1 Imaging defects and - PowerPoint Presentation

Slide1

1

Imaging defects and

contours/effectsSlide2

2

Dynamical diffraction theory

Dynamical diffraction: A beam which is diffracted once will easily be re-diffracted (many times..)

Understanding diffraction contrast in the TEM image

In general, the analysis of the intensity of

diffracted beams

in the TEM is not simple because a beam which

is diffracted

once will easily be

re-diffracted

. We call

this

repeated

diffraction ‘dynamical diffraction.’Slide3

3

Dynamical diffraction:

http://pd.chem.ucl.ac.uk/pdnn/diff2/kinemat1.htm

Assumption:

T

hat

each

individual diffraction/interference

event, from whatever locality within the crystal, acts

independently of the others

M

ultiple

diffraction throughout the crystal; all of these waves can then

interfere

with each other

Ewald

: Diffraction intensity, I, is proportional to just the magnitude of the structure factor, F, is referred to as the dynamical theory of diffraction

We

cannot use

the intensities

of spots in electron DPs (except under

very special

conditions such as CBED) for structure

determination, in

the way that we use intensities in X-ray patterns

.

Ex.

the intensity of the electron beam varies strongly as

the thickness

of the specimen changesSlide4

4

THE AMPLITUDE OF A DIFFRACTED BEAM

The amplitude of the electron beam scattered from a unit cell is:

Structure factor

The

amplitude

in a diffracted

beam:

(

r

n

denotes the position of each unit

cell)

T

he

intensity at some point P, we then

sum over

all the unit cells in the specimen.Slide5

5

If the amplitude

φg changes by a small increment as the beam passes through a thin slice of material which is dz thick we can write down expressions for the changes in

φ

g

and f

0

by using the concept introduced in equation 13.3

but replacing a by the short distance dz

Two beam approximation

Here

χ

O

-χ

D

is the change in wave vector as the

φ

g

beam

scatters into the

φ

0

beam. Similarly

χ

D

-χ

O is the change in wave vector as the φ0 beam scatters into the φg beam.

Now the difference χO-χD is identical to kO

k

D

although

the individual terms are not equal.

Then remember

that kDkO (=K) is g + s for the perfect crystal.Slide6

6

The two equations can be rearranged to give a pair of coupled differential

equations. We say that φ0 and φ

g

are ‘dynamically coupled

.’

The term dynamical diffraction thus means that the amplitudes

(and therefore the intensities) of the direct and diffracted beams are constantly changing, i.e., they are dynamic

Howie-Whelan equations

If we can solve the Howie-Whelan equations, then

we can

predict the intensities in the direct and

diffracted

beamsSlide7

7

I

ntensity

in the

Bragg diffracted beam

T

he

effective excitation error

I

g

, in the diffracted beam

emerging from

the specimen is proportional to

sin

2

(πt

Δ

k

)

T

hus

I

0

is proportional to

cos

2

(

πtΔk)Ig and I0 are both periodic

in both t and seff

Extinction distance , characteristic length for the diffraction vector g

Solving the Howie – Whelan equations , and thenSlide8

8

Intensity related to defects: WHY DO TRANSLATIONS PRODUCE CONTRAST?

A unit cell in a strained crystal will be displaced from its perfect-crystal position so that it is located at

position

r'

0

instead of rn where n is included to remind us that

we are considering scattering from an array of unit cells;

We now modify

these equations intuitively to include

the

effect of adding a displacementSlide9

9

Planar defects are seen when

≠ 0

α

= 2

π

g·R

Simplify by setting:

We see contrast from planar defects because the translation, R, causes a phase

shift

α

=2

π

g·RSlide10

10

Thickness and bending

contoursSlide11

11

Two-beam:

Bend

contours

:

Thickness

–

constant

s

eff

varies

locally

Thickness fringes: seff remains

constant t varies

(Intensity

in the Bragg diffracted

beam)Slide12

12

Thickness fringes

O

scillations

in I

0

or

I

g

are known as thickness

fringes

Y

ou

will only see these fringes when

the thickness of the

specimen varies

locally, otherwise the contrast will be a uniform graySlide13

13Slide14

14

Bending contours

O

ccur

when a

particular set

of diffracting planes is not parallel

everywhere; the planes rock into, and through, the

Bragg

condition

.

