Solid state physics for Nano

Lecture 2: X-ray diffraction

Prof. Dr. U. Pietsch

The mean aim of Max von Laue (1912)X-rays are electromagnetic wave with wave length much smaller

than wave length of visible light. X-rays are diffracted a crystal lattice

1914

Nobelpreis fürPhysik

Original experiment von Laue, Friedrich und Knipping

X-ray tube

Photographic film

CollimatorCrystal

Ausgestellt im Deutschen Museum in München

20.April 1913

Photographic film

Erster Kristall

Cu2SO4 ⋅ 5 H2O

First Laue Experiment

1912: Begin of modern Crystallography

X-rays are electromagnetic waves of very short wavelength ( ~ 1 Å = 10-10 m).

Crystals are periodic structures in 3D : interatomic distances are of similar order of magnitude as

x-ray wave length

X-ray diffraction is a method to determine the geometric structure of solids !

Erklärung durch Interferenz am 3D Gitter

Explanation of Laue pattern

recip. lattice vector

𝑎 = 𝑏 = 𝑐

sin2

sin²²²

2

sin4²²²

2

d

lkh

a

a

lkh

=

++=

=++

Alternative description of Laue-pattern by W.H.Bragg und W.L.Bragg

Interference at dense backed

„lattice planes“

N=2dsinQ

Bragg equation

Reciprocal space, Ewald construction

,....0*,0*

2*

2*

2*

==

=

=

=

cbba

cc

bb

aa

)(*

)(*

)(*

cba

bac

cba

acb

cba

cba

=

=

=

*)**(2 lckbhaGK ++==

)(

2

2

hkldG

kk fi

=

==

Setup of modern diffraction experiment

Qi

Slits

Detector

AbsorberboxSample

Analysator

pre-mirror

Monochro-

mator

Slits

Qf

Storage

ring

X-ray tube and tube spectrum

E = h v = e U

min= h c / e Uamin = 12.4/Ua Brems-

strahlung

Characteristicradiation

ESRF

Generation of Synchrotron radiation

Electron circulates with relatvististic energy in orbit of storage ring with orbitfrequency w= 106 Hz → Ee = 5 GeV

²1

²

²

²1

² 00

ß

cm

c

v

cmEe

−=

−

=

Ee is much larger compared to rest mass energy mc², g = 5GeV/ 0.511 GeV≈ 104

²1

1

² ßcm

E

o

e

−==g

value 1/g is the vertical open angle Using g one can estimate the electron velocity

92/1 10*61²2

11]

²

11[ −−=−−=

ggß cv 910*61 −−

Generation of Synchrotron radiationSupposing the circulating electron emits a photon at point A of orbit. M Elektron proceeds to point C via B on orbit. At point C it emits a secondphoton. For the electron the length

AC (electron) = v* dt .

For the photon emitted in A reaches this distance is

AC(photon = c *dt

The spatial distance between both waves : (c-v) dt

An observer at point C the light pulse has a length

´²)1(´)(

dtc

dtvct −=

−=

Generation of Synchrotron radiation

Considering the opening angle a = 1/g due to electron orbit

´²2

)²(1´)]

2

²1)(

²2

11(1[´)cos1( dtdtdtt

g

aga

ga

+−−−=−=

For a =0 ´10´

²2

1 8dtdtt −=g

Dilatation of time a consequences for spectral distribution of photon emission

Orbit time of electron along AC differs from time of photon

sTphotont

sTelectront

18

10

10³2

11

²2

1)(

101

2

1)´(

−

−

==

==

wggwg

gwg

Generation of Synchrotron radiation

Inverse pulse length determines the spectral with of light emission

The emission spectrum has a critical wave length given by

This is x-ray range. In energy :

Hzc1810³

2

3= wgw

mc

c

10

18

8

1010

10 −===

)()(*655.0³2

3 2 TBGeVEec = wgw

Generation of Synchrotron radiation

Bending radius R of electron orbit is

Emission power is

For ESRF, E=6GeV, B=0.8T → R= 24.8m

Bmv Re=g)(

)(*3.3

² TB

GeVE

eB

mc

mc

E

eB

mcR ee ===

g

)()()²()(*266.1)( 2 mAImRTBGeVEkWP e a=

WmAmradmTGeVkWP e 3.7100*05.08.24)²(8.0)(6*266.1)(2 =+=

wc

Bend magnet spectrum

ESRF

Well predictable

Kinematic X-ray scattering

Dipole scattering of x-ray wave by an electron: Electron excites dipole radiation

+==

2

²cos1

²

1

²

²² 2

Rmc

eEEI o

Superposition of dipole radiation generated by two electrons at positions X1 and X2

