Wave diffraction

Waves and Diﬀraction in Solids
Solution to Chapter Problems
Bruce M. Clemens
Stanford University
Winter 2002

ii
Copyright 2002 Bruce M. Clemens

Contents
1 Waves and Fourier Analysis 1
2 Group Velocity and Dispersion 9
3 Physical Examples of Waves 11
4 Wave Intensity and Polarization 15
5 Maxwell’s Equations: Light Waves 17
6 X-Ray Scattering From Electrons and Atoms 21
7 X-Ray Scattering From Crystals 23
8 Experimental Methods of Diﬀraction 43
9 Phonons: Lattice Vibrations and Modes 61
10 Quantum Mechanics Review 69
11 Quantized Phonon Behavior 77
12 Statistical Mechanics 81
iii

Chapter 1
Waves and Fourier Analysis
1. The Fourier transform of a Gaussian is given by:
F] [f(x)] =
∞
−∞
e−x2/2σ2
eikx
dx =
∞
−∞
exp
−x2
2σ2
+ ikx dx
We can perform this integral by completing the square in the exponent
of the integrand:
−x2
2σ2
+ ikx =
−1
2σ2
(x − g)2
+
g2
2σ2
where g is found by comparing terms which are ﬁrst order in x:
ikx =
2xg
2σ2
giving:
g = ikσ2
Hence our integral becomes:
F] [f(x)] = eg2/2σ2
∞
−∞
exp
−(x − g)2
2σ2
dx
= e−k2σ2/2
∞
−∞
e−z2/2σ2
dz
=
√
2π σ e−k2σ2/2
so that the Fourier transform of a Gaussian is another Gaussian.
1

2 CHAPTER 1. WAVES AND FOURIER ANALYSIS
To compare the widths of the function and its transform we need to
apply a consistent definition of a width. There are several we might
use, including the full-width-at-half-maximum, or the distance between
points where the function has fallen to a particular fraction of its maxi-
mum. Each choice will give similar relationships between the two width
of the function and its transform. For convenience we choose as a mea-
sure of the width the distance between the point where the function is
a maximum and the point where it is 1/e times its maximum value.
This 1/e point we designate as xe. Since the maximum occurs at x = 0
we have:
f(xe) =
1
e
f(0)
Hence:
x2
e
2σ2
= 1 or xe =
√
2 σ
Applying the same definition for the width of the Fourier transform we
find:
k2
e σ2
2
= 1 or ke =
√
2
σ
so that the width of the transform decreases if the width of the original
Gaussian function increases and vise-versa.
2. A periodic function can be written as a Fourier series:
f(x) =
∞
n=−∞
Cn e−iknx
where kn = 2πn/λ. Taking the Fourier transform we find:
F(k) =
∞
−∞
f(x) eikx
dx
=
∞
−∞
∞
n=−∞
Cn e−iknx
eikx
dx
=
∞
n=−∞
Cn
∞
−∞
ei(k−kn)x
dx
From Eqn. 1.15 in the notes, we see that the integral in parentheses
can be written as:
∞
−∞
ei(k−kn)x
dx = 2πδ(k − kn)

3
Hence the Fourier transform is given by:
F(k) =
∞
n=−∞
2πCn δ(k − kn)
We see that the Fourier transform of a periodic function is non-zero only
at k values corresponding to harmonic components of the periodicity.
3. (a) The plot of this pulse is shown in Fig. 1.1.
(b) The frequency components which make up the pulse can be found
by taking the Fourier transform:
F(ω) = F [f(t)]
=
∞
−∞
cos(ω0t)
1 + t2/σ2
eiωt
dt
= 2σ2
∞
0
cos(ω0t) cos(ωt)
σ2 + t2
dt
=
πσe−ω0σ
cosh(ωσ); ω < ω0
πσe−ωσ
cosh(ω0σ); ω > ω0
(c) We see that the frequency components are centered around the
frequency ω0 of the original pulse, but, due to the finite width of
the original pulse, the frequency spectrum is not a delta function,
but is spread out over some finite frequency range. The width
of the range in frequency is about 2/σ. Thus we see that the
product of the width in time (∼ 2σ) with the width in frequency
is a constant. Shorter pulses will have a broader frequency range
and hence a greater uncertainty in frequency.
4. (a) For this case we can see by inspection that the wave has only one
frequency; ωl. We can also see this by taking the Fourier transform
to find
F(ω) = E0
∞
−∞
e−i(ωl−ω)t
dt
= 2πE0 δ(ωl − ω)
(b) In this case the light is allowed through the shutter only for times
between −τ/2 and τ/2. Hence the distribution of frequencies in

1.0
0.5
0.0
-0.5
f
-4 -2 0 2 4
t/σ
Figure 1.1: Pulse amplitude as a function of time.
1.2
0.8
0.4
F(ω)/σ
-4 -2 0 2 4
(ω - ω ) σ
Figure 1.2: Fourier transform of pulse, showing the Fourier components which
make up the original waveform. The width is ∼ 2/σ.

5
this pulse (still given by the Fourier transform) is now
F(ω) = E0
τ/2
−τ/2
e−i(ωl−ω)t
dt
=
E0
i(ωl − ω)
ei(ωl−ω)τ/2
− e−i(ωl−ω)τ/2
=
2E0
ωl − ω
sin [(ωl − ω)τ/2]
This is a central peak of height E0τ centered on ωl surrounded by
oscillations. So due to the finite length, the pulse now has a range
of frequencies rather than just a single frequency. This is plotted
for a couple of cases in Figure 1.3.
20x10
-15
15
10
5
0
F(ω)/(E0)
1086420
ω (10
15
rad/sec)
τ = 10 fs
τ = 20 fs
Figure 1.3: Frequency distribution for a pulse of blue light (λ = 420 nm,
ωl = 4.488 × 1015
rad/sec) for pulse durations of 10 fs and 20 fs.
(c) Perhaps the easiest way to estimate the width is to examine the
separation between the points where the nominator goes to zero for
the first time on either side of the nominal frequency ωl. Defining
these points to be ω+ and ω−we find
(ωl − ω±)τ/2 = ±π
Hence
∆ω ≈ ω+ − ω− =
4π
τ

So that the more we shrink the length in time with wider the
spread in frequencies (and hence wavelengths). For the case shown
in Figure 1.3 we have pulse widths of 10 fs and 20 fs. This corre-
sponds to frequency ranges of 1.26×1015
rad/sec. and 0.628×1015
rad/sec. respectively. More interestingly this gives a spread in
wavelengths given by
∆λ = 2πc
1
ω−
−
1
ω+
= 2πc
∆ω
ω−ω+
≈ λl
∆ω
ωl
This gives a wavelength spreads of 118 nm and 59 nm for the two
pulse widths, both of which are appreciable fractions of the nomi-
nal wavelength. So the shorter the light pulse the more uncertain
its color.
5. (a) The Fourier transform of a delta function can be easily found by
recalling the sifting property of the delta function
∞
−∞
δ(x) f(x) dx = f(0)
Hence we ﬁnd:
i.
F(k) =
∞
−∞
δ(x) eikx
dx = 1
ii.
F(k) =
∞
−∞
[δ(x − a) + δ(0) + δ(x + a)] eikx
dx
= eika
+ 1 + e−ika
= 1 + 2 cos ka
iii.
F(k) =
∞
−∞
[δ(x − 2a) + δ(x − a) + δ(0) + δ(x + a) + δ(x + 2a)] eikx
dx
= ei2ka
+ eika
+ 1 + e−ika
+ e−i2ka
= 1 + 2 cos ka + 2 cos 2ka

