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
25
35
55
75
The Correct Option is B
Solution and Explanation
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 (Zfcc) 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 × Zfcc = 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 (Ztotal) = 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) / 4from 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 = Ztotal × 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)
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Concepts Used:
Packing Efficiency
The percentage of total space in a unit cell that is filled by the constituent particles, such as atoms, ions, or molecules, packed within the lattice is called the packing efficiency. It is the total amount of space engaged by these particles in three-dimensional space. In a simple manner, we can understand it as the specified percentage of the total volume of a solid which is occupied by spherical atoms. Packing Efficiency can be evaluated in three structures with the help of geometry which are:
- Cubic Close Packing and Hexagonal Close Packing.
- Body-Centred Cubic Structures
- Simple Lattice Structures of Cubic
Following are the importance of packing efficiency:
- It represents the solid structure of the object.
- It displays different properties of solids such as consistency, density, and isotropy.
- Different attributes of solid structures can be derived, with the help of Packing Efficiency.