ELECTRON TRANSMISSION MICROSCOPY

HAL Id: jpa-00215870

https://hal.archives-ouvertes.fr/jpa-00215870

Submitted on 1 Jan 1974

HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

ELECTRON TRANSMISSION MICROSCOPY

R. Gevers

To cite this version:

R. Gevers. ELECTRON TRANSMISSION MICROSCOPY. Journal de Physique Colloques, 1974, 35

(C7), pp.C7-129-C7-139. �10.1051/jphyscol:1974714�. �jpa-00215870�

ELECTRON TRANSMISSION MICROSCOPY

R. GEVERS

Rijksuniversitair Centrum Antwerpen, Middelheimlaan, 1 B-2020 Antwerpen, Belgique

R6sum6. - Le but de ce cours est d'expliquer Ia formation des images de defauts et de montrer qu'une bonne connaissance de la theorie peut donner d'importantes informations sur la nature des dkfauts.

Les diffkrents paramittres decrivant la diffraction des electrons par une lame parfaite sont intro- duits et discutes : longueur d'onde, reflexion de Bragg, angle de Bragg, surface de dispersion, dis- tance d'extinction, kcart B l'interfkrence, longueur d'absorption, champ clair, champnoir, contours d'extinction, approximation de la colonne.

On considkre ensuite les interfaces planes separant deux parties d'un cristal, telles que fautes d'empilement, parois d'antiphase, parois de domaines. On montre que les amplitudes des ondes transmises et diffractees peuvent &tre calculees en considerant que les faisceaux issus de la premiere partie du cristal sont incidents sur la seconde partie. L'image est un diagramme de franges, dont l'interfrange depend du cristal et des plans reflecteurs consider& et non pas du type d'interface.

Quelques proprietes de l'image sont prbentees, et particulierement celles quiipermettent de diffk- rencier les fautes intrinsitques et extrinsitques dans les metaux c. f. c., ainsi que celles qui distinguent les fautes d'empilement des parois de domaine.

Dans Ia dernikre partie, on discute du contraste dQ aux d6fauts tels que dislocations, petites boucles, precipitks, cavitks, amas, etc ... Dans ce cas, l'image est une consequence du champ de defor- mation qui affecte de maniere continue les plans reflecteurs. On presente brikvement quelques pro- priCtes des dislocations, qui peuvent &tre utiliskes pour determiner leur vecteur de Burgers.

Abstract. - The purpose of this short lecture is to explain why defects are imaged in the electron microscope, and to emphasize that a good understanding of the theory can lead to important infor- mations about the defects.

First, the different notions and parameters describing the diffraction of electrons by a perfect foil are introduced and discussed to some extent : wavelength, Bragg reflection, Bragg angle, dispersion surface, extinction distance, excitation error, absorption length, bright field, dark field, extinction contours, column approximation.

Next, one considers planar interfaces separating two crystal parts, like stacking faults, antiphase boundaries, domain boundaries. It is shown that the amplitudes of the transmitted and scattered beam can be calculated by considering the transmitted and scattered waves emerging from the first part again as incident on the second part of the crystal foil. The image is a fringe pattern, the fringe spacing depending on the crystal and the active reflection, and not on the type of interface. Some properties of the image are reviewed, in particular those which enable to differentiate intrinsic and extrinsic stacking faults in f. c. c metals, and the property which enables to distinguish stacking faults and domain boundaries.

In the last part of the lecture, the contrast image is discussed of defects like dislocations, small loops, precipitation, voids, clusters a. s. o. The image is in this case a consequence of theelastic strains which deform continuously the reflecting lattice planes. Some qualitative properties of dislo- cations are briefly discussed, which can be used to determine the Burgers vector.

1. Introduction. - I n previous lectures many trans- mission electron micrographs have been shown. There is n o doubt that this type of observations has contribut- ed powerfully t o a better understanding of crystal defects. Therefore the organizers of the school thought it would be useful t o include a short lecture about transmission electron microscopy for those not acquainted with the technique. Of course the most one can hope t o achieve in so short a time is a rather superficial introduction t o the main problems ; namely (I) why and how is a n image formed ? (2) what

type of information can be deduced from the observed image ?

