Master Laboratory Report

Low Energy Electron Diffraction on Surfaces (LEED - Ma2)

Supervisor : Tauqir

Students : Dachi Meurmishvili, Muhammad Khurram Riaz

and Martin Borchert

Date: November 10th 2016

1

Contents

1 Introduction 31.1 Experimental Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Two dimensional lattices . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Application of LEED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Woods Notation and superstructure. . . . . . . . . . . . . . . . . . . . . . . 101.5 I(V) Curve and the kinematic approximation . . . . . . . . . . . . . . . . 10

2 Measurements and Data 12

3 Elaboration 133.1 Horizontal (in Cu(100) plane) lattice constant . . . . . . . . . . . . . . . . 143.2 I(E) curve and vertical lattice constant . . . . . . . . . . . . . . . . . . . . . 143.3 Copper - Oxygen superstructure . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Conclusion 18

5 Appendix 19

2

1 Introduction

In this experiment we want to study the crystalline surface structure of copper(Cu) byLow Energy Electron Diffraction (LEED). Copper forms a Face-Centered Cubic (fcc)lattice and we want to examine the Cu(100) surface. We will use electrons with energiesof E = 20eV to E = 500eV . The lattice constant is about d ≈ A. Here we treat electronsas waves. With the de-Broglie equation (1) we can calculate an according wavelength λ.

λ =h

p(1)

Here h is planck’s constant and p is the momentum of the particle. We start with arelativistic treatment of the problem[2]. The relativistic energy-momentum relation is

E2 =m2c4 + p2c2 (2)

From here relativistic momentum will be

p =

√E2

c2−m2c2. (3)

We denote m as the rest mass of the particle. Remember that the expression for therelativistic energy is given as

E =mc2√

1 − v2

c2

. (4)

Which if treated as low velocity limit

v

c→ 0. (5)

This can be expanded as

E =mc2 +Ek. (6)

3

Here we denoted Ek as a total kinetic energy of our particle. Which in our experimentcorresponds to the potential V , because all the potential energy of the electron at thestart of the flight transforms to kinetic energy at the end, when it reaches the sample.Now if we substitute the above energy expression in the momentum calculation we willget

p =

√(mc2 +Ek)2

c2−m2c2 (7)

=

√2mEkc2 +E2

k

c2(8)

=

√

2mEk +E2k

c2. (9)

Now we see that if the fraction E2k/c2 is small enough we can discard it and leave just

the classical energy momentum correspondence. The error of using the classical formulawill be of order E2

k/c2 which in our case is indeed small enough and can be discarded.

So we will use the classical formula for the electrons’ momentum and calculate theirwavelength.

λ =h

√2mEk

(10)

λ(500eV ) = 0.54874A (11)

1.1 Experimental Setup

We accelerate electrons with an electric field and focus the beam into an evacuatedchamber onto a probe of copper to study the Cu(100) plane. The electrons are scatteredand reflected back into the inside of the chamber with an opening angle of β. As we aredealing with low energy electrons we want to evacuate the chamber as far as possible toavoid the electrons interacting with the air particles and only with the probe. We use anumber of pumps in series to get the pressure down as low as possible.

• Rotary vane pump

The rotary vane pump consists of vanes mounted to a rotor that rotates insidea cavity. By using this pump we attain a minimum pressure of p = 10−6mbar.This pump has a medium vacuum level and gas transfer. For the nitrogen N2 andhydorgen H2 we must use turbomolecular pump after the rotary pump.

• Turbo molecular pump

The turbo molecular pump is a turbine with blades. The gas molecules changetheir motion due to momentum transfer of the molecules when hitting the rapidlyrotating (> 10000 rpm) pump rotor. Because of these high speeds it can only used

4

in a already evacuated atmosphere, as the friction would otherwise be too high.By using this pump we attain a minimum pressure of p = 10−10mbar. This pumphas an ultra high vacuum level and a low gas transfer.

• Ion getter pump

The ion getter pump removes the remaining gases (ions) by attracting them tothe walls of the chamber by electric fields. These then bind into the pump intothe solid compounds. The minimum pressure we obtain by using this pump isp = 10−10mbar.

• Titanium sublimation pump

The titanium sublimation pump has multiple thin film evaporations to keep thesurface active. This pump is only used for hydrogen, oxygen and nitrogen but hasno effect on inert gases and is only run once at the evening before the experimentfor a few minutes.

