Take home Answer - Spring 2009

Transmission Electron Microscopy (TEM)

University of Puerto Rico, PR-00931, USA.

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Topic: Electron diffraction

Question1. ____/20 points.

State the definition of the structure factor (structure amplitude) and use it to calculate the kinematic intensity of the 100 reflection of (i) Nb (A2 structure) and (ii) CsCl (B2 structure) as a function of the atom form factors fNb, fCs, fCl.

Structure factors are the unit cell equivalents of the atomic scattering amplitude.

i) Kinematic intensity for A2 Structure:

Fig. A2 shows that it is a body centered cubic (bcc) with basis (000) & (0.5 0.5 0.5). The structure factor for this structure is

F= 2f ; if h+k+l is even

F= 0 ; if h+k+l is odd.

Here we have h+k+l= 1+0+0=1; odd.

So, our kinematical intensity fNb = 0.

ii) B2 Structure: Fig. B2 shows that it is an bcc structure with cl at the basis (000) and Cs at the other basis (0.5 0.5 0.5).

Rajasekarakumar VadapooDept. of Physics

More Info: http://nanophysics.wordpress.com

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This leads to two values of F, neither of them is zero. F= fCl +fCs if h+k+l is even F= fCl - fCs if h+k+l is odd. In our case, h+k+l= 1; is odd So, the kinematical intensity for the structure B2 is fCl - fCs .

Question 2. ____/20 points The TEM diffraction pattern below was obtained from a face‐centered cubic crystal with an accelerating voltage of 200 kV, thus with an electron wavelength of λ = 2.507 pm, and a camera length of L = 1 m. Indicate the indices of the viewing direction, index at least two (independent) of the three innermost reflections of the diffraction pattern, and determine the lattice parameter a0 of the crystal. (Hint: To identify the viewing direction, consider the pattern symmetry. Mind forbidden reflections).

The indexed pattern is as shown in below with the viewing direction of [-111].

The lattice parameter calculation using the given details and the measured details from the pattern is as follows:

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Q.4) Question 4. ____/20 points Calibration of the camera length (L) of the microscope. The diffraction standard used is an evaporated aluminum film. Al has the FCC structure, with a = 0.4041 nm. d is the lattice spacing for Al, L is the effective camera length, λ is the wavelength of the electron beam and R is the measured diffraction ring adius onto the film. See attached digital files.

Ans:

L = 1 m:

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F123

The corresponding calculation for the d using the fixed camera length is as follows- to aid to index the lines.

Diameter cm

Radius (R) m

Lamda m

camera length in m

Lattice spacing d in m R1/R2 d2/d1

calculted d in nm

4.04 0.0202 2.51E-12 1

1.24E-10 0.12410891

4.34 0.0217 2.51E-12 1

1.16E-10 0.93087558 0.93087558 0.11552995

4.52 0.0226 2.51E-12 1

1.11E-10 0.96017699 0.96017699 0.1109292

4.93 0.02465 2.51E-12 1

1.02E-10 0.9168357 0.9168357 0.10170385

5.22 0.0261 2.51E-12 1

9.61E-11 0.94444444 0.94444444 0.09605364

Table.1. The calculated d value with the fixed camera length used to index the patten.

Radius (R) m

calculted d in nm

given d in nm

camera length (Rd/λ)

0.0202 0.124108911 0.122 0.98 0.0217 0.115529954 0.117 1.01 0.0226 0.110929204 0.02465 0.101703854 0.101 0.99 0.0261 0.09605364

Table.2. The calculated Camera length from the indexed pattern.

The calculation shows that the Camera length is almost 1m. the effective camera length from the calculation is 0.9933 m.

L= 2.65 m:

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Fig. F126

L=2.65m

Diameter (cm)

Radius(R) in m

Latt.spa d(m) lamda (m) R1/R2 D2/D1

Camera Length [Rd/LAmda] (m)

5.690 0.02845 2.34E-010 2.507E-012

0.86 0.86 2.66

6.580 0.03290 2.02E-010 2.507E-012

0.71 0.71 2.65

9.290 0.04645 1.43E-010 2.507E-012

0.85 0.85 2.65

10.960 0.05480 1.22E-010 2.507E-012

0.96 0.96 2.67

11.420 0.05710 1.17E-010 2.507E-012

#DIV/0! 0.86 2.66

1.01E-010

2.507E-012

0

The above calculations shows that the effective camera length (average)= 2.658 m.