2. Bloch wave in an infinite perfect crystal. - As well known a free electron with wave vector k, ( k ,

1/A, A : wavelength), described by the wave function :

$,(r)

exp i 2 nk, . r

cannot move as such in an infinite perfect crystal, where it feels the three-dimensional periodic crystal

Article published online by EDP Sciences and available at http://dx.doi.org/10.1051/jphyscol:1974714

potential V(r) due to the nuclei and the electrons of the crystal. In fact it will suffer Bragg-reflection on the different lattice planes, defined by the reciprocal lattice vectors h. As a consequence a so-called Bloch wave is constructed, described by the wave function :

$(r) = [exp i 2 nk, . r] U,,(r), U,,(r) : periodic (2) or, after the introduction into (2) of the Fourier series for U(r) :

U,,(r) = C $(;ko) exp i 2 nh. r

h

(3)

$(r) ⁼ 1 _t,bikO) exp i 2 n(k0 + h).r . (4)

h

The wave field (4) must be interpreted as follows.

First one assumes a beam of free electrons with wave vector k, and energy :

h2 kt

E(ko)

-- m = electron mass (5) 2 m '

and density N cm-2 s-'. Next, we switch on the crystal field V(r). Bragg reflection occurs and a statio- nary regime establishes itself. The beam propagates through the crystal as a Bloch wave, such that there will be at any moment N I $Po) l 2 cm-2 s-' electrons of the beam moving in the direction obtained by reflection on the h-plane. Each electron, on its way through the crystal, will suffer many reflections ; the total number of electrons moving in a given direction will however remain the same.

3. Dispersion surface. - The electrons passing through the crystal have been accelerated up to a potential of 100 kV to 1 000 kV. Their velocity is then such that relativistic effects must be taken into account.

I t can be proved that this is done in very good approxi- mation by using in all the formulae the relativistic mass :

(m, ^: rest mass, v : velocity of electron) and the relati- vistic definition of k, = 111, i. e. (Eo : accelerating potential)

h eE

p = - = [2 (eEo) (m, + ²⁾ ] ^{= hko}

2 mo c2

The wave length A and wave vector ko have thus values varying from 37 x A to 87 x A, respecti- vely 27.02 to 114.7 A-'. A Bloch wave is charac- terized by its wave vector k,, determined up to a reciprocal lattice vector, since any term in (4) can be chosen as the leading one. The energy of this Bloch

wave depends also on k,. However, a given energy can correspond to an infinite number of k,-vectors. This can be most simply described by introducing the notion of dispersion surface. This is the surface in reciprocal space formed by the origins of all ko-vectors cor- responding to a given energy if these are drawn such that their endpoints coincide with the origin of the reciprocal lattice of the crystal.

The crystal potential V(r) can be expanded in Fourier series as follows :

V(r) = 1 Vh exp i 2 7ch.r (6)

h

where V, and Vh can be calculated from the atomic wave functions. For low order h-lattice planes one can estimate Vh to be of the order of a few times one volt and Vo of the order of magnitude of 10 V. Since one uses electrons accelerated up to 100 ti 1 000 kV, its potential energy eV(r) will remain only a very small fraction of the total energy all along the passage through the crystal. This will enable us to conclude something about the dispersion surface without actual calculation. First, one neglects V(r) ; the dispersion surface is then formed by spheres with radius k, = (2 m ~ / h ~ ) ' / ~ centered around each reciprocal lattice point. The influence of the perturbation V(r) will only be important near the points where two or more spheres intersect. For a point where the sphere centered at the origin and the sphere centered at the reciprocal lattice node h intersect, one has :

The condition (7) is the exact Bragg condition for reflection on the lattice planes defined by the reciprocal lattice vector H. If (7) is satisfied, or nearly satisfied for a single H-vector, one has a two beam situation. I. e. the Bloch wave contains only two important beams : the transmitted k,-beam and the scattered k, + H-beam, apart from many other much weaker beams. In other words : only the term VH in (6) is very active and makes tha't in (4) the terms $(oko) and are much more important than the other ones. Neglecting the latter is known as the two beam approximation and has been proved to be very successful for predicting observations.

In figure 1 the intersection of the dispersion surface with the plane through the reciprocal lattice vector H containing the incident beam k, is shown in the region of the reciprocal space where (7) is nearly satisfied.