1.2 Scattering

We consider an electron matter wave that scatters of a regular crystalline structure,which can be considered to be a bravais lattice in real space. Here we only considerelastic scattering, which means that the wave number is conserved and that the incidentangle is equal to the scattered angle. We are also assuming, that electron wave diffractsonly once. This conditions are know as Kinematic Approximation and we will return tothem more deeply later.1.5 The scattering can be described in two different ways, firstin the real space by the bragg condition eq.(12) and secondly by the Laue conditioneq.(13). We will show, that these two interpretations are equivalent. First we state theBragg condition according to figure (1)

2d sin(θ) = nλ. (12)

5

d

Θ

Θ Θ

Θ

Figure 1: Parallel waves (orange) are scattered of the parallel lattice planes (grey)under an angle of θ.

The Bragg condition simply states that for two incoming waves we find constructiveinterference if the path difference (in this case we see that this is 2d sin θ) is equal to amultiple n of the wavelength λ. The Laue condition involves some more steps, but cantell us more about the reciprocal space. We now show, that the Laue condition can beused to derive the Bragg condition. We therefore look at one incoming wave with wavevector k and a (from the crystal surface) scattered wave with vector k′. ∆k = k − k′ isthe difference between the wave vectors. Again for the elastic scattering k = k′ as themomentum has to be conserved. The Laue condition now states that we find constructiveinterference if ∆k is equal to a reciprocal lattice vector g

∆k = g (13)

k − k′ = g (14)

k = g + k′ (15)

After squaring and putting (k′)2 = (k)2, we get

2k ⋅ g = g2 (16)

In figure (2) we can clearly see, that g is perpendicular to the plane hkl and thereforethe angle between k (or k′) and g is γ = 90○−θ. It can also be shown[1] that the distance

6

between two of these planes d is 2πg n and also knowing that k = 2π

λ . If we insert thesetwo facts into eq. (16) we get

2k ⋅ g = g2 (17)

2kg cos(γ) = g2 (18)

2k cos(90○ − θ) = g (19)

22π

λcos(90○ − θ) =

2π

dn (20)

21

λsin(θ) =

1

dn (21)

And this is the bragg condition.

2d sin(θ) = λn (22)

d

Θ Θ

k k'

k'

g

plane hkl

Figure 2: The Laue condition states that the difference between an incoming andoutgoing waves is equal to a reciprocal lattice vector g which is perpendicular

to the real lattice planes.

1.2.1 Two dimensional lattices

We now want to generalise the scattering process to two dimensions (figure 3). Theangle between the incoming wave k and the scattered wave k′ we called γ. Again thevon Laue condition holds and ∆k = g for all g in the reciprocal lattice, but now g can go

7

in two dimensions and the factor of n becomes (with pythagoras)√n2 +m2. With that

we can write for γ

sinγ =∣g∣

∣k′∣(23)

=

2πa

√n2 +m2

2πλ

(24)

=λ

a

√n2 +m2. (25)

γa1a2

b1b2

S00

S0

S10

S01

k

k' n²+m² g

γ

a1a2

Figure 3: Scattering at a 2D lattice. The left lattice is in the real space and the rightone in the reciprocal space. The beam in the center is marked in red for

better 3D visibility.

1.3 Application of LEED

We now consider an electron beam with energy E that diffracts on crystal with Cu-Cudistance of a = 2.55A and then hits the spherical detector. (Note that a is not the latticeconstant. As we are looking at the Cu(100) surface, the lattice constant l is l =

√2a)

We now combine equation (10) with the just derived equation (25) to get

8

sin(γ) =h

a√

2E ⋅m

√n2 +m2 (26)

which we can rearrange to find the minimum energy needed to find all maxima dependenton n and m

√E =

h

a sin(γ)√

2m

√n2 +m2 (27)

E =h2

a2 sin2(γ)2mn2 +m2. (28)

We are going to use the S0 beam which diffracts on the sphere and gets detected in theopening angle of β and calculate the minimum energy needed to detect the maximumlattice points for each order of diffraction. The sphere has a total opening angle of β =

52○. This means that between the incident beam and the reflected beam the maximumangle is γ = 0.5 ⋅ β = 26○. With

E(0,1) = E(1,0) =h2

a2 sin2(26○)2m= 120.4eV (29)

we can calculate the energy needed for each combination of maxima.

n \m 0 1 2 30 0,00 120,40 481,60 1083,601 120,40 240,80 602,00 1204,002 481,60 602,00 963,20 1565,203 1083,60 1204,00 1565,20 2167,20

Table 1: Energies of different values for n and m in eV.