7
Sketches are shown in Figure 1.4
(b) The result for part i is a constant and so has no maxima. The
peaks in parts ii and iii occure when ka = 2πn, or k = 2πn/a,
which is the one-dimensional Bragg condition.
(c) For the infinite number of delta functions we can generalize to find
F(k) = 1 + 2
∞
n=1
cos nka
It is tricky to do so, but it can be shown that this is also an infinite
string of delta functions. One might deduce this by observing that
in going from the three delta functions in part ii to the five in
part iii the peaks sharpened and the background dropped. Hence
extrapolating to an infinite number of delta functions will produce
another string of delta function peaks.
5
4
3
2
1
0
-1
F(k)
-6 -4 -2 0 2 4 6
ka
i
ii
iii
Figure 1.4: Fourier transform of delta functions.
6. From an above problem we find that the frequency distribution in a
pulse of with τ is given by
∆ω =
4π
τ

Hence the energy range is given by
∆E = ¯h∆ω = h∆ν =
2h
τ
Plugging in numbers we ﬁnd
∆E =
2 · 4.136 × 10−15
eV sec
50 × 10−15 sec
= 0.17eV

Chapter 2
Group Velocity and Dispersion
1. The given equation can be solved for the dispersion relationship ω(k):
ω(k) = ω2
c + k2c2
The phase velocity is then:
vp =
ω
k
=
ωc
k2
+ c2
and the group velocity is:
vg =
dω
dk
=
kc2
ω2
c + k2c2
=
kc2
ω
=
c2
vp
2. (a) The phase velocity is given by:
vp =
ω
k
=
T
ρ
+ αk2
=
T
ρ
1 +
αk2ρ
T
≈
T
ρ
+
αk2
2
ρ
T
The qroup velocity is given by:
vg = vp + k
dvp
dk
≈ vp + αk2 ρ
T
≈
T
ρ
+
3αk2
2
ρ
T
9

10 CHAPTER 2. GROUP VELOCITY AND DISPERSION
(b) For vibrations on a finite string the allowed wave numbers are
given by:
kn =
nπ
L
The allowed frequencies are:
ωn = vpkn ≈
nπ
L
T
ρ
1 +
αρ
2T
nπ
L
2
Thus the higher order (n > 1) frequencies are greater for the
stiff wire than for the flexible wire. This gives the overtones of
the instrument a distinctive sound. However, Dick Dale used stiff
strings because he wailed on them so hard that normal strings
broke too often.

Chapter 3
Physical Examples of Waves
1. We know that the solutions for waves on a finite string of length L are
given by:
ξn(x, t) = Bn sin (knx)
where the wavenumbers are restricted to maintain an integral number
of half wavelengths in the length of the string:
kn =
nπ
L
; λn =
2L
n
The allowed frequencies are just given by:
ωn = vkn =
T
ρ
nπ
L
These solutions are just the Fourier components for a disturbance with
a wavelength λ = 2L. We can find the magnitude of each harmonic
component by finding the corresponding Fourier coefficient for the ini-
tial disturbance. Hence we find;
Bn =
1
2L
L
−L
f(x) sin
2nπx
2L
dx
=
1
L
L
0
f(x) sin(knx) dx
=
1
L
L/2
0
2bx
L
sin(knx) dx +
L
L/2
2b −
2bx
L
sin(knx) dx
=
4b
n2π2
(−1)
n−1
2 ; n = 1, 3, 5, 7, . . .
11

12 CHAPTER 3. PHYSICAL EXAMPLES OF WAVES
So only the odd harmonics are present.
2. The transducer launches a wave which can be represented as
ξ = A ei(ωt−kz)
Hence the phase diﬀerence ∆φ for the wave at the source and transducer
is just
∆φ = kL
The experimental observation that ∆φ is linear with frequency ω means
that
kL = mω
Hence the velocity is given by
v =
ω
k
=
L
m
Therefore we can ﬁnd the modulus from
Y = v2
ρ
=
ρL2
m2
3. (a) The group velocity is:
vg =
dω
dk
=
aη1/2
cos ka
2
if 2nπ < ka
2
≤ (2n + 1)π
−aη1/2
cos ka
2
if (2n + 1)π < ka
2
≤ 2(n + 1)π
(b) For small k, cos ka/2 ≈ 1, so the group velocity is:
vg ≈ ±aη1/2
where the sign is the sign of k. This is a constant, independent
of k, so in this limit the phase and group velocities are equal, and
there is no dispersion.
(c) The group velocities will be zero when:
cos
ka
2
= 0

13
This will occur when:
ka
2
=
(2n + 1)π
2
or:
k =
(2n + 1)π
a
This is the Brillouin zone boundaries for this one-dimensional case.
Hence the group velocity is zero when the wave number of the
phonon satisﬁes the Bragg condition for diﬀraction.

14 CHAPTER 3. PHYSICAL EXAMPLES OF WAVES

Chapter 4
Wave Intensity and
Polarization
1. (a) The power is proportional to the square of the wave:
P ∝ ξ2
= ξ2
0 cos2
(kz − ωt) =
ξ2
0
2
[1 + cos(2kz − 2ωt)]
So the power has a DC component with a magnitude of ξ2
0/2 and
an oscillatory part with a the same magnitude with twice the
frequency of the original wave.
(b) For this wave, in the z = 0 plane, the angle between the x-axis
and a line connecting origin to the wave is given by:
tan θ =
ξy
ξx
=
sin(kz − ωt)
cos(kz − ωt)
so that:
θ = kz − ωt
As t increases, θ decreases, so this is a left handed circularly po-
larized wave. The power is again proportional to the square of the
amplitude:
P ∝ ξ2
= ξ2
0 cos2
(kz − ωt) + sin2
(kz − ωt) = ξ2
0
which means that the power is constant in time.
15

16 CHAPTER 4. WAVE INTENSITY AND POLARIZATION
2. (a) For t = 0, the points at which the two contributions to this wave
are in-phase are given by the solutions to the equation:
k1xp = k2xp ± 2πn
where n is an integer. From this we ﬁnd:
xp = ±
2πn
k1 − k2
(b) The times at which the contributions are in-phase at the position
x = 0 are found by solving:
ω(k1)tp = ω(k2)tp ± 2πn
This yields:
tp = ±
2πn
ω(k1) − ω(k2)
(c) For non-zero time, the positions where the contributions to the
wave are in-phase are a function of time. We denote these posi-
tions as xp(t). They can be found by solving the following equa-
tion:
k1xp(t) − ω(k1)t = k2xp(t) − ω(k2)t ± 2πn
Diﬀerentiating yields:
dxp(t)
dt
=
ω(k1) − ω(k2)
k1 − k2

Chapter 5
Maxwell’s Equations: Light
Waves
1. (a) Substitution of the given solution into the wave equation gives:
−k2
E = −ω2
r 0µrµ0E − iωµrµ0σE
dividing through by −ω2
E yields:
k2
ω2
=
1
v2
p
= r 0µrµ0 +
iµrµ0σ
ω
(b) The square of the index of refraction is given by:
n2
=
c2
v2
p
=
1
0µ0
r 0µrµ0 + i
µrµ0σ
ω
= rµr + i
µrσ
0ω
(5.1)
The complex part of n can be found by writing:
n2
= Meiφ
where:
M = ( rµr)2 +
µrσ
0ω
2
and φ = arctan
σ
r 0ω
Then n is just:
n =
√
n2 =
√
M eiφ/2
17

18 CHAPTER 5. MAXWELL’S EQUATIONS: LIGHT WAVES
which has imaginary part:
[n] = β =
√
M sin(
φ
2
)
so that n is a complex quantity, and we can write:
n = n0 + iβ
where n, β ∈ R.
We can come up with an algebraic relationship between β and the
materials constants by writing n2
as:
n2
= (n0 + iβ)2
= n2
0 − β2
+ 2in0β
comparing this with Eqn. 5.1 above, we equation the real and
imaginary parts to find:
n2
0 − β2
= rµr ; 2n0β =
µrσ
0ω
These two equations can be solved for β to find:
β = −
rµr
2
+
1
2
( rµr)2 +
µrσ
0ω
2
As a side note, we can also solve for n0 to find:
n0 =
rµr
2
+
1
2
( rµr)2 +
µrσ
0ω
2
Hence for nonconducting media, where σ = 0, we find:
β = 0 ; n = n0 =
√
rµr
The relationship between the imaginary part of n and the absorp-
tion coefficient is explored in the notes.
(c) From the notes we find:
α =
2ωβ
c
which, combined with the above gives the desired result.

