Question

Rhodium has an atomic radius of 0.1345 nm and a density of 12.41 g/cm3. Determine whether it has an FCC or BCC crystal structure.

Answer 1

Answer : The crystal structure of Rhodium is, FCC (Z=4)

Explanation :

Nearest neighbor distance, r = 0.1345nm=1.345\times 10^{-8}cm (1nm=10^{-7}cm)

Atomic mass of rhodium (M) = 102.9 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the cubing of edge length of unit cell for BCC and FCC crystal lattice.

For BCC lattice : $a^3=(\frac{4r}{\sqrt{3}})^3=(\frac{4\times 1.345\times 10^{-8}cm}{\sqrt{3}})^3=3.04\times 10^{-23}cm^3$

For FCC lattice : $a^3=(\sqrt{8}r)^3=(\sqrt{8}\times 1.345\times 10^{-8}cm)^3=5.50\times 10^{-23}cm^3$

Now we have to calculate the density of unit cell for BCC and FCC crystal lattice.

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell  (for BCC = 2, for FCC = 4)

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

$\rho=\frac{2\times (102.9g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (3.04\times 10^{-23}Cm^3)}=11.24g/Cm^{3}$

$\rho=\frac{4\times (102.9g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (5.50\times 10^{-23}Cm^3)}=12.42g/Cm^{3}$

From this information we conclude that, the given density is approximately equal to the density of FCC unit lattice.  

So, the crystal structure of Rhodium is, FCC (Z=4)

Answer 2

The Rhodium atom has FCC crystal structure since the density calculation matches the given density value.

The given parameters;

• radius of the Rhodium atom, r = 0.1345 nm = 0.1345 x 10⁻⁹ m.
• density of the atom, ρ = 12.41 g/cm³

The edge length of unit cell for BCC and FCC crystal lattice is calculated as follows;

$BCC: \ \ \ a^3 = (\frac{4r}{\sqrt{3} } )^3 \ \ = \ \ (\frac{4\times 0.1345 \times 10^{-7} }{\sqrt{3} } )^3 \ =2.996 \times 10^{-23} \ cm^3 \\\\FCC : \ \ \ a^3 \ = (\sqrt{8} \ r)^3 \ \ = (\sqrt{8} \times 0.1345\times 10^{-7} )^3= 5.5 \times 10^{-23} \ cm^3$

The density of unit cell for each crystal lattice is calculated as follows;

$\rho = \frac{Z \times M}{a^3 \times N_A}$

where;

• ρ is the density
• Z is number of atom in unit cell
• M is the atomic mass = 102.9 g/mol
• $N_A$ is the Avogadro's number
• a is the edge length

The density of unit cell for BCC crystal lattice (Z = 2);

$\rho = \frac{Z \times M}{a^3 \times N_A}\\\\\rho = \frac{2 \times 102.9}{2.996 \times 10^{-23} \times 6.02 \times 10^{23}}\\\\\rho = 11.41 \ g/cm^3$

The density of unit cell for FCC crystal lattice (Z = 4);

$\rho = \frac{Z \times M}{a^3 \times N_A}\\\\\rho = \frac{4 \times 102.9}{5.5 \times 10^{-23} \times 6.02 \times 10^{23}}\\\\\rho = 12.43 \ g/cm^3$

Thus, we can conclude that the Rhodium atom has FCC crystal structure since the density calculation matches the given density value.

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