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

Explanation :

Nearest neighbor distance, r =

Atomic mass of niobium (Nb) = 92.91 g/mole

Avogadro's number

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

For BCC lattice :

For FCC lattice :

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

Formula used :

.............(1)

where,

= density

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

M = atomic mass

= Avogadro's number

a = edge length of unit cell

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

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)

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