X-ray diffraction laboration
Peter Román
840509-0518
Johan Kjölhede
851206-0370
Hampus Lundén
850129-1416
May 4, 2007
Abstract
Three unknown crystal samples A,B and C of either Si or GaAs are
identified using x-ray diffraction. A and C were found to be Si, while
B was GaAs. Measurement errors are assumed to be below 2%.
1
Introduction
The goal of the laboration was to determine with the help of X-Ray
diffraction which of three crystals, A, B and C was made of Si or GaAs.
This was done by measuring diffraction of x-rays with wavelength λ.
Diffraction is affected by crystal orientations. The x-ray wavelength λ
was unknown. To determine λ a NaCl crystal with known orientation
was given.
An x-ray diffraction machine was used for these purposes, with an
emitter in a fixed position, and sample rotated β degrees, and the
sensor arm at 2β, assuming angle of incidence = angle of reflection
(I = R) angle. This means that only reflections from planes parallel to
the crystal surface will be detected by the sensor. Scattering vectors
q (the change in the x-rays’ wavevector from before the reflection to
after it) are found always perpendicular to the crystal surface, since
I = R and the conservation of wavelength. The size of q will vary as
sinus of β increases.
Bragg’s law is used is used to find the crystal orientations and
determine what material a crystal is made of.
2d sin θ = nλ
Where d, the distances between the crystal planes of reflections are
calculated using
a
d= √
2
h + k2 + l2
where a is the length of the unit cube and (hkl) the Miller indices
of the reflective plane.
In order to determine the wavelength λ and the selection rules for
allowed reflections in the lattices for NaCl, Si and GaAs we calculate
the structure factor SG
1
SG =
fj exp [−i2π(hxj + kyj + lzj )]
(1)
j
where fj is the atomic structure factor for atom j, (hkl) is the
reflection and (xj , yj , zj ) is the atom position in the unit cube.
1.1
NaCl
The sodium chloride lattice is face-centered cubic and therefore has
(xj , yj , zj )
Cl:
000;
11
2 2 0;
1 1
202;
0 12 21 .
Na:
111
2 2 2;
00 21 ;
0 21 0;
1
2 00.
this together with (1) gives us
SG = fCl 1 + e−iπ(h+k) + e−iπ(h+l) + e−iπ(k+l) +
+fN a e−iπ(h+k+l) + e−iπl + e−iπh + e−iπk
because h, k and l are integers e−in = ein we can factorize the
expression
SG = fCl + fN a e−iπh 1 + e−iπ(h+k) + e−iπ(h+l) + e−iπ(k+l)
by examining this expression we arrive at some selection rules. We
can see that the rightmost factor is not equal to zero only when h, k
and l are either all even or all odd.
Furthermore, since fN a and fCl are of comparable size, the reflection when h is even will be much stronger than then it is odd, because
of the leftmost factor
> fCl − fN a
f +f
Cl N a
h even
1.2
h odd
GaAs
GaAs has the zinc blende structure with (xj , yj , zj )
Ga:
000;
0 21 21 ;
1 1
202;
11
2 2 0.
As:
111
4 4 4;
133
4 4 4;
313
4 4 4;
331
4 4 4.
and if we calculate the structure factor in the same way as for NaCl
we get
π
SG = fGa + fAs e−i 2 (h+k+l) 1 + e−iπ(h+l) + e−iπ(k+l) + e−iπ(k+h)
2
1.3
Si
Since Si has the diamond structure we have the same structure factor
as for GaAs only with the atomic structure factor fSi instead of fGa
and fAs
π
SG = fSi 1 + e−i 2 (h+k+l) 1 + e−iπ(h+l) + e−iπ(k+l) + e−iπ(k+h)
As for NaCl we need the Miller indices to be all even or all odd
because of the rightmost factor, but the leftmost factor is different,
which gives us the possible reflections in the cut planes (100), (110)
and (111) according to Table 1.
plane
100
110
111
allowed reflections
200 400
etc.
220 440
660
333 444
555
etc.
etc.
Table 1: Allowed reflection peaks for Si.
