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

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Homework Help Overview

The problem involves the geometry of a diamond unit cell, specifically focusing on the positions of carbon atoms within a cubic structure and the vectors connecting these atoms. The original poster seeks to identify the vectors from a specific atom position to its nearest neighbors and to determine the bond angles between carbon atoms in diamond.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the positions of atoms displaced by a vector and is uncertain about the correct operation (addition or subtraction) to apply. Some participants suggest considering the displacement as an addition operation and propose visualizing the points to identify nearest neighbors.

Discussion Status

Participants are actively engaging with the problem, clarifying the interpretation of displacement and exploring methods to find the nearest neighbors. There is no explicit consensus yet, but guidance has been provided regarding the approach to visualize the atom positions.

Contextual Notes

The original poster expresses difficulty in determining the correct positions based on the displacement vector, indicating a potential gap in understanding the geometric relationships within the unit cell. The discussion is framed within the constraints of a homework assignment, emphasizing the need for careful reasoning.

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.
 
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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.
 

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