MLE2101 Mid-Term Examination Practice Semester I, 2016/2017.
Name:
Matric Number:
INSTRUCTIONS:
1. 2. 3. 4.
This mid-term mid-term examination examination paper contains contains THREE (3) questions and comprises NINE (9) pages. Answer Answer All questions. This This is is a CLOSED CLOSED BOOK examination. A hand-written hand-written summary summary on one side of a piece of A4 paper is allowed. allowed.
1
2
MLE2101
Question 1
Question 2
Question 3
Total Marks
Question 1 (35 marks) The simple orthorhombic lattice is characterized by a = b = c and α = β = γ = 90 .
◦
(a) Draw the directions of the three vectors a , b , and c in the unit cell of the simple orthorhombic lattice. (6 marks) ∗
∗
∗
(b) Consider the plane Γ that is parallel with (112) and contains the point (0, 1/2, 1/2). The three points, namely, (u, 0, 0), (0, v, 0), and (0, 0, w), are on Γ. Determine u, v , and w. (9 marks) (c) Determine the Miller index of the plane Γ. (5 marks) (d) The picture below depicts a unit cell of the simple orthorhombic lattice along the [110] direction. Draw (112) in the XRD analysis. (6 marks) (e) Both (112) and (201) in the simple orthorhombic lattice are found to appear in the same zone in an experiment. Write down the zone axis V and draw a picture to explain the relation among (112), (201), and V. (9 marks)
Solution
(a) The three vectors are defined by a
∗
=
2π (b V
× c),
∗
b
=
2π (c V
× a),
∗
c
=
2π (a V
× b).
(1.1)
Based on the definition, a , b , and c are found to be along [100], [010], and [001], respectively, see Fig. 1. ∗
∗
∗
3
MLE2101
Figure 1: (b) We first calculate the vector v1
v1 from
(0, 1/2, 1/2) to the point (u, 0, 0):
u 0 u = 0 − 1/2 = −1/2 . 0
The vector v1 is on the plane Γ; thus, i.e. v1 G(112) = 0:
1/2
v1 is
(1.2)
−1/2
perpendicular to the normal direction of (112),
·
T
0 = v 1
· G(112)
u 1 3 = 2π −1/2 · 1 = 2π u − . 2 −1/2
(1.3)
2
Solving the equation above determines u to be 3/2. By the same approach, v and w can be obtained to be v = 3/2 and w = 3/4. (c) The three quantities u, v , and w are the interceptions of the plane Γ with the a, b, and c axes of the lattice, and they are 3/2, 3/2, and 3/4, respectively, as shown in part (b). The inverses of u, v, and w are 2/3, 2/3, and 4/3; accordingly, the Miller index of the plane Γ is (224). (d) The plane array (112) in the XRD analysis is depicted in the figure below.
4
MLE2101
(e) The zone axis V of the zone that includes (112) and (201) can be calculated by
V
1 −2 1 = 1 × 0 = −5 . 2
1
(1.4)
2
The calculation indicates the zone axis V is [152]. The zone axis of (112) and (201), see the figure below.
V is
the common direction
Question 2 (35 marks) (a) Draw the CsCl structure along the [110] direction. (7 marks) (b) Use the picture in part (a) to investigate whether the wave fronts of ψ(333) contain all of the ions in the CsCl structure and whether ψ (333) is a Fourier mode of the structure. (10 marks) (c) Determine the wave fronts, the wave length, and the wave vector of ψ(448) . Express your answer in terms of the lattice parameter a. (7 marks) (d) Consider the following three points: A = (1, 1/2, 1/2), B = (1/2, 1, 1/2), and C = (1/2, 1/2, 1). Which one is on the same wave front of ψ (121) at D = ( 9, 5, 2)? (6 marks)
−
(e) Write down all of the Fourier modes ψ (hkl) along the normal direction of (110). (5marks)
5
MLE2101
Solution
(a) The CsCl structure along the [110] direction is shown in the figure below.