Since fGa and fAs could be concidered of comparable size and the
interesting wave peaks of the crystals were given beforehand, GaAs is
assumed to have the same possible reflections as Si.
2
Experimental procedure
The diffraction patterns of the samples were measured using a step
angle ∆β of 0.1◦ . The X-ray machine rotated the sample by angle β
and the sensor arm by angle 2β. Each step took 4 seconds to measure.
The maximum voltage and maximum current was used. The X-ray
machine used was the Leybold-Didactic 554811.
The first sample to be measured was the NaCl. The measurment
was between 2.5◦ and 30◦ .
Later the three other samples A, B and C were each measured.
This time between 2.5◦ and 35◦ .
3
3
Measurement results
The raw data collected by a computer connected to the X-ray machine
is as follows
1500
Relative intensity I [W/m2]
← (200)
1000
500
← (400)
← (600)
0
0
5
10
15
20
Angle of incidence β [°]
25
30
Figure 1: X-ray diffraction in NaCl with the Miller indices (hkl) given for
the reflection peaks. I is the relative intensity [W/m2 ] and β the angle of
diffraction [◦ ].
350
← (111)
Relative intensity I [W/m2]
300
250
200
150
100
← (333)
← (444)
50
← (555)
0
0
5
10
15
20
25
Angle of incidence β [°]
30
35
Figure 2: X-ray diffraction in the sample A with the Miller indices (hkl)
given for the reflection peaks. I is the relative intensity [W/m2 ] and β the
angle of diffraction [◦ ].
4
140
← (400)
Relative intensity I [W/m2]
120
100
80
60
← (800)
40
20
0
5
10
15
20
Angle of incidence β [°]
25
30
35
Figure 3: X-ray diffraction in the sample B with the Miller indices (hkl)
given for the reflection peaks. I is the relative intensity [W/m2 ] and β the
angle of diffraction [◦ ].
60
50
Relative intensity I [W/m2]
← (400)
40
30
20
← (800)
10
0
0
5
10
15
20
Angle of incidence β [°]
25
30
35
Figure 4: X-ray diffraction in the sample C with the Miller indices (hkl)
given for the reflection peaks. I is the relative intensity [W/m2 ] and β the
angle of diffraction [◦ ].
5
4
Discussion of the results
Only the marked peaks were considered in Figures 3 and 4. The error
in β for the peaks are assumed to be less than 2%.
Since the crystal orientation, (100), and the nearest-neighbor distance, a = 5.63 Å, is known for the NaCl crystal, the X-ray wavelength,
λ, can be found by the Bragg law.
2a
sin θ = λ =⇒ λ = 7.0 · 10−11 m
h
4.1
Identification of the unknown crystals
To identify the materials, it is necessary to find how each material is
cut relative its reflective planes. At first the planes (hkl) = (100),
(110) and (111) are tested.
By calculating n from the Bragg equation
sin θ
2a
√
=n
2
2
2
h +k +l λ
(2)
for each relevant peak measured, the crystal orientation can be
decided. Material A, B and C were found to have 4, 2 and 2 relevant
peaks respectively. These peaks correspond to different integer values
of n for the material in question. If for each measured β for the specific
material a (hkl) setup provides (approximetly) integer values of n, the
crystal orientation has been found.
The length of the unit cube a for Si, GaAs and NaCl are known. A,
B and C are all crystals of materials Si or GaAs (order unknown). Since
the variation of a between Si and GaAs (just a few percent difference)
has a smaller influence on n than the orientation of the crystal has,
the orientation can be found before knowing the material.
When the crystal orientations and λ are known, it is just a matter
of testing both values of a against materials A,B and C and checking which result (with aSi or aGaAs ) of (2) provides results closest to
integers. This is likely to be the correct material.
Results can be found in table 2.
sample
A
B
C
orientation
(111)
(100)
(100)
material
Si
GaAs
Si
Table 2: Identification of the crystal orientation and material of the unknown
samples A, B and C.
6
5
Conclusions
It was possible to distinguish between Si and GaAs. The results of
the laboration can be found in Table 2. No major problems were
encountered with the material or equipment.
7