for face centered unit cell, r=6
find edge in terms of r, face diagonal in terms of r, body diagonal in terms of r,
edge=squareroot(8)r
face=4r
but i don't know how to find body diagonal
Isn't the face diagonal
4r = a*21/2
The body diagonal is a*31/2 .
The triangle has sides of a (the upright line on the left side of the unit cell), the diagonal (along the floor of the unit cell which we know is a*21/2 so the body diagonal is the hypotenuse of that triangle. So db2 = a2 + (a*21/2)2
I hope this turns out right.
It looks ok to me. You solve, of course, for db2 then take the square root of it. That should be the db = a*31/2
for body diagonal. take face diagonal and square it and take edge diagonal and square it. add those 2 together than take the SQUARE ROOT of it. Answer. so its basically the a^2+b^2=c^2 equation.
To find the length of the body diagonal in terms of "r" for a face-centered unit cell, we can use the Pythagorean theorem. Let's label the length of the edge as "a".
The body diagonal of a unit cell is a line connecting two opposite vertices, passing through the center of the unit cell. It can be visualized as a diagonal that spans from one corner to the opposite corner, passing through the center of the cube.
To find the length of the body diagonal, we can consider a right triangle formed by one of the edges, the face diagonal, and the body diagonal. The face diagonal is the diagonal of one face of the unit cell.
Using the Pythagorean theorem, we have:
(body diagonal)^2 = (edge)^2 + (face diagonal)^2
Let's solve for the body diagonal (d):
d^2 = (a^2) + (4r)^2
d^2 = a^2 + 16r^2
Since we know "a" in terms of "r" as the edge length, we can substitute that:
d^2 = (sqrt(8)r)^2 + 16r^2
d^2 = 8r^2 + 16r^2
d^2 = 24r^2
Taking the square root of both sides to solve for "d":
d = sqrt(24r^2)
Simplifying further:
d = 2r * sqrt(6)
Therefore, the length of the body diagonal in terms of "r" is given by:
body diagonal = 2r * sqrt(6)