Electron Diffraction

Electron Diffraction

August 1999

1. Wave Character of Electrons

Electrons, like all particles, have a dual character: having both particle and wave characteristics. In this experiment the wave character is used to determine the atomic distances in crystalline matter. The electron wave is described by two important parameters, the wave vector and the frequency. The wave vector is given by the DeBroghli condition

λ = h

p = h

√2mE or k = 2π

λ = p

~ (1.1)

where p = mv is the momentum of the electron, h and ~ are the Planck constants, and E = p²

2m is the energy of the electron. If the electron is accelerated by an electric Þeld and passes a voltage V , then E = eV . The frequency of the electron wave is given by

ω = E

~

(The frequency of a particle wave can not be measured. Only frequency differences have physical relevance).

2. Scattering of a Wave by a Periodic Crystal

If a wave (any wave) with an adequate wave length hits a periodic arrangement of atoms (such as a crystal), then the wave is scattered by each atom of that periodic

crystal. Each atom generates a (spherical) scattering wave. The waves created by each atom interfere and cancel each other in almost all directions. Only in a few direction the interference is constructive. This constructive interference is described in relatively simple terms by the Bragg condition and in a somewhat more sophisticated treatment by the reciprocal lattice of the crystal. (For details see, for example: Kittel, Solid State Physics or Ashcroft and Mermin: Solid State Physics).

2.1. Different scattering methods

There are several methods to investigate the atomic arrangement of the crystal using waves. We will discuss here only the two extremes: (i) the Debye-Scherrer method and (ii) the Laue method.

2.2. Bragg condition

Bragg considered parallel lattice planes of the periodic lattice separated by the distance d. In a simple cubic crystal, for example, the x-y-plane (called the (0,0,1)- plane) would be such a lattice plane. Here the distance d between neighboring planes is the lattice parameter a. There is a whole family of lattice planes parallel to the x-y-plane which are shifted by a multiple of a (or d). An incident (electron) wave whose direction forms the angle Θ with the lattice plane will be reßected by the plane with the angle Θ when the phase difference of the scattered waves, scattered by different planes is a multiple of 2π (see Þgure).

Fig.1: Bragg reßection by parallel planes.

This yields the condition for constructive interference

nλ = 2d sin Θ (2.1)

To obtain constructive interference, the lattice plane has to have the correct angle with respect to the incident wave. This is easily achieved by the Debye-Scherrer method.

2.2.1. Debye-Scherrer method

In the Debye-Scherrer method one uses a poly-crystalline solid or, preferably, a powder of the material to be investigated. In this case the incident wave always Þnds some small crystallites whose lattice planes have the correct angle with respect to the incident wave. The angle between direction of the incident and the scattered wave is 2Θ. The direction of the scattered waves lies on a ring of angle 2Θ about the direction of the incident wave. The ring with the smallest radius gives the distance of the most densely packed lattice planes. (Their distance is the largest in the crystal because the distance of neighboring planes is inversely proportional to the density of atoms in the lattice plane).

The Debye-Scherrer method is not well suited to determine the structure and orientation of a single crystal. To obtain these information one uses the Laue method. The theoretical background is more involved and can only be sketched in this discussion.

2.3. Laue method

An alternative description of the wave diffraction by a crystal uses the reciprocal lattice (see below an example). For each periodic arrangement of atoms one can construct a reciprocal lattice. The position of the lattice points in the reciprocal lattice is given by the lattice plane in the real lattice. First we shall explain this method with the simple example of a (0,0,1)-plane of the primitive cubic lattice.

1. One determines the unit vector n normal to the lattice. For the (0,0,1)-plane this is the vector [0,0,1], i.e. the unit vector in z-direction.

2. one determines the distance d to the closest parallel plane.

3. Then the vectors G given by G = ν2π

d n for ν = ±1, ±2, ±3, ... are all reciprocal lattice vectors (they belong to the same family of lattice planes).

The Laue condition for a constructive diffraction is that the difference between the wave vector k⁰ of the scattered wave and the wave vector k of the incident wave is equal to a reciprocal lattice vector.

