Answer the following for a face centered unit cell.
r = 6.0.
how do i find the following
edge in terms of r, the lattice pt radius
face diagonal in terms of r, the lattice pt radius
body diagonal in terms of r, the lattice pt radius
You need to be able to see this in a text. 4r = a(2)1/2 where a is the edge.
The face diagonal is 4r.
The body diagonal is d = a(3)1/2
You may find something here that will help.
http://www.google.com/search?q=crystal+structure&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a
To find the edge length of a face-centered unit cell in terms of the lattice point radius (r), we can use the concept of closest packing in a face-centered lattice structure.
In a face-centered unit cell, there are lattice points at each of the eight corners and an additional lattice point in the center of each face. The lattice points touch each other along the edges of the unit cell.
To visualize this, imagine a cube with each corner occupied by a lattice point, and one additional lattice point at the center of each face. The distance between adjacent lattice points along an edge is equal to the diameter of a lattice point (2*r).
Now, let's calculate the edge length (a) in terms of r:
In a face-centered unit cell, there are four lattice points that touch along an edge; two at the corners and one on each face. The distance from the center of one lattice point to the opposite face center is equal to the edge length of the unit cell.
Using the Pythagorean theorem, we can find the edge length:
(a/2)^2 + (a/2)^2 = (2*r)^2
2(a/2)^2 = 4*r^2
a^2/2 = 4*r^2
a^2 = 8*r^2
a = sqrt(8*r^2)
a = 2*r*sqrt(2)
Therefore, the edge length of a face-centered unit cell in terms of the lattice point radius (r) is given by a = 2*r*sqrt(2).
Now, let's move on to finding the face diagonal length and body diagonal length.
The face diagonal is the distance from one corner of a face to the opposite corner, passing through the center of the unit cell. In terms of r, we can calculate the face diagonal (d_face) using the Pythagorean theorem:
d_face^2 = a^2 + a^2 + a^2
d_face^2 = 3*a^2
d_face^2 = 3*(2*r*sqrt(2))^2
d_face^2 = 12*r^2*2
d_face = 2*r*sqrt(6)
Therefore, the face diagonal length of a face-centered unit cell in terms of the lattice point radius (r) is given by d_face = 2*r*sqrt(6).
Lastly, the body diagonal is the distance from one corner of the unit cell to the opposite corner, passing through the center of the unit cell. In terms of r, we can calculate the body diagonal (d_body) using the Pythagorean theorem:
d_body^2 = a^2 + a^2 + a^2
d_body^2 = 3*a^2
d_body^2 = 3*(2*r*sqrt(2))^2
d_body^2 = 12*r^2*2
d_body = 2*r*sqrt(6)
Therefore, the body diagonal length of a face-centered unit cell in terms of the lattice point radius (r) is also given by d_body = 2*r*sqrt(6).
To summarize:
Edge length (a) = 2*r*sqrt(2)
Face diagonal length (d_face) = 2*r*sqrt(6)
Body diagonal length (d_body) = 2*r*sqrt(6)