Since k,

20 A 100 A-' and H N lo-' A-1 the radius of the spheres are very large as compared to the mesh of the reciprocal lattice. Therefore the spheres are replaced in very good approximation in figure 1 by their tangent planes xx' and yy'. The line zz' is then the trace of the Brillouin zone boundary.

The perturbation V(r) deforms (xx', y ~ ' ) to a hyperbole-like curve shown as full lines in figure 1. It would be interesting to estimate the distance A' A"

between the two branches of the dispersion surface

without actually going through all the calculations.

Since I k' I

I kt + H I one can note that the elec- trons in the wave field corresponding to the point A' of the upper branch have kinetic energy h2 E 2 / 2 m, whereas those in the wave field corresponding to the points A of the lower branch have the smaller kinetic energy h2 W2/2 m. The very small difference in kinetic energy is compensated by a difference in potential energy. In the considered two-beam situation it is plausible to assume that :

However it follows, if one notes A' A = l / t H , from k' - ^k" - ^k, >>> H, in very good approximation : and thus :

It turns out that formulae (8) is better than a good estimate ; it is the correct expression for t,.

The very important parameter t,, called extinction distance, depends not only on V,, i. e. the material and the reflecting lattice planes, but also on the accelerating potential Eo through m en k,. Introducing (5a, b) into (8) tH becomes :

for VH ⁼ 1 V and that the second factor varies smoothly from 0.55 0.94 if Eo increases from 100 kV to 1 000 kV, and that VH is a few times one volt, one concludes that t , will be of the order of magnitude of several times 100 A.

In figure 1 the incident beam ko satisfies exactly the Bragg condition for reflection on the lattice planes H and the Bragg angle 2 8, between k, and k, + H is also the angle between xx' and yy' (see Fig. 1). Since k, %- H, one has in very good approximation :

leading to very small Bragg angles of the order of magnitude of a degree of arc.

An incident beam k, can however deviate from the exact orientaion for Bragg reflection, as in figure 2.

This misorientation can be described by x, the distance of the origin of k, to the Brillouin zone (see Fig. 2). An associated parameter is :

called the deviation parameter of the exact Bragg condition, or al'so the excitation error.

112

tH = -- hc

2 ( e v ~ ) eEo (9) The interpretation of s is as follows. I t is the distance

1 f -

mo c2 measured in the direction of k, from the endpoint of k, + H to the Ewald sphere. In fact from

Taking into account that the first factor

(k, + ^H + ^{s ) ~ =} kg, -- hc

2(eVH) - 248 A it follows, neglecting in very good approximation s2

and putting in very good approximation k, .s = k , s, one finds :

or - xH + ^{sk, =} 0, leading to (1 1).

Notice that x should be measured in the direction of H, starting from the Brillouin zone boundary. In particular it follows from this that s > 0 means that k, + H is inside the Ewald sphere and s < 0 that k, + H lies outside the Ewald sphere. There is of course a relation between the deviation A8 of 8 from 8, and x or s.

One has obviously in very good approximation (see Fig. 4) :

The distance B' B" corresponding to s becomes larger if I s I increases. Moreover B' B" must tend to llt, for s = 0, to I s I if s increases indefinitely and moreover it has to be an even function of s. The most simple expression satisfying this condition is :

and it turns out that (13) in fact, is exact.

4. Diffraction of electrons by a pIate-shaped crystal foil. - We consider now with a plate-shaped crystal foil with thickness

(Fig. 5). Apart from the elastic

scattering considered until1 now there is also inelastic scattering which reduces the transparancy to thick- nesses

to a few times 1 000 A for E, = 100 kV up t o something like one micron for Eo = 1000 kV. For simplicity we assume further a two-beam illumination situation. Since 0, < < < 1 the incident beam k, is nearly parallel to the reflecting planes H. To avoid some further complications we assume that the foiY surfaces are normal to the reflection planes. Since V, is very small compared to E, it will also be a very good approximation to neglect reflection and refraction.

The wave field describing the passage of the electrons through the crystal is now found by expressing that the wave function, describing the beam outside and inside the crystal, and the projection of its gradient on the normal of the entrance surface must be continuous across the entrance surface. Let us choose the origin in the entrance surface and let r, be the position vector of a point of the entrance surface. It is obvious that one can not satisfy the continuity of the wave function if there is only one wave field excited inside the crystal, since one should then have :

exp i 2 n(k,),, .ro = exp i 2 7ckI1 .ro Uk(rO) (14) where (ko)ll, (k)l, are the projections of k, and k on the entrance surface.