K= k⁰ − k = G (2.2)

A simple graphical construction makes this relation more clear. One draws in k-space the reciprocal lattice and the vector of the incident beam k from the origin. Since the scattered wave has the same energy and wave length as the incident beam one has |k| = |k⁰|. Therefore the vector K lies on a sphere of radius k with its center at k. This sphere is called the Ewald sphere.

Since the diffraction of the wave conserves the energy of the wave its wave length is unchanged. This means that |k⁰| = |k|. The Laue condition (2.2) requires in addition that for a constructive interference the vector K coincides with a reciprocal lattice vector. For a given orientation of the crystal and a given wave vector of the incident wave one obtains any diffraction only by chance.

However, if one maintains the direction of the incident wave but allows a wide range of wave lengths (which is easy in x-ray waves) then one obtain a Þnite number of directions (given by K) with constructive interference. This is shown in Fig.2. These diffraction points (observable on a screen or a photo) contain much more information about the structure and orientation of the crystal then the Debye-Scherrer method. Its evaluation, however, is also considerably more complicated than the Debye-Scherrer method.

Fig.2: Edwald construction of a Laue diffraction experiment 2.4. Hexagonal structure

The hexagonal lattice has primitive vectors a₁ = a (1, 0, 0) a₂ = a

Ã

−1 2,

√3 2 , 0

!

(2.3) a₃ = c (0, 0, 1)

a is the atomic distance in the x-y-plane and c is the distance of neighboring planes in z-direction.

In nature one Þnds the close packed (hcp) structure. Here one has in the x- y-plane a hexagonal lattice with the vectors a1 and a2. The next plane of atoms lies in the valleys of the Þrst layer. For the ideal hcp structure it is shifted by the vector a¡₁

2,¹₆√ 3,¹₃√

3¢

. The following layer lies directly above the Þrst layer and its distance from the Þrst layer is c. In the ideal hcp crystal it is shifted with respect to the Þrst layer by a

µ 0, 0,2

3

√3

¶

, i.e. c = 2 3

√3a. In real hcp crystals

one has often a deviation from the ideal lattice distance and the value 2 3

√3a is replace by c.

2.5. Graphite structure

The structure of graphite is shown below. The left part shows the graphite planes in three dimensions. The right side shows the hexagon structure in one plane (full lines) and the next upper (or lower) one (dashed lines).

The structure is not given by a Bravais lattice but it has a (complicated) basis.

In the x-y-plane it can be described by the primitive vectors a₁ = (0, a, 0)

a₂ =

Ã√3 2 a,−1

2a, 0

!

where a is ”not” the atomic distance in the x-y-plane but but the distance of second nearest neighbors as shown on the left side of the Þgure, a = 2.46A. (The atomic distance in the x-y-plane is a/√

3 ≈ 1. 42A). The primitive vector in z-direction is

a3 = (0, 0, c)

The three primitive vectors describe a hexagonal lattice. However, the graphite lattice has four basis vectors hν. Two basis vectors lie in the x-y-plane

h₀ = (0, 0, 0) h₁ =

µ a

√3, 0, 0

¶

There are two more basis vectors in the intermediate plane h₂ =

µ a 2√

3,a 2,c

2

¶

h₃ =

Ã√3a 2 ,a

2,c 2

!

By combining an arbitrary multiple of a1, a2 and a3 one obtains the position Rj1j2j3 of the Bravais lattice

Rj1j2j3 = j1a₁+ j2a₂+ j3a₃

To obtain also the position of all atoms one has to add the ”basis vector” hν to Rj1j2j3.

