Technology for a better society

Imaging defects and contours/effects

Technology for a better society

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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.’


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Dynamical diffraction:

Assumption: That each individual diffraction/interference event, from whatever locality within the crystal, acts independently of the others

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


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THE AMPLITUDE OF A DIFFRACTED BEAM

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

Structure factor

The intensity at some point P, we then sum over all the unit cells in the specimen.

The amplitude in a diffracted beam:

(r n denotes the position of each unit cell)


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If the amplitude φ g 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

Similarly χ

D is the change in wave vector as the φ g

-χ

O beam scatters into the φ is the change in wave vector as the φ

0

0 beam. beam scatters into the φ g beam.

Now the difference χ

O

-χ

D is identical to k

O k

D equal. Then remember that k

D k

O although the individual terms are not

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


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Howie-Whelan equations

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

We say that φ

0 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

If we can solve the Howie-Whelan equations, then we can predict the intensities in the direct and diffracted beams


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Solving the Howie – Whelan equations , and then

Intensity in the Bragg diffracted beam

The effective excitation error

Extinction distance , characteristic length for the diffraction vector g

• I g

• Thus I

0

• I g

, in the diffracted beam emerging from the specimen is proportional to sin and I

0 is proportional to cos 2 (πtΔk) are both periodic in both t and s eff

2 (πtΔk)


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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 r n 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 displacement


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Simplify by setting:

α = 2πg·R

Planar defects are seen when

≠ 0

We see contrast from planar defects because the translation,

R, causes a phase shift α=2πg·R


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Thickness and bending contours


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Two-beam:

Bend contours: Thickness – constant s eff varies locally

(Intensity in the Bragg diffracted beam)

Thickness fringes: s eff remains constant t varies


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Thickness fringes

Oscillations in I

0 fringes or I g are known as thickness

You will only see these fringes when the thickness of the specimen varies locally, otherwise the contrast will be a uniform gray


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Bending contours

Occur 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.


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Moirè fringes


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Imaging

Dislocations


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Edge dislocation


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Edge dislocation


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22

(u)


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FCC

BCC

Slip plane:

Slip direction:

Burger vector:

Slip plane:

Slip direction:

Burger vector:


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FCC

BCC

Slip plane:

Slip direction:

{111}

<110>

Burger vector: a o

/2[110]

Slip plane:

Slip direction:

{110}

<111>

Burger vector: a

0

/2[111]


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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)?


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Howie-Whelan equations

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

Defects are visible when α ≠ 0

Adding lattice displacement

α = 2πg·R

𝐼 = |𝜑 𝑔

| 2

Intensity of the scattered beam


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Contrast from a dislocation:

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

𝑅 =

1

2𝜋

(𝑏𝜑 +

1

4(1 − 𝑣) 𝑏 𝑒

+ 𝑏 × 𝑢(2 1 − 2𝑣 𝑙𝑛𝑟 + 𝑐𝑜𝑠2𝜑) ) 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


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g · R / g · b analysis

𝑅 =

1

2𝜋

(𝑏𝜑 +

1

4(1 − 𝑣) 𝑏 𝑒

+ 𝑏 × 𝑢(2 1 − 2𝑣 𝑙𝑛𝑟 + 𝑐𝑜𝑠2𝜑) )

Screw:

Edge: b e

= 0 b || u b x u = 0 b = b e b ˔ u 𝜑

𝑅 = 𝑏

2𝜋

= 𝑏

2𝜋 𝑡𝑎𝑛 −1 ( 𝑧 − 𝑧 𝑥 𝑑

) 𝑔 ∙ 𝑅 ∝ 𝑔 ∙ 𝑏 Invisibility criterion: 𝑔 ∙ 𝑏 = 0

𝑅 =

1

2𝜋

(𝑏𝜑 +

1

4(1 − 𝑣) 𝑏 + 𝑏 × 𝑢(2 1 − 2𝑣 𝑙𝑛𝑟 + 𝑐𝑜𝑠2𝜑) )

Invisibility criterion: 𝑔 ∙ 𝑏 × 𝑢 = 0


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Cindy Smith


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Screw dislocation g·b = 2 g·b = 1

Important to know the value of S


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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 g · b = +1 g · b = +2

Gives invisibility

Gives one intensity dip

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 image


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Example:

U = [2-1-1]

B=1/2[101]


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Imaging dislocations with

Two-beam technique


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Imaging dislocations with

Weak-beam technique


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The contrast of a dislocations are quite wide (~ ξ g eff /3)

Two-beam:

Increase s to 0.2 nm -1 in WF to increase S eff

Weak beam Small effective extinction distance for large S

Characteristic length of the diffraction vector

ξ 𝑔 𝑒𝑓𝑓

=

ξ 𝑔

1 + 𝑠 2 ξ 𝑔

2

Narrow image of most defects


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Imaging

Stacking faults in FCC


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Example: Stacking faults in FCC


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Intensity of the fringes depends on α

BF: sin α > 0 : sin α < 0 :

FF (First Fringe) – Bright

LF (Last Fringe) -- Bright

FF – Dark

LF – Dark

α = 2πg·R

DF: sin α > 0 : sin α < 0 :

FF – Bright

LF -- Dark

FF – Dark

LF – Bright

α ≠ ± π


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(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.


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Then there are some rules for interpreting the contrast:

1. 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.

2. 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.

3. The central fringes fade away as the thickness increases.

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

5. 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.


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Images


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Technology for a better society

Imaging defects and contours/effects

Dynamical diffraction theory

Dynamical diffraction:

Howie-Whelan equations

Thickness and bending contours

Moirè fringes

Imaging

Dislocations

Edge dislocation

Edge dislocation

Howie-Whelan equations

Contrast from a dislocation:

g · R / g · b analysis

Imaging dislocations with

Two-beam technique

Imaging dislocations with

Weak-beam technique

Imaging

Stacking faults in FCC

Images

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Technology for a better society

Imaging defects and contours/effects

Dynamical diffraction theory

Dynamical diffraction:

Howie-Whelan equations

Thickness and bending contours

Moirè fringes

Imaging

Dislocations

Edge dislocation

Edge dislocation

Howie-Whelan equations

Contrast from a dislocation:

g · R / g · b analysis

Imaging dislocations with

Two-beam technique

Imaging dislocations with

Weak-beam technique

Imaging

Stacking faults in FCC

Images

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