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Question 5. ____/100 points

Results from prior CBED experiments. All diffraction patterns were obtained from a sample of silicon.

a) Speculate why we used a Si single crystal, and not, for example, a foil of polycrystalline

aluminum for this experiment.

As we know that in polycrystalline the each grains oriented in random direction, contrary to the single crystal. In the case of CBED pattern from a single crystalline we get lot of information about the crystal symmetry, structure, lattice parameter, thickness, etc by analyzing ZOLZ,HOLZ and HOLZ lines. Lets think that if we take a CBED pattern in a region where two distinct oriented grains presents, then we would have a pattern containing both the cyrstallites information together in a pattern along with any systematical absences/ excess according to the dynamical conditions. It would be too difficult to analyse the resultant pattern, unless otherwise we confirm the effect of two crystallites independent and combine with the dynamical effects. So, if we have a polycrystalline sample with N number of distinctly oriented crystallites, then retrieving most of the absolute crystalline details from the CBED pattern would be almost impossible. This is the reason behind to use a Si single crystal and not a foil of polycrystalline aluminium for the CBED experiment. b) Describe all patterns with respect to each other’s when applicable.

a) What is the difference between F157 and F158 or F159? F158 and F159 are the same the patterns, but what is the variable?

Fig. F157 shows that it is a Kossel pattern which we get by adopting large 2α by increasing the C2 aperture size. But Figs. F158 & F159 is Kossel-Mollenstedt pattern which we get by adopting low 2α by selecting smaller C2 aperture size. We can confirm it by seeing that the ZOLZ patterns are not merged together in the later cases.

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Fig. F158 & Fig. F159 are the same the variable is the exposure time. There is no indication of change in 2 α with the given information of constant camera Length (L), since the ZOLZ patterns are not merged in both the cases.

b) What would you say about the pattern F162?

This pattern occurs during the process of focusing the CBED pattern by adjusting the

strength of the objective or C2 lens. This fig. Shows that it is an over/under focus CBED pattern which shows the Bright Field image (BF) in 000 disk ( the one with full circle in this case) & Dark Field images (DF) in other hkl disks.

c) Compare F164 with F165.

Fig. F164 is an CBED pattern under kinematical conditions. This pattern is nothing more than SAD (Selected Area Diff.) but comes from very small region than normal SAD. Fig. 165 is an CBED pattern from thicker area than F164 showing dynamical contrast in the ZOLZ disks which happens when at focus.

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d) Compare F157 with F178. Is F178 oriented on a low index zone axis?

Fig. F157 is a Kossel pattern which we get at large convergence semiangle (2α) and at Zone axis which gives ZAP (Zone axis pattern).

Fig. F178 shows excess & deficient kikuchi lines as noted in the fig., which we can see in non-ZAP pattern [marked in red color]. But in zone axis pattern (ZAP) we would see the Zero Ordered Laue Zone (ZOLZ) kikuchi lines appear as bright bands, which also appear in the same pattern [marked in Green color]. So, as a conclusion we could see ZAP and non-ZAP pattern in

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a single image that could be due to the intermediate conditions for ZAP & non –ZAP or could be due to the double exposure of the ZAP & non-ZAP conditioned patterns in a single photographic plate.

Another interesting observation towards the question is it low index ZAP? Leads to see the structural observation of ZOLZ disks arrangement shows that it is correspond to the Zone axis of [001] and the corresponding reflection disks are indexes as shown in fig.

e) Compare F170 with F172 or F176.

F170 and F176 are the CBED pattern taken under two beam conditions. In both the cases

we can see the 000 disk and the hkl disk. Since the sample is silicon, the possible hkl disks are 220 family planes (<220>) or <111> depends on the direction of e-beam makes the above ZOLZ disk. F170 shows the number of pairs of parallel dark fringes of 5 and F76 shows 6. From that we could conclude that the pattern F176 taken in a thicker region of the specimen than the pattern F170 taken.

f) Compare the patterns F170, F172, F174 and F176

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The details of the patterns F170, F172, F174 and F176 are the same as we discuss in the question (d). While correlating all the four patterns we could conclude the pattern F172 & F174 taken in a region where the specimen having lesser thickness, F170 taken in more thicker region and F176 taken at the thickest region of the sample.

g) Determine the zone axis and index the pattern F185.

Fig. F185 shows that the pattern is indexed with the zone axis [111] for Silicon. The diffraction pattern shows the dynamical contrast within the disks as well as kikuchi lines and other lines.

h) Compare F185 with F186.

