Solid State Physics 2014
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Re: Exams 2014: Problem 1

Posted by Steve Simon on June 20, 2014, 6:59 pm, in reply to "Exams 2014: Problem 1"

 Hi Martin. Think about the conventional unit cell of the original fcc lattice. Now consider the xy plane at the base of this unit cell. In this plane is a square lattice where the square lattice constant is the same size as the conventional unit cell. (Say the points [0,0,0] [1,0,0] [1 1 0] [0 1 0]). HOWEVER, there is also a lattice of smaller squares turned at 45 degrees. This includes the points in the center of these faces. For example, a square is made by the points [0,0 0] [1/2,1/2 0] [0,1 0] [-1/2,1/2 0]. The lattice constant for this square lattice is 1/sqrt[2]. This square will now be the base of the new conventional unit cell of the bcc. In order to make the rest of the new conventional unit cell, we have to squash the old conventional unit cell until its edge length is the same 1/sqrt[2]. The point that was in the center of the face [0,1/2,1/2] now gets squeezed down until its coordinate is [0,1/2,1/(2 sqrt[2]]. However, this is precisely the point in the center of the new conventional unit cell -- so it forms the body center point! Does that help?
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 Message Thread Exams 2014: Problem 1 - Martin June 20, 2014, 12:56 pm Re: Exams 2014: Problem 1 - Steve Simon June 20, 2014, 6:59 pm « Back to index