19
2. (a) The expression for light which is a right-circularly polarized wave
at z = 0 and is traveling in the ˆz direction through media with
indices of refraction nx and ny for the x and y components of the
field respectively is given by:
E = E0 cos ω
nxz
c
− t ˆx − sin ω
nyz
c
− t ˆy
We see that in the x−y plane at z = 0, the vector E would have a
fixed length and rotate in a clockwise direction when viewed from
negative to positive z.
(b) As we move through the media, we see that the phases of the
two components change at a different rate due to the difference
in refractive indices. The light will be linearly polarized when the
two components are in phase. Since they were originally out of
phase by π/2 at z = 0, this will occur at a position zL which
satisfies:
ω
nxzL
c
= ω
nyzL
c
+
π
2
Solving for zL we find:
zL =
πc
2ω(nx − ny)
Inserting this position into our field we find:
E(zL, t) = E0 cos ω
nxzL
c
− t ˆx − sin ω
nyzL
c
− t ˆy
= E0 cos
nxπ
2(nx − ny)
− ωt ˆx
− sin
nyπ
2(nx − ny)
− ωt ˆy
= E0 cos
nxπ
2(nx − ny)
− ωt ˆx
− sin
nxπ
2(nx − ny)
−
π
2
− ωt ˆy
= cos
nxπ
2(nx − ny)
− ωt (ˆx + ˆy)

20 CHAPTER 5. MAXWELL’S EQUATIONS: LIGHT WAVES
If we plot E on the x−y plane at z = zL we ﬁnd that it is linearly
polarized and oscillates with time along a line at 45◦
between the
+x and +y axes.

Chapter 6
X-Ray Scattering From
Electrons and Atoms
1. (a) The scattering factor for a sphere of uniform charge density of
radius R0 is given by:
f =
∞
0
4πr2
ρ(r)
sin(qr)
qr
dr
=
R0
0
4πr2
ρ0
sin(qr)
qr
dr
=
4πρ0
q3
[sin(qR0) − qR0 cos(qR0)]
=
3Z
(qR0)3
[sin(qR0) − qR0 cos(qR0)]
where we have introduced the total charge Ze as:
Ze =
4
3
πR3
0ρ0
Hence the intensity is given by:
I =
E0re
R
2
3Z
(qR0)3
2
[sin(qR0) − qR0 cos(qR0)]2
(b) This is shown in Fig. 6.1, where it is clear that there will be zero’s
of the scattered intensity when:
tan(q0R0) = q0R0
21

10
-8
10
-6
10
-4
10
-2
I(q)
151050
qR
22CHAPTER 6. X-RAY SCATTERING FROM ELECTRONS AND ATOMS
so from successive minima in the scattering, which correspond to
successive roots for q0, we can find the radius of the sphere.
Figure 6.1: Scattered intensity for uniform sphere.
2. The scattering factor can be found by using the expression for spheri-
cally symmetric electron distributions:
f(q) =
∞
0
4πr2
ρ(r)
sin qr
qr
dr
For the present case we find:
f(q) = ρ1
R1
0
4πr2 sin qr
qr
dr + ρ2
R2
R1
4πr2 sin qr
qr
dr
= (ρ1 − ρ2)
R1
0
4πr2 sin qr
qr
dr + ρ2
R2
0
4πr2 sin qr
qr
dr
Recall from the solutions to the above problem we had the integral:
R
0
4πr2 sin qr
qr
dr =
4π
q3
[sin(qR) − qR cos(qR)]
Applying this result to the present case we find the scattering factor:
f(q) =
4π
q3
{(ρ1 − ρ2) [sin(qR1) − qR1 cos(qR1)] +
ρ2 [sin(qR2) − qR2 cos(qR2)]}

Chapter 7
X-Ray Scattering From
Crystals
1. (a) Since the expression:
sin2
(Nx)
sin2
(x)
is invariant with respect to the substitution x → x + nπ, we only
need consider the limit as x → 0. Furthermore we can find the
desired limit by squaring the limit of the ratio:
lim
x→0
sin(Nx)
sin(x)
which can be found using L’Hospital’s rule:
lim
x→0
sin(Nx)
sin(x)
= lim
x→0
N cos(Nx)
cos(x)
= N
so the maximum in intensity is N2
.
(b) In finding the width of a Bragg peak, again the periodicity of the
intensity function allows us to consider only the width of the peak
near the origin. To find this we find the value for x where the
intensity has dropped to half its maximum value:
sin2
(Nxw)
sin2
(xw)
=
N2
2
23

24 CHAPTER 7. X-RAY SCATTERING FROM CRYSTALS
so that the full width at half maximum will be 2xw. Rearranging
we find:
sin(Nxw) =
N
√
2
sin(xw) (7.1)
This equation can be solved numerically for a particular value of
N, or we can expand the sin functions about appropriate values
and obtain an approximate analytical solution. The intensity ex-
pression will have zero’s at x0 = nπ/N, where n is an integer,
so it is reasonable to expect that the half maximum will near to
halfway between the origin and the first zero, that is xw ∼ π/(2N).
Thus we can expand the function sin x around 0, but must expand
sin(Nx) around Nx = π/2. Performing these expansions we find:
1 −
N2
2
xw −
π
2N
2
=
Nxw
√
2
Solving we find:
xw = 0.886
π
2N
The actual numerical factor is 0.890, so this series method comes
within half of a percent of the correct answer.
The width of the peak in x is given by:
w = 2xw = 0.890
π
N
This gives a width of a peak in q of:
∆q =
2w
a
= 0.890
2π
Na
= 0.890
2π
L
To find the width in θ we differentiate the expression:
q =
4π
λ
sin θ
to obtain:
∆q =
4π
λ
cos θ ∆θ
Rearranging we find:
L =
0.89 λ
cos θ ∆(2θ)

25
(c) The subsidiary maximum between the Bragg peaks occur when
the numerator of the intensity function has maxima. This occurs
at:
xsm = n +
1
2
π
N
; n = 1, 2, 3, . . . , N − 1
There are a total of N −2 of these maximum between Bragg peaks.
2. (a) To calculate accurately the thickness of the film using a ruler and
Fig. 8.11, we need to find a dimension in the figure which can be
accurately measured with a ruler and which correlates with the
thickness of the film. We could measure the main peak width and
extract from that the thickness of the film as in the problem in the
homework, but it is difficult to measure the width of such a narrow
peak with just a ruler. The many subsidiary maxima provide us
with a better way. These subsidiary maxima are have spacing in
q of ∆q = 2π/t, where t is the film thickness. We can measure the
width in angle for a given number of oscillations and accurately
extract the average oscillation width in that range by dividing the
angular width for many oscillations by the number of oscillations.
I found that 32 fringes on the left-hand side of the peak covered
a linear distance on the figure of 34.8 mm. I measured also that
there were about 6.45 mm/degree on this figure, so I find the
average angular width for the fringes in this region to be ∆(2θ) =
0.169 degrees, so that the angular width of a fringe in radians is
∆θ = 1.47 × 10−3
radians. Differentiating the expression for q we
find:
∆q =
4π
λ
cos θ ∆θ
So we find:
t =
2π
∆q
=
λ
2 cos θ ∆θ
Which for 2θ ≈ 28.5 gives t ≈ 435˚A.
(b) For crystals with two different thicknesses we have fringe spacing
given by:
∆q1 =
2π
t1
; and ∆q2 =
2π
t2
Near the Bragg peak the subsidiary maxima for the two crystals
will add together, but as we move away from the Bragg peak maxi-
mum we the differences in spacing will add up until the maximum

