Niobium crystallises in body-centered cubic structure. If density is 8.55 \mathrm{gcm}^{-3}. Calculate the atomic radius of niobium using its atomic mass 93 \mathrm{u}.
Niobium crystallises in body-centered cubic structure. If density is 8.55 \mathrm{gcm}^{-3}. Calculate the atomic radius of niobium using its atomic mass 93 \mathrm{u}.

Ans: It is given that the density of niobium, \mathrm{d}=8.55 \mathrm{gcm}^{-3}
Atomic mass, M=93 \mathrm{gmol}^{-1}
As the lattice is bcc type, the number of atoms per unit cell, z=2
Let the edge length of the unit cell be ‘ a^{+}and the length of the face diagonal A C be b.
From \triangle A B C, we have:

    \[\begin{array}{l} \mathrm{AC}^{2}=\mathrm{BC}^{2}+\mathrm{AB}^{2} \\ \Rightarrow \mathrm{b}^{2}=\mathrm{a}^{2}+\mathrm{a}^{2} \\ \Rightarrow \mathrm{b}^{2}=2 \mathrm{a}^{2} \\ \Rightarrow \mathrm{b}=\sqrt{2 \mathrm{a}} \end{array}\]

Let r be the radius of the atom.
Now, from the figure, it can be observed that:

    \[\begin{array}{l} b=4 r \\ \Rightarrow \sqrt{2 a}=4 r \\ \Rightarrow a=2 \sqrt{2 r} \end{array}\]

Now, volume of the cube, a^{3}=(2 \sqrt{2 r})^{3}
We know that the number of atoms per unit cell is 4 .
So, volume of the occupied unit cell =4 \pi \frac{4}{3} r^{3}
Therefore, packing efficiency =\frac{\text { Volumeoccupied by four spheres in the unit cell }}{\text { Total volume of the unit cell }} \times 100 \% 6 =\frac{4 \pi \frac{4}{3} r^{3}}{(2 \sqrt{2})^{3}} \times 100 \% =\frac{3}{16 \sqrt{2 r^{3}}} \times 100 \% =74 \%
Therefore, packing efficiency in face centered is 74 \%.