From (14) should then follow :

and The result is :

The condition (16) is of course impossible to satisfy.

From (15) it follows that the Bloch wave fields are excited which correspond to the intersections of the different blades of the dispersion surface with the normal on the entrance surface. The strength A, of the

excitation of the different Bloch waves is then further and thus also, taking (15) into account : determined by the condition :

S

An uk,,(ro) . (17) cos 2 cp = - - a , sin 2 cp = ot

n

In a two beam situation, one considers only two wave (28)

vectors k' and k (see Fig. 1 or 2). It seems plausible to The expressions (23) for the amplitude express that the expect that the corresponding Bloch waves are ortho- transmitted beam is obtained by the interference of the gonal and exact calculation proves this to be true. two transmitted beams contained within the Bloch After normalisation one has then : wave fields with different wave vectors k' and k", such

that k' - k" = oe,.

+ , ^{= (cos cp} + sin cp exp i 2 nH.r) exp i 2 nk'.r

(18) The same reasoning is valid for the amplitude $, .

+,. ^{= (sincp -} coscpexp i2nH.r)exp i2nkr'.r The prefactor :

and the condition (17) becomes in this particular case : _{sin2 2 cp =} - 1 - - ¹

(at)" l+ (st)2 (29) ,A, cos cp + A, sin cp = 1 A, = cos cp

A , sin cp - A, cos cp = 0 or A2 = sin cp (19) is maximum and equal to one for s = 0, i. e. for the From (18) and (19) follows then for the wave field

inside the crystal, taking into account that (see Fig. 2) (e, : unit vector normal to the surface pointing inwards) :

+ ^{= ((COS naz + i cos 2 q . sin noz) +}

+ (i sin 2 qo . sin noz) exp i2 nH . ^R)

x (exp insz) (exp i 2 nkO .r) . (21) After the crystal there is a transmitted and a scattered beam i. e. the wave field is :

II/ = $,(exp i 2 nkO .r) +

+ $, exp i 2 n(ko + ^H + ^{se,) .r} . (22) From the condition that the wave function (21), (22) must be continuous at the back surface, follow imme- diately the amplitudes $, and

One finds :

$,

(exp insz,) T , ^{T= cos naz,} + i(cos 2 cp) sin nozo (23)

$H

= (exp insz,) S , S = i sin 2 cp . sin naz, . (24) The intensities of the transmitted and scattered beam : I. = I ^{1' = 1 T 1' = 1 -} (sin2 2 cp) sin2 noz, (25)

exact Bragg orientation, and decreases when the inci- dent beam deviates more and more from the Bragg condition.

It should be emphasized that the interaction of the accelerated electron with the crystal is very strong, since (1) the diffracted beam can be very strong if the thickness is not extremely small and s not very large ; (2) the Bragg condition must not be very strictly fulfilled, since sin2 2 cp becomes only small when

I st I ^% I, or following (lo), (12), where :

In particular this means that even in a good two beam situation there will be a large number of weak beams.

5. Absorption effects. - Apart from the elastic

scattering giving rise to the diffraction, there are also

the quasi-elastic scattering events whereby phonons are

annihilated or created and the inelastic events whereby

e. g. x-ray photons and plasmons are created. These

events yield electrons which are not found back in the

I , = I $, 1' ⁼ I S l2 = (sin2 2 cp) sin2 noz, different beams but in the diffuse background, or which

(26) are scattered out of the optical system of the micro-

fluctuate periodically with crystal thickness zo with scope. In transmission r n i c r o ~ c o ~ ~ one looks at the

period llo given by (13). This feature is important enlarged image of the transmitted beam (bright field

since z, is mostly several times 110. The second factor image), or of one of the scattered beams (dark field

in (26) can thus be important. image). The loss of electrons in these beams can be

The expressions for cos cp, sin cp can only be found by described as absorption. It can be shown that this effect

actually solving the Schrodinger equations. can be accounted for by the usual phenomenological

method consisting in replacing V(r) by the complex potential V(r) + iW(r). This amounts to replacing

where z, is called the absorption length. The ratio

z,/t, can be estimated to range from 10 to 30.

