Technology for a better society 1 Imaging defects and contours/effects.

Technology for a better society 1

Imaging defects and contours/effects


Dynamical diffraction theoryDynamical 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.’


Dynamical diffraction:

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

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


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:

(rn denotes the position of each unit cell)

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


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 f0 by using the concept introduced in equation 13.3 but replacing a by the short distance dz

Two beam approximation

Here χO-χD is the change in wave vector as the φg beam scatters into the φ0 beam. Similarly χD-χO is the change in wave vector as the φ0 beam scatters into the φg beam.

Now the difference χO-χD is identical to kOkD although the individual terms are not equal. Then remember that kDkO (=K) is g + s for the perfect crystal.


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

We say that φ0 and φg 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 beams


Intensity in the Bragg diffracted beam

The effective excitation error

• Ig, in the diffracted beam emerging from the specimen is proportional to sin2(πtΔk) • Thus I0 is proportional to cos2(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 then


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 displacement


Planar defects are seen when ≠ 0

α = 2πg·R

Simplify by setting:

We see contrast from planar defects because the translation, R, causes a phase shift α=2πg·R


Thickness and bendingcontours


Two-beam:

Bend contours: Thickness – constantseff varies locally

Thickness fringes: seff remains constant t varies

(Intensity in the Bragg diffracted beam)


Thickness fringes

Oscillations in I0 or Ig are known as thicknessfringes

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


Bending contoursOccur 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 2θB.

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.


Moirè fringes


Imaging Dislocations


Edge dislocation


(u)


FCC BCC

Slip plane:Slip direction:Burger vector:

Slip plane:Slip direction:Burger vector:


FCC BCC

Slip plane: {111}Slip direction: <110>Burger vector: ao/2[110]

Slip plane: {110}Slip direction: <111>Burger vector: a0/2[111]


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)?


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

Adding lattice displacement

α = 2πg·R Defects are visible when α ≠ 0

𝐼=¿𝜑𝑔¿2

Intensity of the scattered beam


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, be 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

𝑅=12𝜋

(𝑏𝜑+1

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


g · R / g · b analysis

Screw:

Edge:

be = 0b || u b x u = 0 𝑅=𝑏 𝜑

2𝜋= 𝑏2𝜋

𝑡𝑎𝑛−1(𝑧−𝑧𝑑

𝑥)

b = be b ˔ u 𝑅=

12𝜋

(𝑏𝜑+1

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

𝑅=12𝜋

(𝑏𝜑+1

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

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

𝑔 ∙𝑏×𝑢=0Invisibility criterion:


Cindy Smith


g·b = 1

g·b = 2

Screw dislocation

Important to know the value of S


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 image


Example:

U = [2-1-1]B=1/2[101]


Imaging dislocations with Two-beam technique


Imaging dislocations with Weak-beam technique


The contrast of a dislocations are quite wide (~ ξgeff/3)

Weak beam Small effective extinction distance for large S

Two-beam:

Increase s to 0.2 nm-1 in WF to increase Seff

Narrow image of most defects

ξ 𝑔❑𝑒𝑓𝑓 =

ξ𝑔√1+𝑠2ξ𝑔2

Characteristic length of the diffraction vector


Imaging Stacking faults in FCC


Example: Stacking faults in FCC


Intensity of the fringes depends on α

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

LF (Last Fringe) -- Bright sin α < 0 : FF – Dark

LF – Dark

DF: sin α > 0 : FF – Bright

LF -- Dark sin α < 0 : FF – Dark

LF – Bright

α ≠ ± π

α = 2πg·R


(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.


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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