Question

The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching its nearest neighbors. Determine the lattice constant and effective radius of the atom.

Answer 1

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

        $5 \times 10^{26} atoms = 1 [tex]m^{3}$

        2 atoms = $\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms$

                     = $4 \times 10^{-27} m^{3}$

Formula for volume of a cube is $a^{3}$. Therefore,

           Volume of the cube = $4 \times 10^{-27} m^{3}$

As lattice constant (a) = $(4 \times 10^{-27} m^{3})^{\frac{1}{3}}$

                                   = $1.59 \times 10^{-9} m$

Therefore, the value of lattice constant is $1.59 \times 10^{-9} m$.

And, for bcc unit cell the value of radius is as follows.

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

Hence, effective radius of the atom is calculated as follows.

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

                   = $\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m$

                   = $6.9 \times 10^{-10} m$

Hence, the value of effective radius of the atom is $6.9 \times 10^{-10} m$.