'

)0(~

)(~

'41'*'*

)0(~

')(

~

''2

121

2

2

22

'

'

))1((1

2

1),(

)0()()()(1

2

1),(

),('

),()ˆˆ('

'

jjT

itijjjj

jN

jN

jjtinuc

Ti

jti

jjj

jjti

mag

nucmag

jj

jj

eeIIbbbbN

dteS

eJetJgFgFN

dteS

Sk

kNS

mc

e

k

kN

dEd

d

RκRκ

RκRκ

κ

κ

κκκκ

Unpolarisierte Neutronen- Van Hove Streufunktion S(κ,ω)

'

'41'*'*

'' ))1((1

)(jj

iijjjj

jN

jN

WWjjelnuc

jjjj eIIbbebbN

S RκRκ

)()(~

ttjjj

uRR

''

''21

'21 )()(

1)( jjjj WWii

TjjTjjj

j

elmag eeJgFJgF

NS RκRκ

Aufteilung von S: elastische – inelastische Streufunktion

inelmag

elmagmag

inelnuc

elnucnuc

SSS

SSS

)(2

.../2

)()(

1)'(

')'(1

)(

...

')'(......)(

0

0

/2)'(

2/

2/

/'2

0

/2

2/

2/

/'2

0

/2

qa

eLxqa

c

xcx

eL

xx

dxexfeL

xf

dxexffwithefxf

n

iqna

n

Lxxin

L

L

Lnxi

n

Linx

L

L

Lnxin

n

Linxn

A shortExcursionto FourierAnd DeltaFunctions ....

it follows by extending the range of x to more than –L/2 ...L/2 andgoing to 3 dimensions (v0 the unit cell volume)

τ

GκGκ τκ

..0

3

'

)()2(

'

lattrezG

kk

ii

vNe kk

ddd

dN

B

d

Wdd

B

dd

WWidd

Blattrez

elnuc

IIbN

ebbN

eebbNv

S

d

dddd

)1(1

)(

1)(

1)(

)2()(

41

2

222

'

)('

*

..0

3''

BBκ

τ

τκ

Neutronen – Diffraktion

'412'*

)1(||11

)( ''

jj jjj

jN

WWiijjelnuc IIb

Neebb

NS jjjj RκRκ

Gitter G mit Basis B: dkjkdj BGR

)........(

„Isotopen-inkoherente-Streuung“

„Spin-inkoherente-Streuung“

unabhängig von κ:

Gitterfaktor Strukturfaktor

ein Element(NB=1): )1()(44 41222

dd

dN

incel

nucinc

elnuc IIbbbN

d

d

i 2||4 bc

ddd

dN

B

W

ddd

B

dd

WWidd

Blattrez

elnuc

IIbN

ebbN

eebbNv

S

d

dddd

)1(1

)(

1)(

1)(

)2()(

41

2

222

'

)('

*

..0

3''

BBκ

τ

τκ

Gitterfaktor Strukturfaktor

The Nobel Prize in Physics 1994

"In simple terms, Clifford G. Shull has helped answer the question of where atoms are, ...“, (Nobel citation)

Inco

min

g

Neutr

on

τ

Oa*

c*

Bragg’s Law in Reciprocal Space (Ewald Sphere)

Scatte

red

Neutr

on

Einkristall-diffraktometrie

E2 – HMI, Berlin

τ

O

k

Bragg’s Law in Reciprocal Space (Ewald Sphere)

Neu

tron

C

τ

O

Reflecting Plane

sin OB τ= sin = O τ /OB = O τ /(2/

sin = (O τ/2)But since τ is a reciprocal lattice point, the length O τ is by definition equal to 2/dhkl

sin = (1/2dhkl)

2dhkl sin =

sin = (1/2dhkl) = (1/2)(1/dhkl) = (1/2)hkl

Pulver- diffraktometrie

D1B

guide hall n°1, thermal guide H22

monochromator

take-off angle 44.22°

crystal pyrolytic graphite (002)

. wavelength 2.52 Å

. flux at sample/n cm-2s-1 6.5 x 106

crystal Germanium (311)

. wavelength 1.28 Å

. flux at sample/n cm-2s-1 0.4 x 106

max beam size 5 x 2 cm2

angular range 2     -20° ... 144°

detector

3He multidetector containing 400 cells

angular range 2     80°

radius of curvature 1.525 m

detector efficiency 60 % at   = 2.52 Å max diameter / mm available around the sample

600

sample environment

cryostat 1.7 ... 300 K

furnace < 800 °C

furnace < 2500 °C by special arr

electromagnet 1 T; 22 mm vertical or horizontal gap

GdCu2In

|κ|[Å-1] 0 1 2 3 4 5 6

I(κ)

[co

unts

]

2||)( 21

SFTLI SF.. StrukturfaktorL ... Lorentzfaktor (betont kleine Winkel) Einkristall: 1/sin2θ Pulver Zyl.:1/(sin2θ.sinθ)

