Vector Algebra for Diamond Unit Cell: Neighbouring Atoms and Bond Angles

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The discussion focuses on determining the vectors connecting the carbon atom at position 0.25A(i + j + k) to its four nearest neighbors within a diamond unit cell. Participants clarify that the displacement from original atom positions should be treated as an addition operation, leading to the correct new positions for the atoms. The user expresses confusion about identifying the nearest neighbors and is advised to sketch the points for better visualization. Ultimately, understanding the geometry of the unit cell and the concept of displacement is crucial for solving the problem. The conversation emphasizes the importance of visual aids in tackling complex vector algebra in crystallography.
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Homework Statement



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.

Homework Equations



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.
 
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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?
 
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?
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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