Answer the following for a body centered unit cell.

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

whats the radius given?

the radius is 5.0

please help!

To answer these questions for a body centered unit cell, we need to understand the characteristics of this type of unit cell.

A body centered unit cell is a type of crystal lattice structure in which lattice points (atoms or ions) are present at the corners of the unit cell as well as at the center of the unit cell. This creates additional points at the body center of the unit cell.

Now, let's find the answers to the questions:

1. Edge in terms of r, the lattice point radius:
In a body centered unit cell, the body diagonal is equal to four times the edge length. Therefore, to find the edge length, we need to divide the body diagonal by 4.

The body diagonal can be calculated using the Pythagorean theorem in the unit cell. Consider the length of the body diagonal as D. The body diagonal connects opposite corners of the unit cell, thus forming a right triangle with two sides equal to the edge length (L) and one side equal to the length from the body center to a corner, which is equal to two times the lattice point radius (2r).

Using the Pythagorean theorem, we have:
D^2 = L^2 + L^2 + (2r)^2
D^2 = 2L^2 + 4r^2

Since the body diagonal is four times the edge length, we have:
D = 4L

Substituting it back into the previous equation:
(4L)^2 = 2L^2 + 4r^2
16L^2 = 2L^2 + 4r^2
14L^2 = 4r^2
L^2 = (4r^2)/14
L = sqrt((4r^2)/14)

So, the edge length (L) in terms of the lattice point radius (r) is given by:
L = sqrt((4r^2)/14) = sqrt((2r^2)/7)

2. Face diagonal in terms of r, the lattice point radius:
The face diagonal is the distance between two opposite faces of the unit cell. It is equal to the square root of three times the edge length (L).

Therefore, the face diagonal (FD) in terms of the lattice point radius (r) is given by:
FD = sqrt(3L)

Substituting the value of L we found earlier, we have:
FD = sqrt(3 * (2r^2)/7)
FD = sqrt((6r^2)/7)

So, the face diagonal (FD) in terms of the lattice point radius (r) is given by:
FD = sqrt((6r^2)/7)

3. Body diagonal in terms of r, the lattice point radius:
The body diagonal is the distance between two opposite corners of the unit cell. It is equal to the square root of four times the edge length (L).

Therefore, the body diagonal (BD) in terms of the lattice point radius (r) is given by:
BD = sqrt(4L)

Substituting the value of L we found earlier, we have:
BD = sqrt(4 * (2r^2)/7)
BD = sqrt((8r^2)/7)

So, the body diagonal (BD) in terms of the lattice point radius (r) is given by:
BD = sqrt((8r^2)/7)