for one crystal will be at the position of the minimum for the
other. At this point the subsidiary maxima will be washed out.
This is exactly the mechanism for washing out of finite thickness
oscillations due to a continuous distribution of thicknesses.
The maxima and minima for the two crystals coincide at Nw oscil-
lations away from the Bragg peak the difference in position of the
maximum for the two crystals is equal to half the average spacing
of the subsidiary maxima, or:
Nw(∆q1 − ∆q2) =
∆q1 + ∆q2
4
Inserting ∆q1 = 2π/t1 and ∆q2 = 2π/t2, we find:
Nw =
∆q1 + ∆q2
2(∆q1 − ∆q2)
=
1
4
2π/t1 + 2π/t2
2π/t1 − 2π/t2
=
1
4
t2 + t1
t2 − t1
=
1
2
t
∆t
where t = (t1 + t2)/2 is the average thickness, and ∆t = t2 − t1.
So we have that the difference in thickness is given by:
∆t =
t
2Nw
since there are about 50 oscillations observed on either side of the
peak, we have that the difference in thickness is given by:
∆t =
435˚A
100
≈ 4˚A
which is about two lattice planes.
3. (a) For the FCC real space lattice we find:
a2 × a3 =
a2
4
ˆx ˆy ˆz
0 1 1
1 0 1
=
a2
4
(ˆx + ˆy − ˆz)

27
so that the cell volume is given by:
Vcell = a1 · a2 × a3 =
a3
8
(1 + 1) =
a3
4
and the reciprocal lattice vector b1 is given by:
b1 =
a2 × a3
a1 · a2 × a3
=
1
a
(ˆx + ˆy − ˆz)
Similarly we find:
b2 =
1
a
(−ˆx + ˆy + ˆz)
b3 =
1
a
(ˆx − ˆy + ˆz)
This is just a BCC lattice with lattice parameter 2/a.
(b) For the BCC real space lattice, we find:
a2 × a3 =
a2
4
ˆx ˆy ˆz
−1 1 1
1 −1 1
=
a2
2
(ˆx + ˆy)
so that the cell volume is given by:
Vcell = a1 · a2 × a3 =
a3
4
(1 + 1) =
a3
2
and the reciprocal lattice vector b1 is given by:
b1 =
a2 × a3
a1 · a2 × a3
=
2
a3
a2
2
(ˆx + ˆy) =
1
a
(ˆx + ˆy)
Similarly we find:
b2 =
1
a
(ˆy + ˆz)
b3 =
1
a
(ˆx + ˆz)
Which is an FCC lattice with lattice parameter 2/a. Hence the
reciprocal lattice of BCC if FCC and vise versa.

(c) For the hexagonal real space lattice we ﬁnd:
a2 × a3 =
ˆx ˆy ˆz
−
√
3
2
a a
2
0
0 0 c
=
ac
2
ˆx +
√
3 ac
2
ˆy
so that the cell volume is:
Vcell = a1 · a2 × a3 =
√
3 a2
c
2
and we ﬁnd for the reciprocal lattice vectors:
b1 =
1
a
√
3
3
ˆx + ˆy
b2 =
1
a
−
√
3
3
ˆx + ˆy
b3 =
1
c
ˆz
This reciprocal lattice is also a hexagonal lattice with a lattice
parameter 2
√
3/(3a), and c lattice parameter 1/c. It is also rotated
90◦
relative to the original lattice.
a1
a2
b1b2
a) b)
Figure 7.1: Schematic of hexagonal lattice a) and its reciprocal b).

29
4. The unit cell of diamond cubic can be taken to be FCC with a basis
of (000), and a/4(111), so there will be eight terms in the sum for the
structure factor, corresponding to the atoms at:
(000) a
4
(111) a
2
(110) a 3
4
3
4
1
4
a
2
(101) a 3
4
1
4
3
4
a
2
(011) a 1
4
3
4
3
4
We ﬁnd:
F(q) =
n
fneiq·rn
= f 1 + eiπ(h+k+l)/2
+ eiπ(h+k)
+ eiπ(3h+3k+l)/2
+
eiπ(h+l)
+ eiπ(3h+k+3l)/2
+ eiπ(k+l)
+ eiπ(h+3k+3l)/2
= 1 + eiπ(h+k)
+ eiπ(h+l)
+ eiπ(k+l)
1 + eiπ(h+k+l)/2
=



8 h, k, l = even and
h + k + l = 4 × integer
4 (1 i) h, k, l odd
0 otherwise
In the h, k, l odd case, the minus sign occurs if (h + k + l + 1)/2 is even
and the plus sign if (h + k + l + 1)/2 is odd.
5. (a) The lattice vectors of the orthorhombic lattice can be taken as:
a1 = aˆx
a2 = bˆy
a3 = cˆz
The cell volume is obviously Vcell = abc and the reciprocal lattice
vectors are:
b1 =
1
a
ˆx
b2 =
1
b
ˆy
b3 =
1
c
ˆz

(b) The reciprocal lattice vectors which correspond to the planes (h1, k1, l1)
and (h2, k2, l2) are:
Gh1k1l1 =
h1
a
ˆx +
k1
b
ˆy +
l1
c
ˆz
Gh2k2l2 =
h2
a
ˆx +
k2
b
ˆy +
l2
c
ˆz
The spacing between sets of (hkl) planes is given by:
dhkl =
1
|Ghkl|
=


h
a
2
+
k
b
2
+
l
c
2


−1/2
The angle θ between these two plane normals is found by:
cos θ =
Gh1k1l1 · Gh2k2l2
Gh1k1l1 Gh2k2l2
= dh1k1l1 dh2k2l2
h1h2
a2
+
k1k2
b2
+
l1l2
c2
6. In general the scattered amplitude from a collection of scatterers is
given by (in the kinematic approximation):
=
E0re
R
exp [i(ωt − kR)]
p
fpeiq·Rp
where the sum runs over all the scatterers in the collection. In this
case, the position vectors can be written as:
Rp =
paˆx −(N − 1)/2 ≤ p ≤ (N − 1)/2
0 |p| > (N − 1)/2
so the sum in the above expression for scattered amplitude becomes:
p
fpeiq·Rp
=
(N−1)/2
p=−(N−1)/2
fpeipqxa
We can perform this sum by making the substitution p = p+(N −1)/2.
This leads to:
p
fpeiq·Rp
= e−iqxa(N−1)/2
N−1
p =0
fpeip qxa
= f
e−iqxaN/2
e−iqxa/2
1 − eiqxaN
1 − eiqxa
= f
sin qxaN/2
sin qxa/2

31
where we have pulled f out of the sum since for this problem they are
identical scatterers.
In this problem you are asked to describe the scattered amplitude, so
you must describe the behavior of the scattered amplitude as a function
of the scattering vector q.
(a) In this part the scatterers are point charge electrons. Hence f = 1.
The scattered amplitude is independent of the y and z compo-
nents of the scattering vector so that the scattered amplitude is
everywhere the same on planes which are perpendicular to the x
direction. The scattered amplitude has a maximum in magnitude
at the condition qxa/2 = nπ, which leads to a condition for the
maximum of:
qmax
x =
2πn
a
which is really the Bragg condition for this one-dimensional crys-
tal. However, instead of Bragg points, we have Bragg planes of
constant amplitude perpendicular to the x axis with spacing 2π/a.
A scattering vector which terminates on one of these Bragg planes
satisﬁes the condition for a maximum in scattered amplitude.
(b) For the case where we have atoms rather than point electrons, the
atomic scattering factor f is no longer a constant independent of
q. Hence the magnitude of the scattered amplitude is no longer
a constant on one of the Bragg planes, and drops oﬀ with the
dependence of f(q) away from the x-axis.
7. (a) The scattering factor for this electron distribution is found by
taking the Fourier transform in the normal fashion:
f = ρ eiq·r
dV
= ρ0
∞
∞
δ(z) eiqzz
dz
a/2
−a/2
eiqxx
dx
a/2
−a/2
eiqyy
dy
= a2
ρ0
sin qxa/2
qxa/2
sin qya/2
qya/2
(b) For q along the z-axis, the scattering factor becomes
f = a2
ρ0

