Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in%) of the resultant solids is closest to

The correct option is 35.

Step-by-Step Explanation:

Step 1: Calculate the effective number of atoms per unit cell (Z)

In a face-centered cubic (fcc) lattice:

• The number of lattice sites occupied by Atom X in the fcc structure (corners + face centers) is:

$8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$

• In an fcc unit cell, there are a total of 8 tetrahedral voids. Atom X occupies alternate tetrahedral voids, which means half of the total tetrahedral voids are occupied:

$\text{Occupied tetrahedral voids}=\frac{1}{2}\times 8=4$

• Therefore, the total number of atoms X per unit cell (Z) is:

$Z=4\text{(from fcc sites)}+4\text{(from tetrahedral voids)}=8$

Step 2: Find the relationship between radius (r) and unit cell edge length (a)

In a tetrahedral void, the atom at the tetrahedral site touches the corner atom along the body diagonal. The distance between a corner atom and a tetrahedral void is one-fourth of the body diagonal ($\sqrt{3}a$):

$2r=\frac{\sqrt{3}a}{4}$

Rearranging for $r$:

$r=\frac{\sqrt{3}a}{8}$

Step 3: Calculate the packing efficiency

Packing efficiency is given by the ratio of the volume occupied by the atoms in the unit cell to the total volume of the unit cell, expressed as a percentage:

$\text{Packing Efficiency}=\frac{Z\times \frac{4}{3}\pi r^3}{a^3}\times 100$

Substituting $Z=8$ and $r=\frac{\sqrt{3}a}{8}$:

$\text{Packing Efficiency}=\frac{8\times \frac{4}{3}\pi {\left(\frac{\sqrt{3}a}{8}\right)}^3}{a^3}\times 100$

$\text{Packing Efficiency}=\frac{\frac{32}{3}\pi \times \frac{3\sqrt{3}{a}^3}{512}}{a^3}\times 100$

$\text{Packing Efficiency}=\frac{\sqrt{3}\pi }{16}\times 100$

Substituting the numerical values $\sqrt{3}\approx 1.732$ and $\pi \approx 3.1416$:

$\text{Packing Efficiency}=\frac{1.732\times 3.1416}{16}\times 100\approx 34.01\%$

Thus, the packing efficiency is closest to 35%.