The expected diffraction patterns look like in figure (4).

m = 0 (or 1)n = 1 (or 0)

m = 1n = 1

m = 2n = 2

Figure 4: Expected patterns for different values of m and n

9

1.4 Woods Notation and superstructure.

In condensed matter physics we sometimes find periodic structures superimposed ona base crystal lattice. This is called a superstructure. In our case we find additionalOxygen molecules superimposed on the copper surface. A model for that can be seen infigure (5). We achieve this by heating the copper and leaking oxygen into the chamber.To mathematically describe the superstructure, we use the woods notation. We use thelattice vectors a1 and a2 for the copper and the vectors b1 and b2 for the oxygen. Wenow compare the lengths between the an and bn vectors and and find a ratio of

√2 for

both pairs. Therefore we call this a√

2×√

2 superstructure. As the b vectors are rotatedby 45○ from the a vectors, sometimes a R45○ is appended.

a1

a2

bb 21

Figure 5: (√

2 ×√

2)R45○ Superstructure of Oxygen (shown in green) on Copper(shown in reddish brown)

1.5 I(V) Curve and the kinematic approximation

Until now we assumed that the electrons scatter only once with the crystal. This iscalled the kinematic approximation. But when the energies get higher the beam entersfarther into the crystal and interferes with more deeper layers. This penetration takesenergy U that is subtracted from the original energy with E = Ekin − U . We can nowrearrange the Bragg condition (equation 22) for λ and set it equal to the de-Brogliewavelength (10) to get

10

2d sin(θ)1

n!=

h√

2m(Ekin −U)(30)

Ekin =h2

8md2 sin(θ)n2 +U (31)

= αn2 +U (32)

α ≡h2

8m sin(θ)d2(33)

d =

√h2

8m sin(θ)α(34)

We now look at the S00 beam (θ = 90○) to make calculations and measurements easier.If we now insert different values for n we can calculate the energies needed to find theinterference with the deeper parallel layers of the copper crystal and compare this to theexperimental data.

2d

n= λ (35)

n =2d

λ(36)

with n = 1,2,3, .... Here we see that in order to have constructive interference fraction2dλ must be integer. Which means that we will have peaks (i.e. brightness of Intensity)at different energies - corresponding to those λ s. That’s why for some energies we havepeaks and for some not.

11

2 Measurements and Data

10:15 Briefing for the experiment10:20 Check that the LEED screen, the ion getter pump and the titanium sublimation

pump (T.S.P.) are off. The vacuum is p = −10−9 mbar.10:25 The position of the sample was already adjusted for use of the sputter gun. We

turned on the sputter gun slowly with the controller slowly increasing the emissionup to 25 mA at a beam voltage of 1.5 kV

10:30 We opened the leak valve for the argon gas, which increases the pressure in thetank. We stopped when the pressure reached 5.0 ⋅ 10−5 mbar. After that we wait 35minutes for the sample to be mostly free of contamination.

11:05 We closed the argon leak valve and decrease the current to 0 mA. Very soon thepressure after turning off the argon beam becomes 2.2 ⋅ 10−7 mbar.

11:15 After some time the pressure is back to 3 ⋅10−9mbar and we start with the annealing.11:17 We turn on the filament (wires on back surface of the copper sample) voltage this

will heat the sample to around 700-800K.11:53 The maximum we could reach is T=728K and we kept this for 17 minutes.12:10 We stopped the heating. And had a break until the probe is cooled down again.13:55 We adjusted the sample position by rotating the probe so it faces the LEED appa-

ratus. We moved the LEED screen from 0th to 6th position (closer to sample).13:57 We turned on the LEED controller, slowly increasing the current of electrons in the

electric potential of up to 7kV. This results in a current of around 25mA.The pressure is around 10−9 mbar. And we start the imaging software. We increasethe energy until the different higher order maxima can be seen on the screen fordifferent values of n and m.The energies we measured can be seen in table (3). The opening angle was foundto be β = 82○.

15:10 We then looked at the S00 beam which we moved a bit off center on the screen byrotating the probe so the beam is not hidden behind the LEED gun.