2.5.1. Reciprocal lattice of graphite

The primitive vectors of the reciprocal lattice are deÞned as

b₁ = 2π a₂× a³ (a1a₂a₃) = 2π

ac Ã√3

2 ,1 2, 0

!

a²c

√3 2

= 2π a

√3 2

Ã√3 2 ,1

2, 0

!

b₂ = 2π a₃× a¹

(a1a₂a₃) = 2πac (0, 1, 0) a²c

√3 2

= 2π a

√3 2

(0, 1, 0) (2.4)

b₃ = 2π a₁× a² (a1a2a3) = 2π

a²

√3

2 (0, 0, 1) a²c

√3 2

= 2π

c (0, 0, 1) An arbitrary point in the reciprocal lattice has the position

G_n1n2n3 = n1b₁+ n2b₂+ n3b₃ The Bragg condition can be rewritten as

λ = 2d

ν sin Θ or ν2π

d = 22π

λ sin Θ or

¯¯G_n1n2n3¯

¯ = 2k sin Θ or (2.5)

Here we utilized that the length of the reciprocal lattice vector is an integer multiple of 2π

d , i.e. ν2π d . 2.5.2. Structure factor

There is one additional complication. Since the graphite lattice has a basis the reciprocal lattice has a structure factor S_n1n2n3. This structure factor accounts for the interference of the scattered wave by the basis atoms.

Sn1n2n3 =X

ν

exp£ i¡

hνG_n1n2n3¢¤

= exp£ i¡

h0G_n1n2n3¢¤

+ exp£ i¡

h1G_n1n2n3¢¤

+ exp£ i¡

h2G_n1n2n3¢¤

+ exp£ i¡

h3G_n1n2n3¢¤

= 1 + exp [ih1(n1b₁+ n2b₂+ n3b₃)]

+ exp [ih2(n1b₁+ n2b₂ + n3b₃)]

+ exp [ih3(n1b₁+ n2b₂ + n3b₃)]

With

h₁b₁ = µ a

√3, 0, 0

¶ 2π

a

√3 2

Ã√3 2 ,1

2, 0

!

= a

√3 2π a

√3 2

√3 2 = 2π

√3

h₁b₂ = 0 h₂b₁ = 2π

√3 h₂b₂ = 2π

√3

h3b1 = 4π

√3

h₃b₂ = 2π

√3

one obtains for the structure factor in the kx− k^y-plane (n3 = 0) Sn1n20 =

= 1 + exp [ih1(n1b₁+ n2b₂)]

+ exp [ih2(n1b1+ n2b2)]

+ exp [ih3(n1b1+ n2b2)]

= 1 + exp

· i2π

√3n1

¸ + exp

· i2π

√3(n1+ n2)

¸ + exp

· i2π

√3(2n1+ n2)

¸

This structure factor is always non-zero. The intensity of a wave scattered by a family of lattice planes is proportional to the absolute square of Sn1n2n3. Therefore one observes all the defraction pattern of the corresponding hexagonal Bravais lattice.

2.6. Face centered cubic structure of aluminum

The structure of aluminum can be described in different ways.

1. One uses the same hexagonal dense packed plane as in the hexagonal lattice.

This plane is often denoted as the A-plane. The next plane is shifted as in graphite by the vector a3 =¡₁

2a,¹₆√

3a,¹₃√ 3a¢

. This is the so-called B-plane.

The next plane is in graphite the same as the Þrst one, so that one has in graphite the sequence ABAB.. In Al, however, the third plane is shifted with respect to the second by the same vector a3. This is called the C-plane. The forth plane in Al is obtained by the same shift a3. However, applying the shift a3 three times yield in the x-y-plane a total shift of 3*¡₁

2a,¹₆√ 3a¢

¡₃ =

2a,¹₂√ 3a¢

. This is equal to 2a1+ a2. This means it is a shift by a lattice vector. Therefore the forth plane is directly above the Þrst plane and the Al has the sequence ABCABCA.. The main difference between a hexagonal lattice and the Al lattice is the sequence of the dense packed hexagonal plane.