If one effectuates the substitution (30) into (13), one finds if higher powers of t/z are neglected :

In (18) one has now to take into account that k' and k"

become complex. Namely :

Furthermore, one has to introduce a factor exp (- ^{n 3)} expressing the normal absorption cor-

z

responding to the mean value of W(r). Moreover 9 will become complex, but the imaginary part will be unimportant and it is a very good approximation to neglect the influence of the absorption on cp. One finds then, for the two wave fields :

$y ^{= ' $ ;} exp - nz i:, - -

where the index 0 means that one takes the expressions neglecting absorption.

The interpretation of (32) is straightforward : the wave field k' is much more strongly absorbed than the wave field k . Moreover the two absorption coefficients depend on the crystal orientation (anomalous absorp- tion). This can be understood as follows : Since k' > k , the electrons of the wave field k' will have larger kinetic energy than those of the other field. They will propagate thus also more in regions of lower potential energy. Since V(r) is attractive, this means that the electrons of wave field kt will travel closer to the nuclei then the electrons of the other field, therefore they will have a much larger probability to create phonons and X-rays. For s > 0 one has, from (27) that sin 9 > cos 9, i. e. one excites more the easily trans- mitted wave field than the strongly absorbed one. For s < 0 the reverse takes place. Thetotalintensity will be larger for s > 0 than for s < 0, in thick absorbing crystals.

From (1 3) and (27) it follows that :

7E

4 s ) = 4- S) and cp(- s) = - 2 - p(s). (33) I t follows then from (33) and (24), that :

Is($ = Is(- 3) (34)

even if absorption is taken into account.

From (33) and (23) will only follow : T(- s) ⁼ T*(s) and thus IT@) ⁼ IT(- s) if o is real, i. e. if one neglects absorption. The latter approximation is however only good for foils thinner than say one extinction distance.

In most circumstances it is a bad approximation t o neglect absorption. One has then :

I@) # I( - $1 -

For s < 0, cos cp > sin cp and one excites strongly the wave field k' which is strongly absorbed. On the contrary if s > 0 one has cos 9 > sin rp and one excites thus more the other, more easily transmitted wave field. One expects thus :

I(s) > I(- s) , for small s . (35) 6. Column approximation. - When a narrow beam enters a crystal foil, say under the two-beam illumina- tion condition described above, two wave fields are created. They describe electronic charges moving through the crystal with a group velocity equal to :

The direction of the movement is thus parallel to the normal of the dispersion surface. For s = 0 this is for the two wave fields the direction parallel to the reflect- ing plane and belonging to plane containing H and k,.

For a reasonable value of s, different from zero, say st = +, 1, 2, ... both wave fields will have different directions. The angle between the two directions will be however smaller than 0, and can be estimated to be of the order of magnitude lod3. For foils of at most a few times lo3 A it can be expected to be a good approxi- mation to neglect this effect. This leads to the so-called column approximation.

Assume a crystal of an arbitrary form as on figure 6 , oriented such that only one family of reflecting planes, defined by the reciprocal lattice plane H, is near the exact orientation for Bragg reflection. The wave field at a point like A', B', of the exit surface is then the same as that calculated for a plate-shaped foil of thickness AA', BB', if AA', BB', . .. is parallel to the component of k, parallel to the reflecting planes.

7. Extinction contours. Thickness contours. - In dark and bright field one observes I,@,) and Is(zo).

If absorption is neglected these functions are periodic

with period 110. When absorption is taken into account,

I, and Is are still pseudo-periodic with this period, i. e.

there are still maxima and minima. The distance between two extrema of the same kind being ^llo,. The contrast i. e. the difference between maxima and the successive minima, decreases with thickness. In fact, for large thickness the strongly absorbed wave field is nearly extinct, and the fluctuating interference of the two wave fields becomes less important.

In a bright or dark field micrograph of a wedge- shaped foil the minima of IT(zo) and I,(z,) are observed as dark extinction contours.

