Question

The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical density (g/cm^3).

Answer 1

Solution:

Given :

atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m

atomic wieght, M = 95.96

Cell structure is BCC (Body Centred Cubic)

For BCC, we know that no. of atoms per unit cell, z = 2

and atomic radius, r =$\frac{a\sqrt{3} }{4}$

so, a = $\frac{4r}{\sqrt{3}}$

m = mass of each atom in a unit cell

mass of an atom = $\frac{M}{N_{A} }$,

where, $N_{A}$ is Avagadro Number = 6.02×10^{23}

volume of unit cell = a^{3}

density, ρ = $\frac{mass of unit cell}{volume of unit cell}$

density, ρ = $\frac{z\times M}{a^{3}\times N_{A}}$

ρ = $\frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}$

ρ = 10.215gm/$cm^{3}$