(b) The wave fronts of ψ (333) are (333) and the plane array does not include all of the ions, see the figure above. However, (333) does include all of the lattice points. Thus, ψ(333) is a Fourier mode in the CsCl structure. (c) The wave fronts of ψ (448) are (448); the wavelength λ (448) of the Fourier mode is equal to the d-spacing of (448), i.e. λ(448) = d (448) = The wave vector of ψ (448) , denoted as
a = 42 + 42 + 82
√
G(448) ,
G(448
√
a 6a = . 24 96
√
(2.1)
can be expressed as
4 2π = 4 . a
(2.2)
8
(d) If any two points are on the same wave front of a Fourier mode, the phase angles of the two points must be identical and equal to multiple integer of 2 π. The phase angles of ψ (121) at A, B, C, and D can be calculated to be T
1 1 α = 2π 2 1/2 = 5π, 1 1/2 1 1/2 α = 2π 2 1/2 = 5π, A
T
C
1
1
T
1 1/2 α = 2π 2 1 = 6π, 1 1/2 1 −9 α = 2π 2 5 = 6π,
B
(2.3)
T
D
1
(2.4)
2
The results show that the point at B is on the same wave front as D. (e) We first investigate the Fourier mode ψ (110) by checking whether its wave fronts (110) contain all lattice points, and the answer is yes. Thus, ψ(110) is the first Fourier mode, and all of the Fourier modes along the normal direction of (110) can be expressed as
ψ
(000) , ψ(110), ψ(220) , ψ(330) . . . ψ(110) , ψ(220) , ψ(330)
,... .
(2.5)
6
MLE2101
Question 3 (30 marks) The figure below plots the [111] zone in the BCC lattice.
(a) Write down the coordinates of the points at of the points.
A, B,
and
C,
and explain the physical meaning (9 marks)
(b) Explain the physical meaning of the line along L 1 . (5 marks) (c) The magnitude of the vector between the origin and point Determine the lattice parameter a of the lattice.
D is
found to be 82.02 nm
1.
−
(6 marks) (d) Explain the physical meaning of the angle θ. (5 marks) (e) Evaluate the value of ψ (101) at (1/4, 1/4, 1). (5 marks)
7
MLE2101
Solution
(a) We first choose the basis vectors in the zone to be G1 = G(101) and G2 = G(110). With G1 and G2 , any reciprocal lattice point G in the zone can be written as G = u G1 + v G2 , where u and v can be determined by the parallelogram rule. For point A, u and v can be found to be -1 and 1, respectively, and the coordinates of A can be expressed as A =
−G1 + G2
1 −1 −2 = − 0 + 1 = 1 = (211).
(3.1)
0 1 −1 For point B, u = − 2 and v = − 1, B = − 2G1 − G 2 = (112). Point C, in comparison, is
opposite of B; thus, C = (112). (b) The line L 1 indicates the Fourier modes of a periodic function in the crystal structure along the direction from (000) to C = (112), i.e. along the direction of G(112) or the normal direction of (112). (c) Using the parallelogram rule determines u and v for point D to be u = 1 and v = 2; therefore, the coordinates of D are D = G1 + 2G2 = (321). The corresponding reciprocal lattice vector G(321) and its magnitude can be expressed as
−
−
G(321)
=
|G(321)|
=
−3 2π 2 , a 1 √ 2π 2 14π 3 +2 +1 = = 82.02 nm 2
2
a
(3.2) 1
−
a
.
(3.3)
Solving the equation above determines a to be
√
2 14 a = = 0.2866 nm. 82.02
(3.4)
(d) The angle θ is between GA and GB , i.e., between G(211) and G(112) . Since the two reciprocal lattice vectors give the normal directions of (211) and (112), the angle θ is equal to the angle between the two planes. (e) We first calculate the phase angle α of ψ (101) at (1/4, 1/4, 1), T
1 1/4 1 3π α = 2π 0 1/4 = 2π − 1 = − . 4 2 −1
(3.5)
1
With the knowledge of α, the value of ψ (101) at (1/4, 1/4, 1) can be evaluated to be ψ(101) = e
iα
= cos α + i sin α = cos
3π −2
+ i sin
3π −2
= i.
(3.6)
