Molybdenum (Mo) has a density of 10.22 g/cm3 and it crystalizes into a fcc unit cell, calculate the radius of a (Mo) atom?
The crystal structure according to webelements is a body centered cubic.
http://www.webelements.com/molybdenum/crystal_structure.html
To calculate the radius of a (Mo) atom, we can use the formula:
Radius of atom (r) = [(3 x Volume of unit cell) / (4 x π x Number of atoms per unit cell)]^(1/3)
First, we need to determine the volume of the unit cell.
In a face-centered cubic (fcc) unit cell, there are 4 atoms at the corners and 1 atom at the center of each face. So, the total number of atoms per unit cell is 4 + 1/2 = 4.5 atoms.
Now, we need to determine the volume of the unit cell. Since Molybdenum is an element that crystallizes into a fcc structure, the unit cell is a cube with atoms at each corner and one at the center of each face.
The volume of the unit cell (V) can be calculated using the formula:
Volume of unit cell = (edge length)^3
To find the edge length, we can use the concept that the diagonal of a face-centered cubic unit cell is equal to four times the radius of the atom (diagonal = 4r).
So, we have:
diagonal of unit cell = 4r
diagonal of unit cell = edge length * √2
Equating the two expressions and solving for edge length:
4r = edge length * √2
edge length = (4r) / √2
Now, substitute the value of the edge length into the volume formula:
Volume of unit cell = [(4r) / √2]^3
Finally, substitute the values into the formula for the radius:
Radius of atom (r) = [(3 x Volume of unit cell) / (4 x π x Number of atoms per unit cell)]^(1/3)
Radius of (Mo) atom = [(3 x [(4r) / √2]^3) / (4 x π x 4.5)]^(1/3)
Now, you can substitute the values and perform the calculations to determine the radius of a (Mo) atom.