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Vector algebra

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A unit cell of diamond is a cube of side A, with carbon atoms at each corner, at
    the centre of each face and, in addition, at positions displaced by 0.25A(i + j + k) from each of those already mentioned; i, j, k are unit vectors along the cube axes. One corner of the cube is taken as the origin of the coordinates. What are the vectors joining the atom at 0.25A(i + j + k) to its four nearest neighbours? Determine the angle between the carbon bonds in diamond.

    2. Relevant equations

    3. The attempt at a solution

    The atoms at the corners are (0,0,0), (A,0,0), (0,A,0), (0,0,A), (A,A,0), (A,0,A), (0,A,A), (A,A,A).

    The atoms at the centre of each face are (0.5A,0.5A,0), (0.5A,0,0.5A), (0,0.5A,0.5A), (0.5A,0.5A,A), (0.5A,A,0.5A), (A,0.5A,0.5A).

    I am having trouble figuring the atoms at positions displaced by 0.25A(i + j + k) from each of those already mentioned.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 14, 2012 #2

    BruceW

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    "positions displaced by 0.25A(i+j+k) from each of those already mentioned" this tells you the displacement vector from the points already mentioned to some new points. You have written down displacement vectors of the original points from the origin, so how would you work out the displacement vectors of the new points from the origin? Hint- what is one of the simplest operations you can do with two vectors?
     
  4. Apr 14, 2012 #3
    Well, the trouble is figuring out if it's addition or subtraction, because both give displacements of 0.25A(i+j+k) from each of the original points. But, now I'm beginning to think 'displaced by' instructs you to add (by convention). So, the displaced positions are (0.25A,0.25A,0.25A), (1.25A,0.25A,0.25A), (0.25A,1.25A,0.25A), (0.25A,0.25A,1.25A), (1.25A,1.25A,0.25A), (1.25A,0.25A,1.25A), (0.25A,1.25A,1.25A), (1.25A,1.25A,1.25A), (0.75A,0.75A,0.25A), (0.75A,0.25A,0.75A), (0.25A,0.75A,0.75A), (0.75A,0.75A,1.25A), (0.75A,1.25A,0.75A), (1.25A,0.75A,0.75A).

    So, do I have to use trial and error to find out the the vectors joining the atom at 0.25A(i + j + k) to its four nearest neighbours?
     
  5. Apr 15, 2012 #4

    BruceW

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    yes, 'displaced by' usually means addition, and you've got all the new points correct. You could use trial and error, but I think it is best to roughly draw out the points, then hopefully you will see which are the nearest neighbours.
     
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