)])(2

[exp(1

²

²210 XXti

Xmc

eEE −−=

w

For n electrons

])[2

exp(][2exp(1

²

²00 SS

Rti

Rmc

eEE

n

−−=

Stotal scattering amplitude of one atom : → atomic form factor

)(2

0SSkk if −=−

Kinematic X-ray scattering

Form factor f[(s-s0)/] result fromquantun mechanical calculationsf(0) = Z ; f[(s-s0)/] decays continuoslyas function of (s-s0)/

])[2

exp(][2exp(1

²

²00 SS

Rti

Rmc

eEE

n

−−=

Scattering by all electrons of one atom : → atomic formfactor

−

=−n

dVrss

iSSrie

][2exp()])(

2[exp(* 00

Kinematic X-ray scattering

Scattering by two atoms :

Debye´sche Streugleichung Debye equation

)])([2exp(]([2exp(]))(

[2exp(][2exp( 02

21

20

21 SSR

iffR

tiRSSR

tifR

tifE

−++=−−

−+−

Scattering by many disordered atoms :

)](2

cos[2* 0212

2

2

1

* SSffffEEI

−++=

nm

nm

mn

nmkr

krffI

)sin(

,

=

2)sin(]][

2exp 0 =→→− k

kr

krrSS

i

nm

nm

nm

Kinematic X-ray scattering

Scattering by „small“ crystal: rnm= Rnm+rn distances between unit cells → shapefactor, distances betwen atoms within one unit cell→ structure factor

Summation over all unit cells→ N1a1 , N2a2 , N3a3

332211 amamamRnm ++=

])[(2

exp()2exp()exp( 3322110 amamamSSiR

itifE n ++−−

w

2|| =K

²²*

)sin(

)sin(

)sin(

)sin(

)sin(

)sin(²

321

3321

221

2221

121

1121

SFEEI

Ka

KaN

Ka

KaN

Ka

KaNFE

=

if kkK −=

FWHM

Kinematic X-ray scattering

I 0 only in near vicinity of the peak maxima

lKa

kKa

hKa

2

2

2

3

2

1

=

=

=h,k,l integer

FWHM of a Bragg peak can be approximated

Q=

Q=

Q=Q

cos

94.0

cos

94.0

cos

/)(ln2)2(

LNaNa

sB

FWHM B(2Q) is inversely proportional to crystal size L

kax =2

1)/)²(exp(²

)²(sin

)²(sin)( xNN

x

xNxI −

Q= cos2

1Bk

2

1)²cos)²

2(

²

²²(exp(² =Q−

BaNN

Structure factor F

))(exp(1

i

n

i

i krifF −= =

czbyaxr iiii ++=

))*)(**(2exp(1

czbyaxlckbhaifF iii

n

i

i ++++−= =

))(2exp(1

iii

n

i

i lzkyhxifF ++−= =

*)**(2 lckbhak ++=

Examples→

bcc- structure: xyz = 000, ½ ½ ½

))(exp(1( lkhifF ++−+=

F= 2f if h+k+l = even;F= 0 if h+k+l odd

fcc- structure: xyz = 000, ½ ½ 0, ½ 0 ½ , 0 ½ ½

))(exp()(exp()(exp(1( lkilhikhifF +−++−++−+=

F= 4f if h,k,l all are even or odd;F= 0 if h,k,l are mixed

hcp- structure: xyz = 000, 1/3 2/3 ½

))2/3/)2((2exp(1( lkhifF ++−+=

F= 2f if h+2k =3n, l even;F= √3f if h+2k= 3n+1, l oddF= 0 if h+2k= 3n+1 l even

Extinktion rules

Not all combinations of h,k,l are „allowed“ , Bragg peak intensity can bevanishing due to crystal (point group ) symmetry

In fcc system, only reflections appear with h²+k²+l² = 2,4,8,11,12….

= 111, 200, 220, 311, 222,…..

In NaCl 111, 311….. Are weak reflectons220, 400….. Are strong reflections

In ZnSe 220, 400…….strong111, 311…….middle200, 222…….weka

Structure (point group) identification considering extinktionrules and Bragg peak intensities

Powder diffraction pattern from CdSe Nanowires

Debye-Waller Factor

)exp()exp())(exp( rikikrrrik =+

)²2

1exp(...²

2

11)exp( xxxiix −=+−+=

)2exp()²²2

1exp()exp( Mrkrik −=−

²

²sin²²162

uM =

)2exp()( MFTF −=

Due to thermal atom oscillations

X-ray intensity becomes damped, Bragg peaks exit

Data evaluation

Electron charge density distribution fromx-ray diffraction data

Electronic charge densityof silicon

Valence charge density difference charge density(bonding charges)

U.Pietschphys.stat.sol.(b) 137, 441,(1986)

−−= dkikrkFr )exp()()(

− − −

−=h

h

k

k

l

l

ii HrHFV

r );2exp()(2

)( )(1

hkla

H =

Electron density distribution of GaAs fromx-ray diffraction data

Valence charge densitydifference charge density(bonding charges)

Ga As

Ga

As