which is just the total number of electrons in each plate. The
diffracted intensity from N of these plates stacked along the z-
axis with spacing d is then just
I =
I0r2
e
R2
(a2
ρ0)2
N−1
n=0
einqzd
2
= I0
reaρ0
R
2 sin2
Nqzd/2
sin2
qZd/2
This, of course, will have peaks at qz = 2Π/d.
8. The general expression for scattered intensity in electron units from
a collection of atoms can be found by squaring and normalizing Eqn.
8.11:
Ieu =
p
fpeiq·Rp
2
where the sum runs over all the atoms in the solid. Note that the
exponent in the sum involves the dot product of q and the position
vector for the atoms. Hence any change in the position of the atoms
which does not have a component along the direction of q will have no
effect on the observed diffraction. Hence, movement of the atoms in a
direction perpendicular to [111] direction will not affect the (111) peak,
so the stacking faults as described in this problem have no effect.
9. The scattered intensity is given by our general formula:
I(q) =
E0re
R
|F(q)|2
3
j=1


sin Nj
q·aj
2
sin q·aj
2


2
In this case N+1 = N2 = 100 and N3 = 10. We have a simple cubic
structure so F(q) is just the atomic scattering factor f(q). There will
be peaks at
qB = 2πGhkl = 2π h
ˆx
a
+ k
ˆy
a
+ l
ˆz
a
The peaks will have widths:
∆qx = ∆qy =
2π
100a
; ∆qz =
2π
10a
A cross section of the x − z plane of this intensity is shown in Fig. 7.2.

33
Figure 7.2: Intensity distribution in qx and qz for a simple cubic crystal with
100 atoms in the x and y direction and 10 atoms in the z direction. The Cu
atomic scattering factor was used.

10. (a) The scattered amplitude in electron units is found by summing
over the scattering centers:
F(q) =
p
fpeiq·rp
where fp is the scattering factor for the point charges, which for
this case is the same for each point charge. We can write the
position vectors as:
rp = m1aˆx+m2aˆy ; m1 = 0, 1, 2, . . . , N−1 ; m2 = 0, 1, 2, . . . , N−1
so the dot product in the exponent becomes:
q · rp = m1aqx + m2aqy
where qx and qy are the x and y components of the scattering
vector. The scattered amplitude then becomes:
F(q) = f
N−1
m1=0
N−1
m2=0
exp [i(qxm1a + qym2a)]
= f


N−1
m1=0
eiqxm1a




N−1
m2=0
eiqym2a


= f
1 − eiqxNa
1 − eiqxa
1 − eiqyNa
1 − eiqya
= feiφ sin(qxNa/2)
sin(qxa/2)
sin(qyNa/2)
sin(qya/2)
The scattered intensity is then:
I(q) =
E0re
R
2
f2 sin(qxNa/2)
sin(qxa/2)
2
sin(qyNa/2)
sin(qya/2)
2
(b) This will have peaks at:
qxa
2
= n1π ;
qya
2
= n2π
with n1, n2 ∈ I. Note that this places no condition on qz, so
that any qz will satisfy the Bragg condition, as long as qx and qy
satisfy the above. Hence the diﬀraction features are continuous
rods along the z-direction.

35
11. (a) The scattering factor is found by integration:
f(q) = ρ(r) eiq·r
dV
= ρA
tA/2
−tA/2
L/2
−L/2
L/2
−L/2
ei(qxx+qyy+qzz)
dx xy dz
= ρAVA
sin qxL/2
qxL/2
sin qyL/2
qyL/2
sin qztA/2
qztA/2
where VA = L2
tA is the volume of A in one bilayer.
(b) The scattered intensity for N layers stacked on top of one another
is given by:
I =
I0r2
e
R2
N−1
n=0
fn(q)eiq·Rn
2
=
I0r2
e
R2
f2
(q)
N−1
n=0
eiqzn(tA+tB)
2
=
I0r2
e
R2
f2
(q)
sin2
(NqzΛ/2)
sin2
(qzΛ/2)
where Λ = tA + tB.
(c) A scan varying qz while holding qx = qy = 0 will have peaks at:
qzB =
2nπ
Λ
with weighting given by:
I0r2
e
R2
(ρAVAN)2 sin qztA/2
qztA/2
2
This weighting will have zeros spaced in qz by 2π/tA. This is
shown for a particular example in Fig. 7.3.
12. (a) The structure factor for this molecule is given by:
F(q) =
n
fn eiq·rn
= f 1 + eiq·r

0.001
0.01
0.1
1
10
I
0.200.150.100.050.00
q
Figure 7.3: Intensity for a multilayer with N = 20 layers with thicknesses
tA = 80 and tB = 20.
where r is the vector connecting the two atoms in the molecule
and the atomic scattering factor is givey by:
f(q) =
1
(1 + a2q2/4)2
as found in the notes in Eqn. 7.11.
The scattered intensity in electron units is then:
Ieu = |F(q)|2
= f2
1 + e−iq·r
+ eiq·r
+ 1
= 2f2
(1 + cos q · r)
i. For q parallel to r this reduces to:
F (q) = 2f2
(1 + cos qr)
ii. For q perpendicular to r this reduces to:
F⊥(q) = 4f2
These cases are shown in Fig. 7.4. Note that both cases converge
to 4 as q goes to zero. Note also for that for q parallel to r the
interference between the two atoms results in zeros in the intensity
at qr = π, 3π, 5π etc.

10
-6
10
-4
10
-2
10
0
I(eu)
121086420
qr
parallel
perpendicular
37
Figure 7.4: Intensity for q parallel to r and q perpendicular to r for the
hydrogen molecule discussed in the problem.
(b) For a gas of these molecules, we must average over all possible
orientations which the vector r can be relative to the scattering
vector q for the experiment. In our expression for intensity, the
only term which depends on this orientation is the term containing
q · r, so we ﬁnd:
|F(q)|2
= 2f2
(1 + cos q · r )
where the notation · · · means the average of the quantity en-
closed.
To ﬁnd this average, we must integrate over all possible values of
the orientation of r. Since r can point in any direction with equal
preference, the endpoints of all the vectors r for all the molecules
in the gas uniformly cover the surface of a sphere of radius r. The
number of endpoints per area on the sphere is just N/4πr2
, where
N is the number of molecules in the gas. The average of a cos q·r
can then be found by integrating over the area of the sphere and
dividing by the density, as shown schematically in Fig. 7.5:
cos q · r =
1
4πr2
π
0
cos(qr cos φ) 2πr2
sin φ dφ
=
1
2
1
−1
cos(qrx) dx
=
sin qr
qr

10
-5
10
-4
10
-3
10
-2
10
-1
10
0
I(eu)/N
121086420
qr
r
q
φ
Figure 7.5: Schematic of sphere of radius r with vectors q and r and angle
φ.
So the scattered intensity from the gas is given by:
|F(q)|2
= 2Nf2
1 +
sin qr
qr
This is shown plotted in Fig. 7.6. Note that there are no real
distinctive features in this graph, so to find r we would have to
fit the observed intensity with our expression using r as a fitting
parameter.
Figure 7.6: Intensity from a gas of hydrogen molecules.
13. (a) We start with the general expression for the scattered intensity

39
for a collection of atoms
Ieu =
p
fpeiq·rp
2
For this case the positions of the atoms can be written as a sum
of the position of the cell plus the position within the cell:
rp = 2maˆx + rn
where rn is the position of the atom within the cell and takes the
values:
r1 = 0 and r2 = (a + δ)ˆx
for the two atoms within each cell. Hence the intensity becomes:
Ieu = f2
N−1
m=0
eiqx2am
2
1 + eiqx(a+δ) 2
= 2f2 sin qxaN
sin qxa
2
[1 + cos qx(a + δ)]
where qx is the component of q along the x-direction, and where
we have used the sum we performed earlier in this class. This is a
series of planes of intensity perpendicular to the x-axis and with
spacing π/a.
For small δ we find
Ieu ≈ 2f2 sin qxaN
sin qxa
2
1 + cos qxa − qxδ sin qxa −
(qxδ)2
2
cos qxa
(b) At the point q = πˆx/a we find
Ieu = 2f2
N2
[1 + cos π(1 + δ/a)]
≈ f2
N2 πδ
a
2
(c) At the point q = 2πˆx/a we find
Ieu = 2f2
N2
[1 + cos 2π(1 + δ/a)]
≈ 4f2
N2