In this description the Al lattice is a Bravois lattice with the primitive vectors a₁, a2 and a3. The disadvantage of this description is that the human mind gets easily confused in stapled hexagonal lattices. Therefore one generally prefers to described the Al as a face centered cubic lattice

2. The face centered cubic (fcc) lattice is built from a cubic lattice with the lattice paramater a. In the centers of each of the six surface squares of the cube another atom is located. Therefore the unit cell of the cube contains four atoms (each of the 8 corner atoms is to 1

8 in the cube and each of the 4 face centered atoms is to 1

2 in the unit cell).

In the following we discuss the second discription in more detail. The primitive vectors of the unit cell are

a1 = a (1, 0, 0) (2.6)

a₂ = a (0, 1, 0) a₃ = a (0, 0, 1) 1. The basis of the fcc lattice is

h₁ = a (0, 0, 0) h₂ = a

√2(1, 1, 0) h3 = a

√2(1, 0, 1) h₄ = a

√2(0, 1, 1)

The reciprocal latticed of the cubic lattice is itself a cubic lattice with the primitive vectors

b1 = b (1, 0, 0) (2.7)

b₂ = b (0, 1, 0) b₃ = b (0, 0, 1)

b = 2π a

Because the fcc lattice has a basis one has to calculate the structure factor of the reciprocal lattice. The calculation yields that the structure factor Sn1n2n3 only non zero when the n1, n2, n3 are either all odd or all even. (As a consequence the reciprocal lattice is a body centered lattice in k-space with the lattice parameter

π

a.) These points of the reciprocal lattice are

indices length |b| d=2π

|b|

(1, 1, 1) 2π a

√3 ¹₃a√ 3

(2, 0, 0) 2π a

√4 a

2 (2, 2, 0) 2π

a

√8 ¹₄a√ 2

(3, 1, 1) 2π a

√11 ₁₁¹ a√ 11 ect

This are the lattice distances which one should Þnd in a Debye-Scherrer experi- ment. They are characteristic for an ffc lattice.

2.7. Evaluation of experimental data 2.7.1. Laue diffraction pattern

For the Laue diffraction in our experiment the direction of the incident k vector is perpendicular to the hexagonal plane and therefore parallel to the z-direction. In k-space the Ewald sphere touches the (kx, ky)-plane at the origin. The direction of the scattered wave vector k⁰ is in forward direction. We restrict ourselves to small angle forward scattering. Then the vector k⁰ is almost parallel to k. That means the vector K = k − k0 which has to lie on the Ewald sphere has almost zero component in kz-direction. The only reciprocal lattice vectors which are very close to this part of the Ewald sphere lie in the (kx, ky)-plane with

K≈n¹b₁+ n2b₂

The diffraction pattern reproduce essentially the (kx, ky)-plane of the reciprocal lattice.

We describe the position of the Laue diffraction spots on the screen of the tube by its (x,y)-coordinates ρ_n1n2=(x,y)n1n2. Then the position of the spots is

given by

ρ_n1n2 = L

k (n1b₁ + n2b₂)

(Here we used that K = 2k sin Θ and tan (2Θ) = _L^ρ, where L is the distance between the target (the graphite foil) and the screen. For small Θ we use the replacement sin Θ ≈ Θ and tan (2Θ) ≈ 2Θ. This yields for the absolute values of K and ρ

K = 2kΘ = kρ L

ρ = L

kK

Returning to the vectors one obtains the above relation).

Note that in this experiment for small angle forward scattering one obtains diffraction spots even for a mono-chromatic electron beam.

1. For the smallest Θ the Bragg-relatio (2.1) has the form λ = 2d sin Θ since n = 1. From the DeBroghli relation (1.1) one obtains λ = h

√2meV. Plot sin Θ versus V^−1/2 and you check the Debroghli relation.

2. Draw the reciprocal kx-ky-plane and give the units of the drawing.

3. Calculate the atomic distances of the C-atoms.

2.7.2. Debye-Scherrer rings

1. For the smallest Θ plot sin Θ versus V^−1/2.

2. Calculate from the diffraction rings the distances of the Bragg planes.

3. Determine the lattice parameter from rings.

4. Verify the lattice structure by verifying the permitted indices.

5. Determine the atomic distances in the lattice.