In general the crystal foil under observation is slightly buckled. This means that s varies from point to point. Small orientation differences of the order 10-3-10-2 radians result already in variations of the parameter st corresponding to important variations of I,(s), Is($. The geometrical loci of the points for which IT(s), Is(s) are minima (as function of s) are inclination extinction contours. They will move rapidly over the image if the foil in slightly tilted ; thickness contours on the contrary will remain nearly in place.

8. Image of a stacking fault. - We consider a plate- shaped crystal foil under the two-beam illumination condition described above. We cut the crystal along a lattice plane parallel to the crystal foil at the depth z, under the entrance surface. Let z, be the total thickness and let z, =

- z,.

Next, we displace the second part over a vector R, where R is not a lattice vector (Fig. 7).

The reflecting planes H are the same in the two parts of the crystal, but they are displaced with respect to each other. It is obvious that one expects that the dis- placement of the planes measured in interspacing distance d, = 1/H will come into the theory, i. e. H.R (see Fig. 8).

It is now easy to calculate the amplitudes of the transmitted and diffracted wave as follows. The beams transmitted and scattered by the first part, can be considered separately as incident on the second part.

The beams transmitted and scattered by part I are given by

(exp insz,) T(s, 2,) exp i 2 nkO .r ; (exp izsz,) S(s, z,) (exp i 2 nH .r) (exp i 2 nk, . r) if the origin is chosen in the entrance surface of part I.

In order to apply these formula we have to rewrite them with the origin now in an equivalent point of the entrance surface of part I1 i. e. we have to substitute :

r -+ r + R + 1 , 1 : lattice vector.

Taking into account that H.1 : integer, and noting :

one finds :

(exp insz,) T(s, 2,) exp i 2 nk, .r ;

(38) (exp insz,) S(s,

(exp iol) exp i 2 n(ko + H) .r The total transmitted beam is obtained by the inter- ference of (1) the doubly transmitted beam :

(exp ixsz,) T(s, z,) (exp insz2) T(s, z,) exp i 2 nk, . r (39) (2) the doubly diffracted beam.

One has now to take into account that for the inci- dent beam k, + ^M + s the diffraction vector is - H and the excitation error - s (see Fig. 9).

One obtains then for the doubly diffracted beam : (exp insz,) (exp ia) S(s, z,) x

x (exp - insz,) S( - S, z2) exp i 2 n(ko + H + s - H) . r, .

(40) Taking into account that s.r, = sz,, one obtains finally for the transmitted beams :

t,bo = (exp insz,) T , T ⁼ T(s, z,) T(s,

+

+ ( e x p i ~ ) S ( s , z , ) S ( - s,z,). (41)

For a perfect crystal one has a = 0, and finds then : To($, zo) = T(s, 21) T(s, ~ 2 ) + S(s, zl) S(- s, z,) . (42) From (41) and (42) follows then :

T ⁼ To - (1 - exp ia) S(s, z,) S(- s, z2j . (43) In the same manner, the scattered beam is found by the interference of two beams, one finds :

S ⁼ T(s, zl) S(s, z2) + (exp ia) S(s, zl) T(-s, 2,) (44) S= So - (I - exp ia) S(s, 2,) T(-s, z,) ; So = S(s, z,).

(45) In general the stacking fault plane is inclined with respect to the foil surface (see Fig. 10). The for- mulae (43), (45) can however still be applied by using the column approximation. One has then to substitute into (43), (45) :

For the meaning of (z,, x, 1) see figure 10.

The expressions (43) and (45) are pseudo-periodic in z,, z, with period llo,. From this and (46) follows that the image of the inclined stacking fault will be a fringe pattern. The fringes are parallel to the intersection of the stacking fault plane and the foil surface ; the distance between two successive bright or two succes- sive dark fringes will be l/o, cotg 9.

9. Some properties of the stacking fault image. -

9.1 EXTINCTION. - The image will disappear for a two beam illumination if one uses reflecting planes H such that :

exp ia = e x p i 2 n H . R ⁼ 1 (46)

H.R ⁼ integer .

From the different H-reciprocal lattice vectors satisfy- ing (46) it is mostly possible to deduce f R.