1 −
πδ
a
2



So compared to a structure where the spacing between the atoms
is the same (δ = 0) we see extra peaks at π/a, 3π/a, 5π/a, . . ..
These peaks have intensity which scales with the displacement δ
squared, so that as δ → 0 they go away.
14. (a) The fcc lattice has four points per conventional cell. The ZnS
structure has two atoms (one Zn and one S) for each of the fcc
lattice points. Hence there are 8 atoms per conventional unit cell.
(b) The structure factor for ZnS can be found by summing over the 8
atoms in the unit cell
F =
n
fne2πi(hxn+kyn+zln)/a
= 1 + eiπ(h+k)
+ eiπ(h+l)
+ eiπ(k+l)
fZn + fSeiπ(h+k+l)/2
(c) The ﬁrst 6 peaks are listed below
peak hkl F/4
1 111 fZn − ifS
2 200 fZn − fS
3 220 fZn + fS
4 311 fZn + ifS
5 222 fZn − fS
6 400 fZn + fS
(d) The calculated d-spacings (using Bragg’s law) for the patterns are
shown below. By examining ratios of these d-spacings the patterns
can be indexed as follows
Ambient Pressure
peak 2θ d (˚A) hkl
1 23.72 3.75 111
2 27.46 3.24 200
3 39.23 2.29 220
4 46.36 1.96 311
5 48.55 1.87 222
6 56.68 1.62 400
High Pressure
peak 2θ d (˚A) hkl
1 26.04 3.42 111
2 30.15 2.96 200
3 43.17 2.09 220
4 51.11 1.79 311
5 53.56 1.71 222
6 62.70 1.48 400

41
(e) Any of the peaks for the two phases can be used to find the lattice
parameter through
d =
a
√
h2 + k2 + l2
We find for the ambient pressure phase a = 6.49 ˚A, and for the
high pressure phase a = 5.92 ˚A.
(f) The structure factors for the peaks for the ambient pressure phase
are consistent with the ZnS structure, while those of the high-
pressure phase are consistent with the NaCl structure (from the
midterm). Hence we see that increasing pressure drives the phase
transition ZnS → NaCl
15. (a) The structure factor is found by summing over the atoms in the
unit cell
Fhkl =
Nb
n
fnei2π(hxn/a+kyn/a+lzn/a)
= f 1 + eiπ(h+k+l+lδ)
= f 1 + eiπ(h+k+l)
eiπlδ
= f 1 ± eiπlδ
where the plus sign is for h + k + l ∈ even, and the minus sign is
for h + k + l ∈ odd.
(b) The square of the structure factor is
|Fhkl|2
= f2
1 ± eiπlδ
1 ± e−iπlδ
= 2f2
(1 ± cos πlδ)
(c) To expand this for small δ we recall that
cos φ ≈ 1 −
φ2
2
Applying this we find
|Fhkl|2
≈ 2f2
1 ± 1 −
(πlδ)2
2
= f2 4 − (πlδ)2
h + k + l ∈ even
(πlδ)2
h + k + l ∈ odd

A change in δ only makes a small relative change in the intensity
of the h+k+l ∈ even peaks, while a change in δ makes a relatively
large change in the intensity of the h+k+l ∈ odd peaks. Hence to
examine changes in δ we will examine the corresponding changes
in the h + k + l ∈ odd peaks.

Chapter 8
Experimental Methods of
Diffraction: Adventures in
Reciprocal Space
1. (a) The reciprocal lattice vector for the (200) planes is given by:
G200 = 2b1 =
2
a
ˆx
So when aligned to observe the (200) peak, q is given by:
q = 2πG200 =
4π
a
ˆx
(b) The path length difference ∆L for the x-rays scattered by planes
separated by a distance a along the direction of q is just:
∆L = 2a sin θ
when on the (200) Bragg peak, the scattering angle satisfies:
λ = 2d sin θ200
where the plane spacing d200 is just given by:
d200 =
a
2
43

44 CHAPTER 8. EXPERIMENTAL METHODS OF DIFFRACTION
solving for sin θ200 we find:
sin θ200 =
λ
a
which gives:
∆L = 2λ
2. (a) The general reciprocal lattice vector for this cubic lattice is given
by
Ghkl =
1
a
(hˆx + kˆy + lˆz)
Therefore, the structure factor for CsCl is
Fhkl = fCs + fCleiπ(h+k+l)
For the (100) peak this becomes
F(100) = fCs − fCl
and for the (200) peak we find
F(200) = fCs + fCl
(b) The structure factor for the chemically disordered BCC lattice is
FBCC
hkl = f 1 + eiπ(h+k+l)
As we have seen in class, this gives no peak for the case h+k+l =
odd, while the CsCl structure gives the difference between the
atomic scattering factors for this condition. Hence by examining
the intensity for the k + k + l = odd reflections we can determine
if the structure is ordered into the CsCl structure or is chemically
disordered and so is the BCC structure.
3. The reciprocal lattice vectors for the orthorhombic system are given
by:
b1 =
ˆx
a
; b2 =
ˆy
b
; b3 =
ˆz
c

45
The Bragg condition becomes:
qB = 2πGhkl
= 2π
hˆx
a
+
kˆy
b
+
lˆz
c
For the primitive lattice there is only one atom at position r1 = 0 so
the the structure factor is:
F(q) = f(q)
For the base-centered lattice, there are two atoms at positions:
r1 = 0 ; and r2 =
aˆx
2
+
bˆy
2
so the structure factor becomes:
F(q) = f(q) 1 + eiq·r2
= f(q) 1 + eiπ(h+k)
= f(q) ×
2 h + k ∈ even
0 h + k ∈ odd
For the body-centered lattice the atom positions are:
r1 = 0 ; and r2 =
aˆx
2
+
bˆy
2
+
cˆz
2
F(q) = f(q) 1 + eiπ(h+k+l)
= f(q) ×
2 h + k + l ∈ even
0 h + k + l ∈ odd
For the face-centered lattice, the atom positions are:
r1 = 0 ; r2 =
aˆx
2
+
bˆy
2
; r3 =
aˆx
2
+
cˆz
2
; and r4 =
bˆy
2
+
cˆz
2
F(q) = f(q) 1 + eiπ(h+k)
+ eiπ(h+l)
+ eiπ(k+l)
= f(q) ×
4 hkl unmixed
0 hkl mixed

The structure factors for these structures for the ﬁrst several peaks
are summarized in Table 8.1. One way to distinguish these structures
would be to look for the (100), (001), (110), (101), and (111) peaks.
We see from Table 8.1 that the primitive lattice would exhibit all of
these peaks, the base-centered lattice would exhibit the (001), (110)
and (111) peaks, the body-centered lattice would exhibit the (110) and
(101) peaks, while the face-centered lattice would exhibit the (111)
peak. Hence the structures could be distinguished by looking at the
occurrence of these peaks.
Structure Factor
hkl q/2π primitive base body face
100, 010 ˆx/a, ˆy/b f 0 0 0
001 ˆz/c f 2f 0 0
110 ˆx/a + ˆy/b f 2f 2f 0
101, 011 ˆx/a + ˆz/c, ˆy/b + ˆz/c f 0 2f 0
111 ˆx/a + ˆy/b + ˆz/c f 2f 0 4f
200, 020, 002 2ˆx/a, 2ˆy/b, 2ˆz/c f 2f 2f 4f
Table 8.1: Structure factor and scattering vector for the ﬁrst several peaks
for the four orthorhombic lattices.
4. The structure factor F contains the information to distinguish these
two phases. In general F is given by:
F(q) =
n
fneiq·rn
where fn and rn are the atomic scattering factor and the position of
the nth
atom. We are interested in the structure factor Fhkl evaluated
at a Bragg peak where:
q = 2πGhkl = 2π(hb1 + kb2 + lb3)
For the ordered compound, the atoms are at:
Pt atoms a(0, 0, 0), a/2(1, 1, 0)
Fe atoms a/2(1, 0, 1), a/2(0, 1, 1)

