PHY 140A: Solid State Physics
Solution to Midterm #1
TA: Xun Jia1
October 24, 2006
1 Email:
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Fall 2006
c Xun Jia (October 24, 2006)
°
Physics 140A
Problem #1
(20pt)Calculate the packing fraction of the body-centered cubic lattice.
Solution:
(Please refer √
Problem #3 in Homework #1.) For bcc lattice, consider the body
diagonal, we have 3a = 4r, thus
√
a = 4r/ 3 . . . . . . (4pt)
(1)
Moreover, there are 8 spheres at the corners, and 1/8 of each is contained in the
cube; there are 1 spheres at the center, which is totally contained in the cube; thus
the number of spheres in the cube is
n=8×
1
+ 1 × 1 = 2 . . . . . . (4pt)
8
(2)
The volume of the cube and the volume of each sphere are:
Vc = a3 . . . . . . (4pt)
4π 3
r . . . . . . (4pt)
Vs =
3
(3)
(4)
Therefore, the packing fraction is:
nVs
=
Vc
√
3π
= 0.68 . . . . . . (4pt)
8
(5)
Problem #2
(40pt)Experimentalists have created a room-temperature superconductor, and you
(a theorist) think its structure is simple cubic with a 6.0Å lattice constant. The
experimentalists have a powder Debye-Scherrer camera with a 1.5Å X-ray source.
Calculate the angular positions φ of the first 3 diffraction rings for a share of the
Nobel prize.
(a). Draw a picture that illustrates the diffraction condition relating G (a reciprocal
lattice vector), k (the wave vector of the incident X-ray), and k0 (the wave
vector of the scattered X-ray).
(b). Use the picture in part (a) to derive a relationship between |G|, |k|, and the
scattering angle φ.
(c). Write a symbolic expression for |G|.
(d). Use the results of parts (a)-(c) to calculate the angular positions φ of the first
3 diffraction rings. Your answer should be numeric, but please leave any nontrivial trigonometric functions, square roots, etc. unevaluated.
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Fall 2006
c Xun Jia (October 24, 2006)
°
Physics 140A
Solution:
(a). As in the Fig. :(8pt)
k'
G
k
Figure 1: Illustration of the diffraction condition.
(b). In the case of elastic scattering, |k| = |k0 |. From above figure, the relationship
among them are:
φ
2|k| sin = |G| . . . . . . (8pt)
(6)
2
(c). The reciprocal lattice for sc is also an sc. The expression of |G| is:
2π q 2
|G| =
v1 + v22 + v32 . . . . . . (8pt)
a
(d). Since |k| =
2π
, from above, we have the expression for φ:
λ
λq 2
φ = 2 arcsin
v + v22 + v32 . . . . . . (4pt)
2a 1
(7)
(8)
For the simple cubic in this problem, the reciprocal lattice is also simple cubic.
The first three rings of diffraction peak correspond to the case (v1 , v2 , v3 ) =
(100), (110), and (111). Substitute into Eqn. (8) with the value of wave length
λ, and lattice constant a, we have:
1
φ1 = 2 arcsin( ) . . . . . . (4pt)
8√
2
φ2 = 2 arcsin(
) . . . . . . (4pt)
√8
3
φ3 = 2 arcsin(
) . . . . . . (4pt)
8
(9)
Problem #3
Copper oxide layers: The common building blocks for most high temperature superconductors are copper oxide layers, as depicted below. Assume the distance between
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Fall 2006
c Xun Jia (October 24, 2006)
°
Physics 140A
the copper atoms (filled circles) is a. Sor simplicity, let us also assume that in the
third dimension these CuO2 layers are simply stacked with spacing c, and there are
no other atoms in the crystal. In the first approximation the layers have four-fold
symmetry; the crystal is tetragonal.
(a). For the crystal in Fig. 2:
• sketch the Bravais lattice.
• indicate a possible set of primitive vectors.
• and describe the basis.
• On the diagram, indicate a primitive cell.
(b). In some compounds closely related to the high-temperature superconductors
one finds, at closer inspection, that the CuO2 lattice is actually not flat, but
that oxygen atoms are moved a small amount out of the plane (’up’ or ’down’)
in an alternating fashion (see Fig. 3, ’+’ means up and ’-’ means down.) On
the diagram, indicate:
• the primitive cell.