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Fig. F185 & F186 are the same pattern the variable is the exposure time/increase in 2α/change in thickness since the later one shows more intensity. The camera Length of both patterns are the same as 1.9 meters, and the distance between 000 and other hkl also constant. Moreover the no of concentric rings in 000 disks is the same as 2 in both the cases, which shows that both of them have almost same thickness within the error of extinction distance (ξ) . So, we can conclude that the variable is the exposure time.

i) Describe F187

Fig.187 shows the CBED pattern contains ZOLZ pattern and the FOLZ ring of silicon. It shows the 3 fold symmetry as marked in yellow lines. The diameter of the HOLZ ring measured as 148.2 mm and we can calculate the lattice parameter with this knowing the Zone axis. Here the zone axis must be [111] as an fcc with 2 basis crystal structure. More we would discuss in the section 5.c.

j) Describe F167

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Fig. F167 shows the HOLZ line in the central 000 disk of a low symmetry ZAP taken at high camera length (L).

5.C) For the CBED patterns obtained in a “random” orientation, describe the effect of changing the condenser aperture, defocusing the pattern and varying the specimen thckness.

The above patterns F162,F164, F165 are the CBED patterns obtained in a random

orientation. The pattern F162 occurs during the process of focusing the CBED pattern by adjusting the strength of the objective or C2 lens. This fig. Shows that it is an over/under focus CBED pattern which shows the Bright Field image (BF) in 000 disk ( the one with full circle in this case) & Dark Field images (DF) in other hkl disks. Fig. F164 is an CBED pattern under kinematical conditions. This pattern is nothing more than SAD (Selected Area Diff.) but comes from very small region than normal SAD. Fig. 165 is an CBED pattern from thicker area than F164 showing dynamical contrast in the ZOLZ disks which happens when at focus.

5.d) For the CBED patterns obtained with the primary beam parallel to a low‐indexed zone‐axis, determine the zone‐axis from the ZOLZ pattern. Describe the symmetry of the HOLZ‐line pattern. Calculate the spacing of the lattice planes normal to the viewing direction from the

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radius of the FOLZ ring and compare the result with the value you would expect given that aSi = 0.543 nm

.

F185.

Fig.185, F186 and F187 are the CBED patterns obtained with the primary beam parallel to a low indexed zone axis. Here we use F185 pattern to determine the zone axis and index the ZOLZ pattern and F187 to find the radius of the FOLZ ring and retrieve the lattice parameter from it. From the pattern we could explain the symmetry too.

The ZOLZ disks are indexed as shown in F185 and with the corresponding Zone axis indexed as [111]. From F187 we found that the diameter of the FOLZ ring is 148.2mm. From that the lattice parameter would be find as follows:

�

��= ��

�(λL/r)2 (1)

Where,

�

�� - measured real lattice spacing

L - Camera Length = 500 mm

r- - radius of the FOLZ ring = 74.55 mm.

And λ - wavelength of the electron = 2.507x 10-3 nm.

For an FCC crystal,

�

� = �� ��� �������.�

� (2)

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Where,

�

� - Calculated real lattice spacing

�� - Lattice parameter = 0.543 ��

P= 1 for U+V+W is odd and

P= 2 for U+V+W is even

UVW – the index of Zone axis.

�

� = � � ���

(3)

and n should be integer for the correct indexing.

In our case, UVW is [111]

�

��= � �

�.-�./��01 23(2.507x 10-3 nm x 500 mm/74.55 mm) = 0.22554

�� = 0.543 �� 4 √1� + 1� + 1�/1 = 0.9405

‘n’ = 0.9405/0.22554 = 4.1699

To be n as an integer this value is within the error of 4%. So, we could say that our indexing is within 4% of error.

The yellow line in the pattern F187 shows that it has a 3-fold symmetry. Which matches with the silicons crystal structure (fcc with 2 basis) when we see in [111] we would get 3- fold symmetry.

5.e)For the CBED patterns obtained in the two‐beam condition, describe the effect of the specimen thickness on the fringe pattern. Estimate the local specimen thickness for at least one location.

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The details of the patterns F170, F172, F174 and F176 are the CBED pattern taken under

two beam conditions. In all the cases we can see the 000 disk and the hkl disk. Since the sample is silicon, the possible hkl disks are 220 family planes (<220>) or <111> depends on the direction of e-beam makes the above ZOLZ disk. While correlating all the four patterns we could conclude the pattern F172 & F174 taken in a region where the specimen having lesser thickness, F170 taken in more thicker region and F176 taken at the thickest region of the sample by counting the pairs of parallel dark fringes.