15:15 On the computer program we selected the spot and drew two boxes around it: asmall one measures the peak intensity of the beam and bigger one that measuresthe total intensity.

15:25 With another box we measure the background intensity to subtract it from the datafrom the smallest box.We closed the aperture from the camera far enough so the CCD sensor will notsaturate. And started the measurements.The computer automatically increases the energy of the LEED beam between 0eVand 600eV and records the intensity inside the three boxes.

15:40 Every 0.5 seconds the energy increases by 1eV . The results can be seen in 6.16:35 The measurements are finished and we prepared the oxidation of the copper surface.17:30 We can not seem to get the apparatus working again. And will use the measurements

from the previous group.

12

Table 2: Measurements and Data

0 100 200 300 400 500 600Energy of the electrons in eV

0

10

20

30

40

50

60

70

80

Inte

nsi

ty in a

rbit

rary

unit

s

Intensity for the S00 beam

Peak spot intensity

Total spot intensity

Background ( noise intensity) intensity

Figure 6: Intensity for the S00 beam depending on the energy of the LEED electrons.

Energy in eV n m70.0 0 193.2105.9 1 1111.9112.8129.6169.0193.1 2 0270.0 2 1287.3301.0 2 2

Table 3: Energies for visible higher orders for n and m

3 Elaboration

From this data we can now determine the lattice constant a of the surface of the Cu(100)plane and the distance between two of these parallel planes. Furthermore with the resultsfrom the previous group we try to characterise the superstructure.

13

3.1 Horizontal (in Cu(100) plane) lattice constant

As mentioned in the measurement protocol, we determined the opening angle of thesphere to be β = 82○ = 2 ⋅ γ by rotating the probe until the S00 beam hits the left sideof the detector and then the right side. From the difference we get β. Solving formula(28) for a we get

a2 =h2

E sin2(γ)2m(n2 +m2) (37)

a =

√h2(n2 +m2)

2m´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

c

1

E sin2(γ)(38)

∆a = c

¿ÁÁÁÀ⎛

⎝−

∆E

2√E

3sin(γ)

⎞

⎠

2

+ (−cos(γ)∆θ

sin2(γ)√E)

2

(39)

= c

¿ÁÁÀ (∆E)2

2E3 sin2(γ)+

cos2(γ)(∆γ)2

sin4(γ)E. (40)

We assume uncertainties1 of ∆E = 10eV and ∆γ = 10○. We get the copper-copper-distance of a = (2.6 ± 0.3)A and with the above mentioned factor between the latticeconstant l and the copper-copper-distance a, we get l =

√2 ⋅ a = 3.7 ± 0.4. This perfectly

matches the literature value of l = 3.61A. The individual values can be seen in table (4).

n m E in eV a in A0 1 70.00 2.23 ± 0.281 1 105.90 2.57 ± 0.282 0 193.10 2.69 ± 0.282 1 270.00 2.54 ± 0.262 2 301.00 3.05 ± 0.31

Table 4: Lattice constants with errors

3.2 I(E) curve and vertical lattice constant

We rotated the sample so that the S00 beam is not in the centre of the detector, but a bitto the side. With the given CCD camera and the computer program we measured theintensity of the beam while varying the energy E of the beam between 0ev < E < 600eV .We also measured the intensity of the background to subtract them from each other.The results can be seen in figure(6). We used python’s scipy.signal.find peaks cwtfunction to find the rough position of the peaks and took these as initial values for the

1∆γ needs to be converted to radiants, of course.

14

following fit. We fitted the sum of n non-normalised Gaussian curves, where n is thenumber of peaks found by find peaks. We used python’s curve fit function with thetrust region reflective (trf) method for that. This will vary all 3 ⋅n parameters (ai, σi, µi)from equation (41).

Gi = aie−(x−µi)

2

2⋅σ2i (41)

The result is shown in figure (7).

0 100 200 300 400 500 600Energy in eV

10

20

30

40

50

60

70

Inte

nsi

ty

Fit as a sum of N gaussian curves for the N peaks.

94.1

125.

4

157.

9

184.

1

224.

1

278.

3

338.

5

408.

3

472.

3

509.

7

556.

