Download Crystal Structure Problems 1.Nickel has a facecentered cubic unit

Crystal Structure Problems
1.Nickel has a face­centered cubic unit cell with a length of 352.4 pm along an edge. What is the density of nickel in g/cm3?
d=
m
V
We need the mass of the atoms in the unit cell and the volume of the unit cell. The mass of all of the atoms in a unit cell is:
m=
n× MW
N
where n is the number of atoms per unit cell, MW is the molar mass and N is Avogadro's number. For aluminum this is:
m=
4 atoms / unit cell×26.98g /mol
6.022×1023 atoms/mol
=1.792×10-22 g/ unit cell
The volume of the unit cell is just the length of the edge cubed:
3
1cm
3
V =352.4pm×
 cm
10
1×10 pm
= 4.376×10
- 23
cm
3
And finally, the density:
- 22
1.792×10 g
d=
4.376×10- 23 cm3
= 4.095g /cm
3
2.Polonium metal crystallizes in a simple cubic arrangement, with the edge of a unit cell having a length l=334 pm. What is the radius (in picometers) of a polonium atom?
For the simple cubic cell the relationship between the length of the unit cell and the radius of the component atoms is d = 2r. So:
r=
=
l
2
334 pm
2
=167 pm
3.Calcium metal has a density of 1.55 g/cm3 and crystallizes in a cubic unit cell with an edge length of 558.2 pm. How many atoms are in one unit cell?
We can calculate the mass of a unit cell in two ways:
m=d×V
or
m=n x
MW
N
So:
d×V =n×
MW
N
Solving for n (the number of atoms in the unit cell):
n=
Plugging the appropriate numbers in:
d×V ×N
MW
3
558.2
23
1.55×
 ×6.022×10
10
1×10
n=
40.078
=4
4.Aluminum has a density of 2.699 g/cm3 and crystallizes with a face­centered cubic unit cell. What is the length of the edge of the unit cell in picometers?
If we work backwards from the volume we can get the length of the edge of the unit cell:
V=
=
n×MW
d× N
4×26.98
2.699×6.022×1023
= 6.640×10- 23 cm3
The length of the edge of the unit cell is the cube root of the volume:
3
l=  V
3
= 6.640×10-23
¿
= 4.049×10- 8 cm
= 404.9pm