37. The description in the problem statement implies that an atom is at the centerpoint
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
h
0
=
1
2
a
√
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
0
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
0
is a distance
AC
0
=
a/
√
3. If we draw
a line from
C
0
all the way to farthest point on the tetrahedron (this will land on an atom we label
B
),
then this new line is the altitude
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details