For the detailed study of thickness we have taken F170. In two beam condition the central

bright fringe is at the exact Bragg condition, where s=0. The deviation Si for the ith fringe can obtain from the following equn.

9: = �;<=�<>?� (4)

Where, @A - Bragg angle for the diffracting hkl plane ‘d’ – Interplanar spacing B@: - Fringe spacing correspond to angles.

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Fig. F170 shows the measured values of B@: (d1,….d5) and the 2@C (2theta) along with the 000 indexed disk with the possible hkl reflection in the sample given as any one of the reflections of the family of planes <220> or <111>. So, we would carried out the calculation for the two family of planes reflections while measuring the thickness of the sample.

The following eqn. is used to determine the foil thickness.

D=

�

EF� + �

EF�GH � = �

I� (5)

When we plot D=

�

EF� Vs

�EF

�, the slope would be �

GH � and the y-intercept would be �I�. From this

we could determine the thickness of the sample. During the plotting if the plot is not straight line when we index the first fringe as n=1 then, plot with starting from n=2 for the first fringe.

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lamda (nm)

2theta-B (cm) d^2: 220 d^2: 111

a (nm)

2.5070E-03 6.69 0.036856 0.098283 0.543

for 220

del-theta S n 1/n^2 S^2/n^2 n-k 1/n-k^2 S^2/n-k^2

0.23 0.00233855 1 1 5.47E-06 2 0.25 1.3672E-06

0.47 0.00477877 2 0.25 5.71E-06 3 0.1111111 2.5374E-06

0.67 0.00681229 3 0.111111 5.16E-06 4 0.0625 2.9005E-06

0.86 0.00874414 4 0.0625 4.78E-06 5 0.04 3.0584E-06

1.06 0.01077766 5 0.04 4.65E-06 6 0.0277778 3.2266E-06

for 111

del-theta S n 1/n^2 S^2/n^2 n-k 1/n-k^2 S^2/n-k^2

0.23 0.00087696 1 1 7.69E-07 2 0.25 1.9226E-07

0.47 0.00179204 2 0.25 1.28E-05 3 0.1111111 3.5682E-07

0.67 0.00255461 3 0.111111 5.87E-05 4 0.0625 4.0788E-07

0.86 0.00327905 4 0.0625 0.000172 5 0.04 4.3009E-07

1.06 0.00404162 5 0.04 0.000408 6 0.0277778 4.5374E-07

Table. XX shows that the calculation of parameters D=

�

EF� and

�EF

� from the pattern to aid plotting for

the <220> & <111> family of planes with for the first fringe the n starts from 1 & 2.

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Graph: 1. D=

�

EF� Vs

�EF

� plot for <220> with nk starts from1 for the first fringe.

AS graph1 shows, we haven’t get the straight line so we proceed with nk starts from 2 for the first fringe and we get the graph.2. as follows.

Graph: 2. D=

�

EF� Vs

�EF

� plot for <220> with nk starts from 2 for the first fringe.

As we see the graph, the y-intercept show the value of �

I�= 3e-6 nm-2 which yield the

thickness of the sample t= 577.35 nm.

For the case of <111> the calculations as follows:

Graph: 3. D=

�

EF� Vs

�EF

� plot for <111> with nk starts from1 for the first fringe.

0

0.000001

0.000002

0.000003

0.000004

0.000005

0.000006

0 0.5 1 1.5

Series1

y = -8E-06x + 3E-06

0

0.000001

0.000002

0.000003

0.000004

0 0.05 0.1 0.15 0.2 0.25 0.3

0

0.0001

0.0002

0.0003

0.0004

0.0005

0 0.5 1 1.5

Series1

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Graph:3 shows that the arbitrary assignment for i and k is not good. So, we have changed the arbitrary value of k as below.

Graph: 4. D=

�

EF� Vs

�EF

� plot for <111> with nk starts from2 for the first fringe.

As we see the graph, the y-intercept show the value of �I�= 5e-7 nm-2 which

yield the thickness of the sample t= 1.414 µm.