10. Type of stacking fault. - It is impossible to determine the sign of R from extinctions alone, since this sign does not inffuence (46). This sign however contains much information about the stacking fault. A stacking fault e. g. in a f. c. c. meQ1 can be intrinsic or extrinsic ; the displacement vectors of the two types are obviously opposite. It can be shown that the sense of R can be determined if the sign of sin a can be found. One must thus look for a property of the image depending on the sign of sin a. From (43) follows immediately that :

IT(zI, ~ 2 ) = IT(z2, ~ 1 ) (47) since S(- s, z) = S(s, z), also if absorption is taken into account. The relation (47) means clearly that the bright field fringe image is symmetric. The nature of the outer fringes will be the same, and calculations prove that this nature depends only on sin a, i. e.

bright for sin a > 0 , dark for sin a < 0 The same is not true for I, = I S 12, as foIlows from (45).

Calculations show on the contrary that the nature of the outer fringes will be different, but will only depend on the sign of sin a.

One finds :

Front surface Back surface

- -

sin a > 0 B D

sin a < 0 D B

From the bright field one deduces the sign of sin a ; from the bright and dark field the sense of the slope of the fault and on the diffraction pattern one finds the active reciprocal lattice vector H. From these infor- mations one can then deduce the sense of R, i. e. the type of the stacking fault.

11. Images of other types of planar defects. - We

assume a plate-shaped crystal foil formed by the super-

position of two crystal parts I and I1 (see Fig. 1 I), such that the interface is a common lattice plane. Moreover, one supposes that the lattice of I and I1 are nearly the same. Boundaries of this type arise on fragmentation by

ordering of e. g. spins (antiferromagnetic domains in NiO), electrical dipoles (ferroelectric domains in BiTiO,), or impurities like hydrogen in b. c. c. metals.

In all these cases I and I1 are twin related ; the twin vector is however very small : something like one percent or less of a lattice vector. In such case one can tilt the crystal such that only one reciprocal lattice vector becomes active in the two parts : H, and Hz.

In other words : the same family of reflecting planes is active, but its orientation changes slightly on passing the interface : this means that the excitation errors will be different in I and I1 : s,, s,. The parameter describing the boundary will then be :

A reasoning completely analogous to the one used. for the stacking fault, will give now :

The bright and dark field ipage I T l2 and I S 1' will again be a fringe pattern, somewhat more complicated than in- the previous case, since T, S(s,) and T, S(s2) have somewhat different depth periods t/ Jl + ^{(s, t)'}

and t/d-, if I s1 1 # I s2,1. Computing the images from (49), one finds that there will be already a good contra~tedima~e if I 6 1 is only a few times 10".

From this follows that orientation digerences of a small fraction of a degree will already be detected.

In order to avoid misinterpretations, one looks for image properties enabling to distinguish clearly a stacking fault from a domain boundary as considered here. Starting from (49) one can prove now that the nature of the outer fringes of the bright field will be different, whereas the outer fringes have similar nature in the dark field. It is clear that this is the property one needs for making the difference.

More complicated planar defects are possible, but one can still calculate in these cases the amplitude T and S in the way used for the calculation of the a- and the b-fringe pattern images.

12. Strain contrast-dislocation image. - In this section we assume that there is a defect in the crystal foil that introduces long range elastic strain, as e. g.

dislocation lines, small dislocation loops, small preci- pitates, voids, clusters of point defects, etc.

Except maybe very close to the site of the defect this strain can be described by the displacement function :

In the neighbourhood of a point one can still recognize lattice planes. The direction of these planes, however, vary continuously from point to point. In other words : the lattice planes are curved.

If a surface with equation off (r) = C becomes after deformation the surface with equation q(r) = C, one has (see Fig. 12) :

or, since R is small :

The family of lattice planes defined by the,reciprocd lattice vector g has the equation :

The equation of the deformed lattice planes in the deformed crystal is then :

the normal is given by the gradient, i. e. by :

g - grad (g . ^{R) = g} + Ag , Ag ⁼ - grad (g . ^R) . ⁽⁵⁰⁾

The change Ag(r) of g induces, a variation in s given by (see Fig. 13) :

When the electrons travel drown along the column defined by the point (x, y) of incidence (see Fig. 14), the diffraction conditions will change continuously, accord-

10

ing to (51). In different columns the strain will be different and thus also the number of electrons coming out of the column in the transmitted and diffracted directions. The strain is imaged ; the defect is made visible by strain contrast.

