For a primitive cubic unit cell, r=6.0
what is the body diagonal in terms of r, the lattice pt radius
For a primitive unit cell, isn't the side a and the other side is a. Therefore, the diagonal is a^2 + a^2 = diagonal^2. Check me out on this but isn't a the diameter of an atom?
Drbob222 is talking about face diagonal not body diagonal
for body diagonal its c^2=a^2+b^2. B you find from a^2+a^2=b^2 and a is a=2r. Hope that helps.
To find the body diagonal in terms of the lattice point radius (r) for a primitive cubic unit cell, we need to understand the key properties of a cube.
A primitive cubic unit cell has lattice points at each corner of the cube. It also has one lattice point at the center of the cube. The body diagonal is the line that connects two opposite corners of the cube, passing through the center.
In a cube, the body diagonal can be calculated using the Pythagorean theorem. The length of the body diagonal (d) can be found by considering the side length (a) of the cube:
d^2 = a^2 + a^2 + a^2
Since all sides of a cube are equal, we can substitute a with 2r, where r is the lattice point radius:
d^2 = (2r)^2 + (2r)^2 + (2r)^2
d^2 = 4r^2 + 4r^2 + 4r^2
d^2 = 12r^2
Taking the square root of both sides, we get:
d = sqrt(12r^2)
Simplifying further, we have:
d = sqrt(4 × 3 × r^2)
d = 2√3r
Therefore, the body diagonal in terms of the lattice point radius (r) for a primitive cubic unit cell is given by 2√3r.