ch41-p045

# ch41-p045 - 45 The description in the problem statement...

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45. The description in the problem statement implies that an atom is at the center point C of the regular tetrahedron, since its four neighbors are at the four vertices. The side length for the tetrahedron is given as a = 388 pm. Since each face is an equilateral triangle, the “altitude” of each of those triangles (which is not to be confused with the altitude of the tetrahedron itself) is ha ' = 1 2 3 (this is generally referred to as the “slant height” in the solid geometry literature). At a certain location along the line segment representing “slant height” of each face is the center C' of the face. Imagine this line segment starting at atom A and ending at the midpoint of one of the sides. Knowing that this line segment bisects the 60° angle of the equilateral face, then it is easy to see that C' is a distance AC a '/ = 3 . If we draw a line from C' all the way to the farthest point on the tetrahedron (this will land on an atom we label B ), then this new line is the altitude h of the tetrahedron. Using the Pythagorean theorem,

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## This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch41-p045 - 45 The description in the problem statement...

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