Respuesta :
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
[tex]5 \times 10^{26} atoms = 1 [tex]m^{3}[/tex]
2 atoms = [tex]\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms[/tex]
= [tex]4 \times 10^{-27} m^{3}[/tex]
Formula for volume of a cube is [tex]a^{3}[/tex]. Therefore,
Volume of the cube = [tex]4 \times 10^{-27} m^{3}[/tex]
As lattice constant (a) = [tex](4 \times 10^{-27} m^{3})^{\frac{1}{3}}[/tex]
= [tex]1.59 \times 10^{-9} m[/tex]
Therefore, the value of lattice constant is [tex]1.59 \times 10^{-9} m[/tex].
And, for bcc unit cell the value of radius is as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
Hence, effective radius of the atom is calculated as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
= [tex]\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m[/tex]
= [tex]6.9 \times 10^{-10} m[/tex]
Hence, the value of effective radius of the atom is [tex]6.9 \times 10^{-10} m[/tex].
