Answer:
Explanation:
The density of the unit cell of a material, Iron in this case, has to be approximately equal with its experimental value of 7.87 g/cm³.
The density d = m/v, so what we need to do is calculate the volume of the unit cell and its mass and perform the calculation.
For a BCC crystal structure the length of the side of the cube is given by:
a = 4r/√3
where a is the atomic radius of Iron
first we will convert this radius to cm since we want the density in g/cm³:
0.124 nm x 1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm
a = 4 x 1.24 x 10⁻⁸ cm /√3 = 2.86 x 10⁻⁸ cm
the volume of the cubic cell is:
v = a³ = ( 2.86 x 10⁻⁸ cm )³ =2.35 x 10⁻²³ cm³
The mass of iron in the body centered cubic cell is obtained from the mass of the atoms in it:
BCC = 2 atoms / unit cell ( 1/8 from the 8 corners + 1 in the center)
m = 2 atoms/unit cell x 1 mol/ 6.022 x 10²³ atoms x 55.85 g/mol
= 1.85 x 10⁻²² g
Therefore,
d = m/v = 1.85 x 10⁻²² g / 2.35 x 10⁻²³cm³ = 7.88 g/cm³
An excelent agreement which confirms that the density of the BCC unit cell agrees with the experimental value.
