GATE ME 2018 Set 1 — Question 27

This question requires knowledge of crystal structures, specifically the Face-Centered Cubic (FCC) structure. An FCC unit cell has atoms at two types of positions:

• 8 atoms at the corners of the cube.
• 6 atoms at the center of each of the 6 faces.

The contribution of each atom to a single unit cell is as follows:

• Each corner atom is shared by 8 unit cells, so its contribution is $1/8$.
• Each face-centered atom is shared by 2 unit cells, so its contribution is $1/2$.

Total number of atoms per unit cell = (Number of corner atoms $\times$ Contribution per corner atom) + (Number of face atoms $\times$ Contribution per face atom)
Total atoms = $(8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 1 + 3 = 4$. A slip system is a combination of a slip plane and a slip direction. For plastic deformation to occur, dislocations move along these slip systems.

• Slip Planes: In FCC crystals, slip occurs on the most densely packed planes, which are the {111} family of planes. There are 4 unique, non-parallel planes in this family: (111), (11$\bar{1}$), (1$\bar{1}$1), and ($\bar{1}$11).
• Slip Directions: Slip occurs in the most densely packed directions, which are the <110> family of directions. Within each {111} plane, there are 3 possible slip directions that are not parallel to each other.

The total number of slip systems is the product of the number of slip planes and the number of slip directions per plane.
Number of slip systems = (Number of slip planes) $\times$ (Number of slip directions per plane)
Number of slip systems = $4 \times 3 = 12$.Therefore, for an FCC crystal, the number of atoms per unit cell is 4, and the number of slip systems is 12.Comparing this with the options:
(A) 3, 3 - Incorrect.
(B) 3, 12 - Incorrect number of atoms.
(C) 4, 12 - Correct.
(D) 4, 48 - Incorrect number of slip systems.