47
The structure factor F is then:
Fhkl = fPt 1 + eiπ(h+k)
+ fFe eiπ(h+l)
+ eiπ(k+l)
=



2 (fPt + fFe) h, k, l unmixed
2 (fPt − fFe) h, k even and l odd or h, k odd and l even
0 h, k mixed
For the disordered compound, the average atom is on each FCC site
so:
Fhkl =
fPt + fFe
2
1 + eiπ(h+k)
+ eiπ(h+l)
+ eiπ(k+l)
=
2(fPt + fFe) h, k, l unmixed
0 h, k, l mixed
Thus there are some peaks, for example (001), (223), and (112), where
the ordered phase has a structure factor of 2(fpt − fFe) and the disor-
dered phase has a structure factor of 0.
5. (a) The substrate has a cubic structure so the reciprocal lattice vectors
are:
bs
1 =
1
as
0
ˆx ; bs
2 =
1
as
0
ˆy ; bs
3 =
1
as
0
ˆz
so that the diffraction vector at the Bragg condition is given by:
qs
B = 2πGs
hkl = 2π(hbs
1 + kbs
2 + lbs
3) =
2π
as
0
(hˆx + kˆy + lˆz)
The observed substrate peak is at a diffraction vector given by:
qs
B = qs
ˆx + qs
ˆz
Equating components of these two vectors we find:
qs
=
2πh
as
0
=
2πl
as
0
For the tetragonally distorted film the reciprocal lattice vectors
are given by:
bf
1 =
1
af ˆx ; bf
2 =
1
af ˆy ; bf
3 =
1
af
⊥
ˆz

and the diffraction vector at the Bragg condition is given by:
qf
B = 2πGf
hkl = 2π(hbf
1 + kbf
2 + lbf
3 ) =
2π
af (hˆx + kˆy) +
2π
af
⊥
lˆz
The observed film peak is at a diffraction vector given by:
qf
B =
qs
ˆx
1 − δx
+
qs
ˆz
1 − δz
Equating the x-components of these vectors we find:
2πh
af =
qs
1 − δx
=
2πh
as
0
1
1 − δx
Rearranging we find:
af
= as
0(1 − δx)
Similarly, by equating the y-components and rearranging we find:
af
⊥ = as
0(1 − δz)
(b) As we found in this homework set, the width in q of a diffraction
peak is related to the length L of the sample through:
∆q = 0.9
2π
L
Hence the thickness of the film is:
t ≈ 0.9
2π
∆qz
and the in-plane grain size is:
lg ≈ 0.9
2π
∆qx
6. • Sample A
To begin indexing we calculate the d-spacing for each of the peaks.
To try to find a pattern, we divide the d-spacing for each peak into

49
that for the ﬁrst peak and square it, since this will give us the ratio
of the squares of the indices:
d1
di
2
=
h2
i + k2
i + l2
i
h2
1 + k2
1 + l2
1
The result of this is shown in Table 8.2, along with a consistent
indexing of the peaks and the associated lattice parameter. Since
all the indices are unmixed, the peak pattern is consistent with
an FCC structure.
Peak 2θ d (˚A) (d1/di)2
hkl a (˚A)
1 43.4 2.085 1 111 3.612
2 50.5 1.807 4/3 200 3.615
3 74.5 1.278 8/3 220 3.615
4 90.0 1.090 11/3 311 3.616
5 95.3 1.043 12/3 222 3.614
6 117.1 0.904 16/3 400 3.615
7 136.8 0.829 19/3 331 3.615
8 145.1 0.808 20/3 420 3.614
Table 8.2: Indexing of peaks for Sample A.
• Sample B
In this case the peaks can be indexed in two ways as shown in
Table 8.3. So from the positions of the peaks, we cannot tell if
this sample is BCC or simple cubic. However, we do have the
intensities, and from this information we can get a better idea.
As shown in Table 8.4, the sequence of multiplicities are quite
diﬀerent for the simple cubic and BCC cases. If we divide the
scaled intensity by the multiplicity and by the Lorentz polarization
factor:
1 + cos2
2θ
sin θ sin 2θ
we should get a smoothly varying function of q which will be
proportional to the atomic scattering factor squared. Figure 8.1
shows the result of this for the intensity for Sample B. It is clear
that the BCC indexing gives more plausible behavior. In fact the
intensities are those calculated for BCC Fe.

SC BCC
Peak 2θ (deg) d (˚A) hkl a (˚A) hkl a (˚A)
1 44.71 2.027 100 2.027 110 2.867
2 65.09 1.433 110 2.027 200 2.866
3 82.4 1.171 111 2.027 211 2.867
4 99.06 1.013 200 2.027 220 2.867
5 116.5 0.907 210 2.027 310 2.867
6 137.4 0.828 211 2.027 222 2.867
Table 8.3: Listing of indexing of peaks for Sample B as either simple cubic
(SC) or body-centered cubic (BCC).
Peak 1 2 3 4 5 6
SC hkl 100 110 111 200 210 211
SC mhkl 6 12 8 6 24 24
BCC hkl 110 200 211 220 310 222
BCC mhkl 12 6 24 12 24 8
Table 8.4: Listing of multiplicities for peaks for Sample B as indexed as
simple cubic or BCC.

51
2.5
2.0
1.5
1.0
0.5
ScaledIntensity
7654
q (A-1
)
SC
BCC
Figure 8.1: Observed intensity for Sample B scaled by the Lorentz polariza-
tion factor and the multiplicity factor as indexed as BCC or simple cubic
structure.
7. We use the expression in the notes for diffracted intensity from a powder
specimen:
I =
I0r2
eλ3
mhkl
16πRV 2
c
A0
2µ
F2
hkl
1 + cos2
2θ
sin θ sin 2θ
e−2M
We are only interested in the angle-dependent parts, so we define:
Iθ = mhkl F2
hkl e−2M 1 + cos2
2θ
sin θ sin 2θ
For the allowed peaks in an fcc structure, the structure factor is just
16f2
. We use the expressions in the notes for calculating M and use
the given expression for f. The results are tabulated in Table 8.5. The
calculate values agree pretty well with those in the JCPDF card in
Fig. 9.8 in the notes.
8. (a) A labeling of the diffraction spots consistent with the information
given is shown in Fig. 8.2.
(b) The sketches of the planes which are associated with the marked
diffraction spots are shown in Fig. 8.3.

