# MCQs on Miller Indices

1 - Question

Stacking sequence in hexagonal close packed (HCP) structure is?
a) AAAAA
b) ABABAB
c) ABCABC
d) AABBAA

Explanation: The geometry of close packed hexagonal unit cell can be understood from the below figure, which indicates a plane view of atoms.

Figure: Stacking sequences of close packed layers of atoms. A-first layer (with outlines of atoms shown in gold colour); B-second layer.
A is the first layer (with the circular outlines of the atoms drawn in gold colour) and B is the second layer (outlines of the atoms not shown for clarity). In hexagonal close packed unit cell, the third layer of atoms go directly above the A layer, the fourth layer over the B layer, and so on; the sequence becomes ABABAB.

2 - Question

Stacking sequence in face centered cubic (FCC) close packed structure is?
a) AAAAA
b) ABABAB
c) ABCABC
d) AABBAA

Explanation: The stacking sequence in FCC can be best understood from the below figure.

Figure: Stacking sequences of close packed layers of atoms. A-first layer (with outlines of atoms shown in green colour); B-second layer; C-third layer.
A is the first layer (with the circular outlines of the atoms drawn in green colour) and B is the second layer (outlines of the atoms not shown for clarity). The third layer of atoms goes above the interstices marked C and the sequence only repeats at the fourth layer, which goes directly above the first layer. Thus, the stacking sequence is now ABCABC.

3 - Question

For pane (1 1 1) of FCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 2.31/a2
b) 2.31/a3
c) 1.31/a2
d) 1.31/a3
Explanation: Upon visualizing (1 1 1) plane of FCC, one can identify that there the equilibrium triangle (1 1 1) of FCC consists 2 atoms (1⁄6×3 + 1⁄2×3=2). Also, the area of the equilibrium triangle (1 1 1) is 0.886 a2. Therefore, the number of atoms per square inch which is nothing but its planar density = 20.866a2=2.31a2.

4 - Question

For pane (1 1 1) of BCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 1.07/a2
b) 0.58/a2
c) 2.07/a2
d) 0.78/a2
Explanation: From the geometry of the triangle (1 1 1), it is clear that it has 0.5 atoms in it (1/6×3 = 0.5) and the area of the triangle (1 1 1) is 0.866 a2. Therefore, the number of atoms per square inch which is nothing but its planar density = 0.50.866a2=0.58a2.

5 - Question

For pane (1 0 0) of BCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 1/a3
b) 2/a2
c) 3/a4
d) 1/a2
Explanation: The atomic arrangement on square (1 0 0) of BCC indicates that it has 1 atom and area of this square (1 0 0) is a2. Thus, its planar density = 1/a2.

6 - Question

For pane (1 1 0) of BCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 3.690/a2
b) 2.312/a2
c) 1.414/a2
d) 0.580/a2
Explanation: From the geometry of the rectangle (1 1 0), it is clear that it has 2 atoms in it (1⁄4×4+1=2) and the area of the rectangle (1 1 0) is 0.866 a2. Therefore, the number of atoms per square inch which is nothing but its planar density = 21.414a2=1.414a2.

7 - Question

For pane (1 1 0) of SC having a lattice parameter ‘a’, planar atomic density is given by?
a) 0.508/a2
b) 0.707/a2
c) 0.707/a3
d) 0.508/a3
Explanation: The atomic arrangement on (1 1 0) plane in simple cubic (SC) unit cell indicates that it has 1 atom (1⁄4×4=1) in this plane having an area = 2–√a2. Therefore, the number of atoms per square inch which is nothing but its planar density = 11.414a2=0.707a2.

8 - Question

For pane (1 1 1) of SC having a lattice parameter ‘a’, planar atomic density is given by?
a) 0.58/a2
b) 0.78/a3
c) 0.68/a2
d) 0.88/a2
Explanation: The atomic arrangement on triangle (1 1 1) in simple cubic (SC) unit cell indicates that it has 0.5 atoms (1⁄6×3=0.5) in this plane having an area = 0.866 a2. Therefore, the number of atoms per square inch which is nothing but its planar density = 0.50.866a2=0.577a2.

9 - Question

Which of the following equation describes Bragg’s law of diffraction? (Assume that all symbols have their usual meaning.)
a) 2d sinθ = λ
b) 2d = nλ
c) 2d = nλ sinθ
d) 2d sinθ = nλ