Remembering Bragg’s law, the (2h 2k

2l) planes diffract strongly

when y has increased to 2θ

B. So we’ll see extra contours because of the higher-order diffraction. As θ

increases, the planes rotate through the Bragg condition more quickly (within a small distance Δx) so the bend contours become

much narrower

for

higher order

reflections

.Slide15

15Slide16

16Slide17

17

Moirè

fringesSlide18

18Slide19

19

Imaging

DislocationsSlide20

20

Edge dislocationSlide21

21

Edge dislocationSlide22

22Slide23

23

(u)Slide24

24Slide25

25

FCC

BCC

Slip plane:

S

lip direction:

Burger vector:

Slip plane:

S

lip direction:

Burger vector: Slide26

26

FCC

BCC

Slip plane: {111}

S

lip direction: <110>

Burger vector:

a

o

/2[110]

Slip plane: {110}

S

lip direction: <111>

Burger vector: a

0

/2[111]Slide27

27Slide28

28

Important questions to answer:

Is the dislocation interacting with other dislocations, or with other lattice defects?Is the dislocation jogged, kinked, or straight?

What

is the density of dislocations in that region

of the

specimen (and what was it before we

prepared

the specimen)?Slide29

29

Howie-Whelan equations

Modify the Howie-Whelan equations to include a lattice distortion R. So for the imperfect crystal

Adding lattice displacement

α

= 2

π

g

·

R

Defects are visible when

α

≠

0

Intensity of the scattered beamSlide30

30

Isotropic elasticity theory, the lattice displacement R due to a straight dislocation in the u-direction is:

Contrast from a dislocation:

b is the Burgers vector, b

e

is the

edge component of the Burgers vector, u is a unit vector

along the dislocation line (the line direction), and ν is Poisson’s ratio.

g·R

causes the contrast and for a

dislocation

 Slide31

31

g ·

R / g · b analysis

Screw:

Edge:

b

e

= 0

b

|| u b x u = 0

b = b

e

b ˔ u

Invisibility criterion:

Invisibility criterion:Slide32

32

Cindy SmithSlide33

33Slide34

34Slide35

35

g·b = 1

g·b = 2

Screw dislocation

Important to know the value of SSlide36

36

Edge dislocation

Always remember:

g

·

R

causes the contrast and for

a dislocation

, R changes with

z.

We

say that

g·b

= n. If we know g and we determine n, then we know b.

g · b = 0 Gives invisibilityg · b = +1 Gives one intensity dipg · b = +2 Gives two intensity dips close to s=0

Usually set s > 0 for g when imaging a dislocation in

two-beam conditions

. Then the dislocation can appear

dark against

a bright background in a BF imageSlide37

37Slide38

38Slide39

39

Example:

U = [2-1-1]

B=1/2[101]Slide40

40

Imaging dislocations with

Two-beam techniqueSlide41

41Slide42

42Slide43

43

Imaging dislocations with

Weak-beam

techniqueSlide44

44

The contrast of a dislocations are quite wide (~

ξg

eff

/3)

Weak beam

Small effective extinction distance for large S

Two-beam:

Increase s to 0.2 nm

-1

in WF to increase

S

eff

Narrow image of most defects

Characteristic length of the diffraction vectorSlide45

45Slide46

46Slide47

47Slide48

48

Imaging

Stacking faults in FCCSlide49

49Slide50

50Slide51

51Slide52

52

Example: Stacking faults in FCCSlide53

53Slide54

54

Intensity of the fringes depends on α

BF: sin α > 0 : FF (First Fringe) – Bright LF (Last

F

ringe

) -- Bright

sin α < 0 : FF – Dark

LF – DarkDF: sin α > 0 : FF

–

Bright

LF -- Dark

sin α < 0 : FF – Dark LF – Bright α ≠

± πα = 2πg·RSlide55

55

(A–D) Four strong-beam images of an SF recorded

using ±g

BF and

±g

DF. The beam was nearly normal to the surfaces;

the SF fringe intensity

is similar at the top surface

but complementary at the bottom surface. The

rules

are summarized

in (E) and (F) where G and W

indicate that the first

fringe is gray or white; (T, B) indicates top/bottom.Slide56

56

Then there are some rules for interpreting the contrast:

In the image, the fringe corresponding to the top surface (T) is white in BF if g·R

is > 0 and black if

g·R

<

0.Using the same strong

hkl reflection for BF and DF imaging, the fringe from the bottom (B) of the fault will be complementary whereas the fringe from the top (T) will be the same in both the BF and

DF

images.

The

central fringes fade away as the

thickness increases

. The reason it is important to know the sign of g is that you will use this information to determine

the sign of R.For the geometry shown in Figure 25.3, if the origin of the g vector is placed at the center of the

SF in the DF image, the vector g points away from the bright outer fringe if the fault is extrinsic and toward it if it is intrinsic (200, 222, and 440 reflections); if the reflection is a 400, 111, or 220 the reverse is the case.Slide57

57

Images Slide58

58Slide59

59Slide60

60Slide61

61Slide62

62Slide63

63Slide64

64