7

Figure 7: Fit for the intensities with Gaussian distribution

The values of the peaks with their error (σ) are shown in 5. A big problem we noticedduring the experiment is that even though the angle of S00 beam should not change withincreasing energy, it actually did change. And it did move far enough to get partiallyblocked by the electron gun. This does not effect the peak positions in figure (7), butit decreases the intensity with increasing E. We did not repeat this experiment as wethought that all peaks would correspond to consecutive values of n, but they actuallydo not.

15

Energies with error of the peak in eV94.1 ± 3.3125.4 ± 6.4157.9 ± 3.6184.1 ± 6.5224.1 ± 7.0278.3 ± 8.0338.5 ± 10.1408.3 ± 9.5

472.3 ± 469.4509.7 ± 7.4556.7 ± 7.5

Table 5: Peak positions with error as standard deviation of the Gaussian curves.

Unfortunately for a reason that we can not explain, we do not see any information below90eV . Therefore we do not know what n0 the first peak corresponds to. There is reallynot much else to do except for trial and error and see which n fits best. We thereforetook the peaks at 94eV.1, 184.1eV , 278.3eV , 408.3eV and at 556.7eV (so basically allthe high peaks except for the peak at 125eV and tried to fit them according to (32)with an non-linear (quadratic) fit. We found n0 = 3 to fit the best α = (11.56 ± 0.19)eVand U = (−9 ± 6)eV because this results in inter-planar lattice constant of d = l/2 =

(1.803 ± 10−11)A. This is around half the lattice constant of l = 3.61A.

For n0 = 2 we get l = (3.96± 10−11)A and for n0 = 4 we get l = (3.21± 10−11)A. Thereforewe are quite certain to have found the correct match n0 = 3.

0 10 20 30 40 50n2

100

0

100

200

300

400

500

600

eV

Fit with α = 11.56 +/- 0.19 and U = -8.53 +/- 5.57

Fit

Data with errorbars from gaussian curves

Extrapolation of peaks 1 and 2

Figure 8: Fit for the Energy of the even numbered peaks of figure 7 plotted withrespect to n2.

16

3.3 Copper - Oxygen superstructure

The filament from the LEED gun broke during the experiment so we got the imagesfrom the previous week (= previous group). These can be seen in 9 and 10. Here wecan see the

√2 ×√

2R± R45○ superstructure of Cu2O.

Figure 9: Fit for the intensities with Gaussian distribution

17

Figure 10: Fit for the intensities with Gaussian distribution

4 Conclusion

Given the energy of the monochromatic electron beam we estimated the relativisticcorrection and deduced that the classical treatment of the problem is completely ac-ceptable. We used Bragg’s condition to show how constructive interferences are formedand then we used the Laue condition to derive Bragg’s condition in one and two dimen-sions. The main goal of this experiment was to calculate the planar lattice constanta which we found to be l = 3.7 ± 0.4. This perfectly matches the literature value ofl = 3.61A. We also determined the vertical lattice constant with the the I(V ) curveof S00 beam using the kinematic approximation. From the peaks of the I(V ) curve weget the distance between to consecutive planes of the crystal. We found this distanceto be d = lvert/2 = (1.803 ± 10−11)A. If we double the number we get the vertical latticeconstant of lvert = (3.61± 10−11)A. This also matches the same literature value of 3.61A.We stumbled on some difficulties on our first measurement, namely we chose a spot tooclose to the camera, so when increasing energy, the angle of the S00 beam was shiftedand was partially blocked by the camera. It caused the intensities to decline in powerand thus we have lower consecutive peaks. We then used trial and error approach tofind the right number corresponding for the first peak and matched it to the orthogonal

18

lattice constant. Unfortunately our experiment suffered from technical difficulties whenwe were trying to measure the superstructure and we were forced to end without takingthose pictures. That is why we incorporated the results of the previous group, who didthe experiment before us and used those results to show how the superstructure formmanifested itself corresponding to woods notations. We can clearly see the

√2 ×√

2R±R45○ superstructure of Cu2O.

All in all this experiment can be counted as a success even though the apparatus wasbroken after we were finished.

5 Appendix

References

[1] Uni Augsburg. Solid State Physics, WS 04/05 Reciprocal space. Nov. 2016. url:http://www.physik.uni-augsburg.de/~ferdi/fkp_04/reziprokes_gitter.

pdf.

[2] L.D.Landau and E.M.Lifshitz. The Classical Theory of Field, Volume 2, FourthRevised English Edition (pages 25-27). 1996.

19