,q,r being integers} and find another such plane. Naturally, these two planes are parallel to each other.

    On the other hand, no two planes in a unit-cell belonging to the same family will be parallel to each other, and hence, a "distance between them" would be meaningless.
     
    Last edited: Oct 4, 2005
  5. Oct 5, 2005 #4
    Thanks a lot for that. I am actually quite new to this concept of crystal planes and miller indices, that's why I have trouble not only for solving problems but also to understand crystal structures.

    I solved the Q that was troubling me, and hopefully I'll be able to go through my exercise now, without any trouble.

    Thanks again guys!
     
  6. Oct 6, 2005 #5
    I just got the same question asking to find the the nearest distance between, say parallel (110) or (100) planes. But to my understanding there are infinite such parralel planes within the unit cell. And the distance should theoretically be zero. Unfortunately I do not believe this topic is covered in our book. Can anyone explain what the question really means? Thanks
     
  7. Oct 6, 2005 #6

    Gokul43201

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    No perry, there is only ONE such plane per unit cell. Where do you get an infinite number of planes from ? I hope you understand what a unit cell is.

    Consider the unit cell for a simple cubic. The (001) plane is the xy plane that makes up a cube face. The next such plane is the next parallel cube face, and is found a distance 'a' away.

    In general, for a cubic lattice, the distance between (hkl) planes is given by [itex]d = a/ \sqrt{ h^2 + k^2 + l^2} [/itex]. The derivation for this will make things clear.

    http://www.chem.qmw.ac.uk/surfaces/scc/scat1_1b.htm

    http://www.mrl.ucsb.edu/~seshadri/2004_100A/100A_MillerBragg.pdf
     
  8. Oct 6, 2005 #7
    Thank you very much. I reallize now what I was describing could have been ([any integer] 0 0).
     
    Last edited: Oct 6, 2005
  9. Apr 6, 2009 #8
    How can I select lattice planes in orthorombic crystal lattice along 100,110 and 001 direction
     
  10. Apr 10, 2009 #9
    I want to select crystal planes in wurtzite and orthorombic crystal along 100,110 and 001 direction. can you help me.
     
  11. Apr 14, 2009 #10
    In Wurtzite crystal structure, this three-digit notation is not used. Instead, four-digit notation is used for the hexagonal unit cell with coordinated a1, a2, a3, and c. For example, the stacking direction is (0001) and the side face belongs to the family of {1-100} planes.


     
  12. Apr 15, 2009 #11
    Dear sir
    Thank u for ur reply. Now I want to select the 110 plane in Orthorombic structure which three coordinates in the unit cell should I select. Please send me the view of it.
     
  13. Apr 18, 2009 #12
    Dear sir
    Thank u for ur reply. Now I want to select the 110 plane in Orthorombic structure which three coordinates in the unit cell should I select. Please send me the view of it.
     
  14. Apr 18, 2009 #13
    Dear sir
    Thank u for ur reply. Now I want to select the 110 plane in Orthorombic structure which three coordinates in the unit cell should I select. Please send me the view of it.
     
  15. Apr 18, 2009 #14
    I sent you a PM several days ago. Did you get it? Anyway, orthorhombic is not very different from cubic. Compare these two and you will find the answer.
     
  16. Apr 23, 2009 #15
    Thank you very much.
     
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