• the lattice spacing.
• and the basis for the distorted crystal.
(c). What is the reciprocal lattice the the new (distorted) Bravais lattice?
• Calculate G(v1 , v2 , v3 ).
• Draw the reciprocal lattice plane with v3 = 0.
• Indicate the primitive reciprocal lattice vector b1 and b2 of the undistorted
lattice on the same drawing.
• Describe (qualitatively) what happens in the reciprocal lattice, and the
X-ray diffraction pattern, as the distortion is decreased gradually to zero.
Solution:
(a). For the crystal:
• Sketch the Bravais lattice. As shown in the dash line in Fig. 2. The Bravais
lattice in this 2d plane is a square structure.(2pt)
• Construct the coordinate as in the figure, then the primitive vectors are:
a1 = ax,
a2 = ay,
a3 = cz . . . . . . (3pt)
(10)
• The basis can be chosen as the copper atom at (0,0) and the oxygen atoms
at (0, a/2) and (a/2, 0), as marked out by a solid curve in the figure.
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Fall 2006
c Xun Jia (October 24, 2006)
°
Physics 140A
y
a
x
a2
a1
Figure 2: CuO2 plane of the superconductor.
• One primitive cell is the square with four copper atoms at its corners.
There are many other equally correct choices, for example, the square
with a copper atom at the center. Both are illustrated in the figure.
(b). For the distorted crystal:
• The primitive cell is shown as the shaded area in Fig. 3:(3pt)
• Now the primitive vector are:
a01 = a(x + y),
a02 = a(x − y),
a03 = cz
(11)
then the lattice spacing is:
a0 = |a01 | = |a02 | =
√
2a . . . . . . (3pt)
(12)
• The basis for the distorted lattice can be chosen as marked out by the solid
curve in the figure, that is, the copper atoms at (0, 0) and (a, 0), and the
oxygen atoms at (a/2, 0), (3a/2, 0), (a, a/2), and (a, −a/2).(4pt)
(c). For the new reciprocal lattice:
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Fall 2006
c Xun Jia (October 24, 2006)
°
Physics 140A
-
+
-
+
y
-
-
+
a
+
-
x
+
-
+
-
'
a
+
1
-
-
+
-
-
+
-
+
-
+
+
+
-
'
a
2
-
+
-
+
Figure 3: The distorted CuO2 plane of the superconductor.
• From the expression of a0i , we have:
(2π)a02 × a03
= 0
=
a1 · (a02 × a03 )
(2π)a03 × a01
0
b2 = 0
=
a1 · (a02 × a03 )
(2π)a01 × a02
b03 = 0
=
a1 · (a02 × a03 )
b01
π
(x + y)
a
π
(x − y)
a
(13)
2π
z
a
therefore, the reciprocal vector:
π
2π
π
G(v1 , v2 , v3 ) = v1 (x + y) + v2 (x − y) + v3 z . . . . . . (5pt)
a
a
a
(14)
• At v3 = 0, the reciprocal lattice plane is shown as in the Fig. 4:(5pt)
• For the undistorted lattice, from Eqn. (10) it is easy to get:
b1 =
2π
x,
a
b2 =
2π
y
a
(15)
and they are drawn in the Fig. 4.(5pt)
• In Fig. 4, the lattice points in the center of each square belong to the reciprocal lattice of the distorted lattice, but do not belong to the undistorted
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Fall 2006
c Xun Jia (October 24, 2006)
°
Physics 140A
y
b'
1
x
/a
b'
2
b
2
b
1
Figure 4: The reciprocal lattice in v3 = 0 plane.
lattice. Since each reciprocal lattice point corresponds to a diffraction peak
in the X-ray diffraction pattern, as the distortion is decreased gradually
to zero, the lattice points in the center of each square disappear, and the
diffraction peak corresponding to those points will get weaker and weaker,
and finally disappear. In other words, the structure factor of the reciprocal
lattice points in the centers of the squares goes to zero as the distortion
decreases. For the peaks corresponding to the rest of reciprocal lattice,
which belong to both distorted and undistorted lattice, the peak will get
stronger. (5pt)
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