T Transmissionskoeffizientγ Korrektur für Extinktion

Pulver-diffraktometrie

)sin(4

2θ.... Streuwinkel Detektor

dIdI )()(

7C2, LLB Saclay

GdCu2In#lambda= 0.58 A#thetamax=18#nat=4 nonmagnetic atoms in primitive crystallographic unit cell:#[atom number] x[a] y[b] z[c] dr1[r1] dr2[r2] dr3[r3] [Gd] 0 0 0 0 0 0[Cu] 0.25 0.25 0.25 0.25 0.25 0.25[Cu] 0.25 0.25 0.75 0.75 0.75 -0.25[In] 0.5 0.5 0.5 0.5 0.5 0.5# a=6.62 b=6.62 c=6.62 alpha= 90 beta= 90 gamma= 90# r1x= 0 r2x= 0.5 r3x= 0.5# r1y= 0.5 r2y= 0 r3y= 0.5 primitive lattice vectors [a][b][c]# r1z= 0.5 r2z= 0.5 r3z= 0# nofatoms=1 number of atoms in primitive unit cell

)(1

)()2(

'

)('

*

..0

3''

dd

WWidd

Blattrez

cohel

nucdddd eebb

NvS BBκ

τ

τκ

Gitterfaktor Strukturfaktor

  h k l d[A] |kappa|[A^-1]2theta Ikern imag itot |sf| lpg } 1.000 1.000 1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 -1.000 -1.000 -1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 -1.000 1.000 1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 1.000 -1.000 -1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 1.000 -1.000 1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 -1.000 1.000 -1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 1.000 1.000 -1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 -1.000 -1.000 1.000 3.82206 1.64389 8.703 11.661 0.000 11.661 24.966 87.101 0.000 2.000 0.000 3.31000 1.89820 10.053 0.025 0.000 0.025 1.336 65.389 0.000 -2.000 0.000 3.31000 1.89820 10.053 0.025 0.000 0.025 1.336 65.389 0.000 0.000 2.000 3.31000 1.89820 10.053 0.025 0.000 0.025 1.336 65.389 0.000 0.000 -2.000 3.31000 1.89820 10.053 0.025 0.000 0.025 1.336 65.389 2.000 0.000 0.000 3.31000 1.89820 10.053 0.025 0.000 0.025 1.336 65.389 -2.000 0.000 0.000 3.31000 1.89820 10.053 0.025 0.000 0.025 1.336 65.389 0.000 2.000 2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 0.000 -2.000 -2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 2.000 2.000 0.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 -2.000 -2.000 0.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 -2.000 0.000 2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 2.000 0.000 -2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 2.000 0.000 2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 -2.000 0.000 -2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 -2.000 2.000 0.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 2.000 -2.000 0.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 0.000 -2.000 2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822 0.000 2.000 -2.000 2.34052 2.68446 14.235 83.558 0.000 83.558 109.804 32.822

Beispiel 2

• In einem elastischen Streuexperiment beträgt die Einfallsenergie 63 meV. Die Gitterkonstante der kubischen Probe beträgt 0.314 nm. Kann der (430) Reflex in diesem Streuexperiment vermessen werden ?

Neutron – PhotonStreuquerschnitte

Vorteile von Neutronen:1. Kontrast bei benachbarten Elementen – man sieht z.B. Überstrukturen2. Leichte Elemente besser nachweisbar3. Isotope können unterschieden werden

Laue Methode• Einkristalle •„weißer“ Strahl• Film oder Flächendetektor hinter der Probeschnelles Erkennen der Symmetrie -wird zum Orientieren von Einkristallen benutzt

VIVALDI

very-intense vertical-axis Laue diffractometer

4-Kreismethode•Einkristalle •monochromatischer Strahl• ein Detektor• EK in beliebige Richtungen orientierbar (Eulerwiege)

D10ILL

φχ

ω

Flugzeitmethode

Spallationsquelle (gepulst)

Probe

Det

2θ

•Streuwinkel fest (Vorteil z.B. bei Druckzellen)• |k| wird variiert (kein Monochromator) über die Zeit (zuerst kommen die raschen, dann die langsameren Neutronen) • bessere Nutzung der Quelle (keine Monochromator-verluste)•Auflösung umso besser, je größer Abstand zur Quelle (HRPD: 90m)

Time-of-flight

Bragg equation - 2dhklsin =

Two basic equations:

mv

h

t

Lv

where m,v = mass, velocity of neutron

L = length of flight path t = time of flight of neutron

Time-of-flight equation

Combine: sind2

mL

ht

sindh

mL2t

L is a constant for the detector, h, m are constants so:

t d

d-spacings are discriminated by the time of arrival of the neutrons at the detector

The biggest error in the experiment is where the neutrons originate

This gives an error in the flight path, L

typical value ~5cm

d

d

t

t

L

L

Hence as L increases, error in d is reduced - resolution of the instrument is improved

e.g. instrument at 10m compared to instrument at 100m

100m = HRPD, currently highest resolution in the world

HRPD, GEM

Sample area collimators and detectors on HRPD.  Neutrons enter via the yellow flight tube on the left.

GEMGeneralpurposeMaterialsDiffraktometer

p-dichlorobenzene (DCB)

refined structure

.

a/h

c/l

b/k

A

C

dhkl

C = ah

bk

-

h ka b -C . hkl= ( ( . (h a * + k b * + l c *)=

bh ka -

. (h a * + k b * + l c *) (h a * + k b * + l c *) =.

hh

+ 0 + 0 – (0 + kk

+ 0) = 1 – 1 = 0

Therefore hkl is perpendicular to C . In the same way one can showthat it is perpendicular to A, therefore perpendicular to the plane

hkl = h a * + k b * + l c *

|hkl| n = ha* + kb* + lc* ha* + kb* + lc*

|hkl|n =

ahdhkl = cos = a

h. n =

ah

ha* + kb* + lc*

|hkl|. =

|hkl|