Sufficiently far from the defect the strain will be too small to be imaged ; this determines the width of the image. One can obtain a rough estimate by expressing the condition :

Or, taking (51) into account

A typical value for the second term is Strains around the defects are mostly larger than this value and the image will be fairly broad.

As an example we consider a screw dislocation in an isotropic medium parallel with the foil surface (see Fig. 15 for the meaning of x and z).

FIG. 15.

The condition is then :

x 2 n

n--N- , (n : integer)

X2

+ ^z2 - ^t or for z = 0, x E nt/2 n.

The width will be of the order of magnitude of the extinction distance. This is very favourable : far above the resolution of the microscope, but small enough to be imaged as fine well separated lines.

13. Extinction. - In a good two-beam illumination condition the images will be absent if one satisfies :

The condition (54) means that the displacement must be parallel to the reflecting planes in order to expect no influence upon the diffraction conditions, and thus no image. Clearly (54) can be only satisfied in very exceptional cases.

This displacement R of a pure screw dislocation, in the isotropic assumption, is parallel to the Burgers vector b. There will thus be extinction if one has :

i. e. the reflection planes contain the Burgers vector. It is thus sufficient to find two extinctions for determining the direction of the vector. Its length can also be determined but not in a simple way, by using very refined features of the images and which cannot be discussed in this short lecture.

If anisotropy is taken into account and also if the dislocation is not pure screw, the condition (54) is not fulfilled for (55). There remains some deformation of the reflecting planes. The image is however poor, and one can mostly still recognize this as a nearly extinction.

14. Image of a dislocation line in a thick foil. - In a thick crystal most dislocation lines shown up as darker lines on a brighter background as well in the bright field as in the dark field. This can be predicted by the following argumentation. By a thick crystal is meant a foil with a thickness not small compared to the absorp- tion length, say from five to ten extinction distances.

The electrons leaving a column of the background far from the dislocation are then in good approximation all in the easily transmitted wave field. Let us consider now a column not far from a dislocation lying near the middle of the foil. In the first part of the column the strain remains so small that one can consider this part as perfect, in good approximation. Due to the absorp- tion effect most of the electrons arriving at the central part of the column, where the strain is important, are travelling in the easily transmitted Bloch wave field.

The strain in this central part of the column induces

transitions of the electrons from this easily transmitted

wave field to the strongly absorbed one. The strongly

strained part of the column is rather thin compared to

an extinction distance and this prevents that many of

these electrons will return through multiple diffusion to the easily transmitted wave field. The electrons transferred in this way to the strongly absorbed wave field will be lost eventually when travelling down through the last part of the column, since here again the strain is so unimportant that the crystal can be in good approximation considered as perfect. The total number of electrons leaving such a column will thus be smaller than the number leaving a column far away from the dislocation. This rough reasoning explains why the image, in bright and dark field, is a dark line and why it is necessary to take the absorption into account when calculating the images. The argumenta- tion developed here is obviously only valid if the first and last parts of the column are not small compared to the absorption length. The images of dislocations close to the surfaces are in fact different. They have a doublet structure, formed by a bright and dark line.

Actual calculations predict such images if absorption is taken into account and also surface relaxation. In favourable circumstances one can use this feature for determining the sign of the Burgers vector, by looking carefully at the side where the bright line is situated.

15. Sense of the Burgers vector:- To end this section we give a short description of a property of the dislocation image which can lead to the determination of the sense of the Burgers vector. This has enabled, e. g. in radiation damage studies, to find out if small dislocation loops found after annealing, where formed by the condensation of vacancies or interstitials.

Assume one has tilted the foil away from the exact Bragg orientation, such that ^s > 0 (Fig. 16a ; G inside the Ewald sphere). One can expect that the image will be formed by columns for which A g given by (50) is

such that g + Ag lies very close to the Ewald sphere.

This means that the image will not coincide with the exact position of the dislocation line. Moreover its position will vary with s. If the crystal is tilted such that s < 0 (Fig. 16b ; G outside the crystal), the A g corresponding to the image region will be roughly opposite to the A g for s > 0. This means that the image will ly at opposite sides of the line for s > 0 and s < 0.

It is clear that methods can be developed with the help

of this property for determining the sign of the

Burgersvector, since changing b into - b corresponds

to changing R into - R, and thus also Ag into - Ag,

following (5 1).