Question

Calculate the radius of a vanadium atom, given that V has a BCC crystal structure, a density of 5.96 g/cm3, and an atomic weight of 50.9 g/mol.

Answer 1

To calculate the edge length of metal, we use the equation:

Where

$\rho = density of metal = 5.96g/cm^3$

Z = number of atom in unit cell = 2 (BCC

M = atomic mass of metal = 50.9 g/mol

$N_{A} = Avogadro's number = 6.022\times 10^{23}$

a = edge length of unit cell = ?

Putting values in above equation, we get:

$5.96g/cm^3=\frac{2\times 50.9}{6.022\times 10^{23}\times (a)^3}\\\\a^3=2.836\times 10^{-23}cm^3\\\\a=\sqrt[3]{2.836\times 10^{-23}}=3.05\times 10^{-8}cm^3=305pm$

To calculate the radius, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

Where

R = radius of the lattice = ?

a = edge length = 305 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 305}{4}=132.07pm$

Hence, the radius of vanadium atom is 132.07 pm