Question

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

• A. 25
• B. 35
• C. 55
• D. 75

Answer 1

The Correct Option is B

Let the radius of atom X be 'r'.

1. Number of atoms X in the unit cell:

• Atom X forms an fcc lattice. In an fcc lattice, the number of atoms per unit cell (Z_fcc) is 4. (Calculated as 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/face = 1 + 3 = 4 atoms).
• In an fcc lattice, there are 8 tetrahedral voids (TVs) per unit cell (2 × Z_fcc = 2 × 4 = 8).
• Atom X occupies alternate tetrahedral voids. This means half of the TVs are occupied by atom X. 
  Number of atoms X in TVs = 8 / 2 = 4 atoms.
• Therefore, the total number of atoms X per unit cell (Z_total) = Atoms from fcc sites + Atoms from TVs = 4 + 4 = 8 atoms.

2. Relationship between edge length (a) and atomic radius (r):

• Since the same atom X (with radius r) occupies both the fcc lattice sites and the tetrahedral voids, the lattice must adjust to accommodate these atoms.
• A tetrahedral void is located at a distance of (√3 a) / 4 from a corner atom (where 'a' is the edge length of the unit cell).
• If an atom X (radius r) is at the corner and another atom X (radius r) is in the tetrahedral void, and they touch each other, then the distance between their centers is r + r = 2r.
• So, we have the relation: 2r = (√3 a) / 4.
• From this, the edge length of the unit cell 'a' can be expressed in terms of 'r': a = (8r) / √3.

3. Packing Efficiency (PE):

Packing Efficiency is defined as the ratio of the volume occupied by atoms in the unit cell to the total volume of the unit cell.

PE = (Volume occupied by atoms in the unit cell) / (Volume of the unit cell)

• Volume occupied by atoms = Z_total × Volume of one atom 
  = 8 × (4/3)πr³ = (32/3)πr³
• Volume of the unit cell = a³ 
  = ((8r) / √3)³ = (8³r³) / (√3)³ = (512r³) / (3√3)

Now, substituting these into the PE formula: 
PE = [ (32/3)πr³ ] / [ (512r³) / (3√3) ] 
PE = (32πr³ / 3) × (3√3 / 512r³) 
The r³ terms and the 3s cancel out: 
PE = (32π√3) / 512 
PE = (π√3) / 16

4. Numerical Calculation:

Using the approximate values: π ≈ 3.14159 and √3 ≈ 1.732: 
PE ≈ (3.14159 × 1.732) / 16 
PE ≈ 5.44130468 / 16 
PE ≈ 0.34708

To express this as a percentage: 
PE % ≈ 0.34708 × 100% = 34.708%

The Correct Option is B (35)