(000)
(001)
*
*
*
*
(002) (112)
(111)
(110)
(222)
(221)
(220)
Figure 8.2: Labeled TEM diﬀraction spots.
x
y
z
(001) (002)
(110) (111)
Figure 8.3: Sketches of several planes associated with diﬀraction spots in Fig.
1

53
hkl 2θ (deg) mhkl f M LPF Iθ I/Imax
111 39.8 8 62.4 0.030 7.30 3.4 × 106
1
200 46.3 6 59.9 0.041 5.20 1.65 × 106
0.48
220 67.5 12 52.5 0.081 2.23 1.00 × 106
0.29
311 81.3 24 48.0 0.112 1.59 1.13 × 106
0.33
Table 8.5: Table of quantities for calculation of Pt peak powers.
9. (a) The diffraction spots are labeled in Figure 8.4.
(b) If the crystal is fcc, the fcc extinction rules apply so that only
peaks with unmixed hkl survive. For the peaks in the figure it is
only the (0¯22) and the (2¯20).
(000)
(011)
(110) (220)
(022)
(121) (231)
(132)(112)
(101)
Figure 8.4: Schematic of TEM diffraction pattern obtained from simple cubic
crystal with the electron beam in the [111] direction.
10. (a) Neither the polarization factor (Eqn. 9.4 from the notes) nor the
integration factor (Eqn. 9.5) will be affected. However, now ev-
ery crystallite in the sample will be oriented to diffract, so the
factor from Eqn. 9.7 should not be included. Furthermore, the in-
coming x-ray beam will not produce a cone of diffracted intensity,
but rather a single beam, so the factor representing length of the

diffraction cone (Eqn. 9.8) should also not be included. Hence the
Lorenz-polarization factor should just be:
1 + cos2
2θ
sin 2θ
(b) The multiplicity factor should not be included, since each crystal-
lite is automatically oriented for the set of planes which is parallel
to the surface and not oriented at all for other sets of planes.
(c) We must integrate the intensity produced by an increment of film
thickness over the thickness of the film. The intensity increment
is the same as that for the infinite sample, but so we have:
I =
tf
0
dI
=
kI0A0
sin θ
tf
0
e−2µz/ sin θ
dz
=
kI0A0
2µ
1 − e−2µtf / sin θ
Taking the small tf limit we find:
I ≈
kI0A0
2µ
1 − 1 −
2µtf
sin θ
=
kI0A0tf
sin θ
so that the effective volume is:
A0tf
sin θ
11. Referring to the figure in the problem statement, the angle between k
and q is π
2
− θ, so that δ, the angle between q1 and q2, is given by:
δ =
π
2
− θ1 −
π
2
− θ2 = θ2 − θ1
12. From the solutions to a previous homework problem, the square of the
structure factor is given by:
F2
= f2
×



64 h, k, l = even and
h + k + l = 4 × integer
32 h, k, l odd
0 otherwise

55
The observed diffraction patterns are shown in Fig. 8.5.
a)
(000)
(400)
(220)
(040)
(440)
4/a
b)
(000)
(400) (422) (444)
(311) (333)
(111) (133)
(022) (044)
4 2/a
Figure 8.5: Schematic of TEM diffraction patterns for diamond cubic when
the incident electron along the a) [00¯1] and b) [0¯11] directions.
13. (a) The peaks can be indexed by noting that for a cubic crystal we
have
qhkl =
2π
dhkl
=
2π
a
√
h2 + k2 + l2
Hence if we assign
q0 =
2π
a
then the peaks can be indexed as shown in the following table.
Peak 1 2 3 4 5 6 7
q/q0
√
3 2
√
8
√
11
√
12 4
√
19
hkl 111 200 220 311 222 400 331
Since all indices are unmixed the structure is fcc.
(b) From our assignment of q0 we can find the lattice parameter:
a =
2π
q0
= 4.19 ˚A

14. (a) To index the peaks we first calculate d-spacings, as shown in the
table below. Then the ratio of the square of d-spacings is calcu-
lated. It is found that this is consistent with a simple cubic lattice
with lattice parameter a = 3.26 ˚Aor a bcc structure with lattice
parameter a = 2.305 ˚A. However, in the peak sequence for bcc,
there is a peak with indices 321 which appears between the 222
and 400 peaks. This peak was not observed for this sample, so it
is concluded that the sample is simple cubic.
Peak 2θ d(˚A) (dn/d1)2
hklsc hklbcc
1 27.32 3.26 1 100 110
2 39.03 2.31 2 110 200
3 48.30 1.88 3 111 211
4 56.38 1.63 4 200 220
5 63.76 1.46 5 210 310
6 70.70 1.33 6 211 222
7 83.83 1.15 8 220 400
(b) The density is given by
ρ =
n(MCu + MZr)
NAa3
where n is the number of formula units per cell and NA is Avo-
gadro’s number. Solving for n we find
n =
ρa3
NA
MCu + MZr
Plugging in numbers we find that n = 1 so there is one formula
unit per cell.
(c) We know we have a cubic cell with one Cu and one Zr atom in
each cell. Although there are several possibilities for the positions
of the atoms in the cell, we can at least think of the simplest,
which is to have one atom at the center of the cell and one at the
corner. This is the CsCl structure which has structure factor
Fhkl = fA + fBeiπ(h+k+l)
Thus for peak 1 which is the (100) peak we have
F100 = fA − fB

57
and for peak 2, the (200) peak, we have
F200 = fA + fB
Inserting numbers for the atomic scattering factors for Cu and Zr
we have
F100
F200
2
=
33.55 − 23.75
26 + 17.9
2
= 0.05
in agreement with the experiment. While this is not a conclusive
test, it is suggestive so our guess for the crystal structure is that
it is CsCl.
15. (a) The structure factor can be found by summing over the 8 atoms
in the conventional cell
Fhkl =
n
fne2πi/a(hxn+kyn+lzn)
= 1 + eiπ(h+k)
+ eiπ(h+l)
+ eiπ(k+l)
fCl− + fNa+ eiπ(h+k+l)
(b) Since the first factor in brackets is the same as that found in the
structure factor for fcc, we see that the structure factor is zero
unless h, k, l are either all odd or all even. The structure factor
then becomes
Fhkl =
4 (fCl− + fNa+ ) h, k, l ∈ even
4 (fCl− − fNa+ ) h, k, l ∈ odd
(c) The sketch is shown in Figure 8.6. The path length difference for
corresponding to the (200) peak is ∆L = 2d200 sin θB where θB is
half the scattering angle associated with the (200) peak. We can
find this angle through its relation to the scattering vector, which
at the (200) peak is given by:
qB = 2πG200
=
4π
a

Hence
sin θB =
λ
4π
qB
=
λ
a
Hence the path difference is
∆L = 2d200 sin θB
= 2
a
2
λ
a
= λ
q
k
a
k'
Figure 8.6: Schematic of diffraction condition for the (200) peak in NaCl
structure.
(d) For the case of two identical atoms we let fCl− → fNa+ → f and
we find that
Fhkl =
8 h, k, l ∈ even
0 otherwise
This results in a reciprocal lattice which is just simple cubic with a
reciprocal lattice parameter of 2/a (Figure 8.7). This suggests that
we could simplify the problem by noticing that the real lattice has
become a simple cubic lattice with a new lattice parameter half
that of the original.

59
2
a
Figure 8.7: Schematic of reciprocal lattice for the NaCl structure. If the
atoms are the same, the lighter spots disappear and the reciprocal lattice
becomes simple cubic.
16. (a) The reciprocal lattice vectors are given by
b1 =
1
a
√
3
3
ˆx + ˆy
b2 =
1
a
−
√
3
3
ˆx + ˆy
b3 =
1
c
ˆz
so we find
Ghkl = hb1 + kb2 + lb3
=
h
a
√
3
3
ˆx + ˆy +
k
a
−
√
3
3
ˆx + ˆy +
l
c
ˆz
=
ˆx
a
√
3
3
(h − k) +
ˆy
a
(h + k) +
lˆz
c
(b) To find the spacing between planes we need the length of Ghkl.
Taking the sum of the square of the vector components we find
G2
hkl =
(h − k)2
3a2
+
(h + k)2
a2
+
l2
c2
=
1
a2
4
3
(h2
+ k2
+ hk) +
l2
c2

Hence the plane spacing is given by
dhkl =
1
Ghkl
=
4
3a2
(h2
+ k2
+ hk) +
l2
c2
−1/2
(c) If c = 8/3 a we ﬁnd:
dhkl = a
4
3
(h2
+ k2
+ hk) +
3
8
l2
−1/2
Hence
d002 = 2/3 a = 0.816 a
d110 = a/2 = 0.5 a
d101 = 24/41 a = 0.765 a
Since the d-spacing and scattering angle are related through λ =
2dhkl sin θ, the largest d will have the smallest scattering angle so
the peaks occur in the order: (002),(101),(110)

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