- #1

spaghetti3451

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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.