Question

Niobium (Nb) has an atomic radius of 0.1430 nm and a density of 8.57 g/cm3 . Determine whether it has an FCC or a BCC crystal structure. (40 pts.). Avogadro’s number (NA) = 6.022 x 1023, atomic weight of Nb is 92.91 g/mol and ???? = nA �

Answer 1

Answer : The crystal structure of Niobium is, BCC (Z=2)

Explanation :

Nearest neighbor distance, r = $0.1430nm=1.430\times 10^{-8}cm$ $(1nm=10^{-7}cm)$

Atomic mass of niobium (Nb) = 92.91 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 : $a^3=(\frac{4r}{\sqrt{3}})^3=(\frac{4\times 1.430\times 10^{-8}cm}{\sqrt{3}})^3=3.60\times 10^{-23}cm^3$

For FCC lattice : $a^3=(\sqrt{8}r)^3=(\sqrt{8}\times 1.430\times 10^{-8}cm)^3=6.62\times 10^{-23}cm^3$

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

$\rho=\frac{2\times (92.91g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (3.60\times 10^{-23}Cm^3)}=8.57g/Cm^{3}$

$\rho=\frac{4\times (92.91g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (6.62\times 10^{-23}Cm^3)}=9.32g/Cm^{3}$

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

Therefore, the crystal structure of Niobium is, BCC (Z=2)