Engineering Questions with Answers - Multiple Choice Questions
Materials Science MCQ’s – FCC and HCP Metallic Crystals
1 - Question
Coordination number of HCP and FCC lattices respectively are:
a) 12, 12
b) 4, 4
c) 12, 8
d) 8, 8
Explanation: Coordination number is the number of atoms that are in direct contact of any particular atom or it is the number of nearest neighbours
2 - Question
Number of particles in one unit cell of HCP lattice is:
Explanation: Consider the top and bottom layers. One-sixth of each of the 12(=6×2) atoms located at the vertices of the two hexagons belong to the cell. Also half of the atoms at the centre of each of the two hexagons are also part of the cell. Three more atoms are present between two hexagonal planes.
3 - Question
Standard axial ratio for metallic HCP lattice is 2√( 2/3). It is the ratio of
a) Atomic radius to hexagon edge length
b) Hexagon height length to atomic radius
c) Atomic radius to hexagon height
d) None of the mentioned
Explanation: Ratio of the height of the hexagonal unit cell to its edge length is called the axial ratio, usually expressed as (c/a).
4 - Question
The void fraction is the ratio of unfilled volume to total volume of a structure. For ideal metallic FCC crystal, it is:
Explanation: Void fraction = 1 − packing efficiency Packing efficiency for FCC lattice is 0.74 and hence, void fraction is 1 − 0.74.
5 - Question
If N is the number of tetrahedral voids in a close-packed structure, then the number of octahedral voids is:
Explanation: Tetrahedral voids are formed when triangular voids of one close-packed layer do not overlap while octahedral ones when they overlap. The above relation is valid for both ABAB type(HCP) and ABCABC type(FCC) packing.
6 - Question
An octahedral void is present at each edge of the FCC lattice. How much of each of these voids belong to each unit cell?
Explanation: In a cubic lattice, each edge is shared by four adjacent cubes. Hence, the octahedral voids present at these edges are equally divided among the four cells.
7 - Question
The edge length of an FCC lattice is X times the atomic radius. Value of X is:
Explanation: In FCC unit cell, the atoms at face diagonal touch each other; thus diagonal is four times the atomic radius(r). Diagonal of the square faces is also equal to √ times of the edge length(a). Hence, √2a = 4r.
8 - Question
Density is the ratio of the mass of crystal to its volume. For a perfect FCC metallic crystal, the mass of a unit cell is 4 times M0. M0 is:
a) specific mass
b) molar mass
c) atomic mass
d) none of the mentioned
Explanation: Mass of an ideal FCC metallic crystal unit cell is four times the atomic mass of the metal since there are four atoms per unit cell.
9 - Question
A maximum of 74% packing efficiency can be achieved for crystalline solids.
Explanation: 74% is the maximum packing efficiency for pure metallic crystals. However, if there are particles of two or more different sizes, greater packing efficiency is possible by filling the tetrahedral and octahedral voids.
10 - Question
Which of the following quantities is larger in HCP as compared to FCC if the constituting atoms are similar?
a) Number of particles per unit cell
b) Volume per unit cell
c) Mass per unit cell
d) All of the mentioned
Explanation: An HCP unit cell contains 6 atoms rather than only 4 present in FCC. Also, both structures are assumed to be comprised of similar atoms thus HCP being heavier. However, the HCP unit cell is also larger in volume than an FCC one such that both have same packing efficiency.