So, as a conclusion the thickness of the sample is 577.35 nm if the hkl disk corresponds to <220> and 1.414 µm if the hkl disk corresponds to <111>. So, it is important to know the index of the disk before go for the thickness measurements.

y = -1E-06x + 5E-07

0

0.0000001

0.0000002

0.0000003

0.0000004

0.0000005

0 0.05 0.1 0.15 0.2 0.25 0.3

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Question 6. ____/10 points Copper (Cu) is FCC with a = 3.65 Å. A schematic of the crystal is shown below. Completely index the corresponding diffraction pattern. Assume the pattern to be recorded at 200 kV with a camera length equal to 1 meter.

The above indexed pattern’s ratio of distance of different pts like A,B,C and the angle between them is not matching with any standard FCC zone axis in the book. So, we have decided to index

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from the scratch by finding the corresponding 1/d values for the hkl as tabulated below with the corresponding index in the above fig. By reverse calculation from the unit bar.

Table.4. possible hkl reflection with 1/d value for the fcc with given lattice parameter.

1/dhk

l (nm-

1)

111 200 220 113 222 400 133 420 422 511 333 440 442 620 533 622

1/dhk

l(nm-

1)

4.75 5.48 7.75 9.09 9.49 10.96 11.94 12.25 13.42 14.24 14.24 15.5 16.44 17.33 17.97 18.17

111 4.75 1 1.15 1.63 1.92 2 2.31 2.52 2.58 2.83 3 3 3.27 3.46 3.65 3.79 3.83

200 5.48

1 1.41 1.66 1.73 2 2.18 2.24 2.45 2.6 2.6 2.83 3 3.16 3.28 3.32

220 7.75

1 1.17 1.22 1.41 1.54 1.58 1.73 1.84 1.84 2 2.12 2.24 2.32 2.34

113 9.09

1 1.04 1.21 1.31 1.35 1.48 1.57 1.57 1.71 1.81 1.91 1.98 2

222 9.49

1 1.15 1.26 1.29 1.41 1.5 1.5 1.63 1.73 1.83 1.89 1.91

400 10.96

1 1.09 1.12 1.22 1.3 1.3 1.41 1.5 1.58 1.64 1.66

133 11.94

1 1.03 1.12 1.19 1.19 1.3 1.38 1.45 1.5 1.52

420 12.25

1 1.1 1.16 1.16 1.27 1.34 1.41 1.47 1.48

422 13.42

1 1.06 1.06 1.15 1.22 1.29 1.34 1.35

511 14.24

1 1 1.09 1.15 1.22 1.26 1.28

333 14.24

1 1.09 1.15 1.22 1.26 1.28

440 15.5

1 1.06 1.12 1.16 1.17

442 16.44

1 1.05 1.09 1.11

620 17.33

1 1.04 1.05

h k l X=sqrt(h2+k2+l2) sqrt(x)/a [nm-1] 1 1 1 1.73 4.75 2 0 0 2 5.48 2 2 0 2.83 7.75 1 1 3 3.32 9.09 2 2 2 3.46 9.49 4 0 0 4 10.96 1 3 3 4.36 11.94 4 2 0 4.47 12.25 4 2 2 4.9 13.42 5 1 1 5.2 14.24 3 3 3 5.2 14.24 4 4 0 5.66 15.5 4 4 2 6 16.44 6 2 0 6.32 17.33 5 3 3 6.56 17.97 6 2 2 6.63 18.17

22

533 17.97

1 1.01

622 18.17

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Table.5. Possible interplanar distance between two different planes for the fcc with the given lattice parameter.

From the measured value and from the table.4 we can get :

the pt B corresponds to the family of planes <220>

Pt A corresponds to <620>

Pt C corresponds to <422>

The indexed pattern is shown in the following fig., the details calculations of angle and zone axis follows:

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Question 7. ____/10 points Nickel (Ni) is FCC with a = 3.52 Å. A schematic of the crystal is shown below. Completely index the corresponding diffraction pattern. Assume the pattern to be recorded at 200 kV with a camera length equal to 2 meters.

The above pattern is readily matched with the standard indexed pattern by verify the angles and M/N= 1.618 & L/N =1.911 which corresponds to B= <112> (family of directions). The calculation of exact viewing direction is as follows.

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Question 8. ____/10 points Silicon (Si) with a = 5.43 Å. A schematic of the crystal is shown below. Completely index the corresponding diffraction pattern. Assume the pattern to be recorded at 200 kV with a camera length equal to 5 meters.

The above pattern is readily index with fcc with the viewing direction of <111>, the pattern here indexed with [-111].