Aperiodic crystal workshop 2013: TEM

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Московский Государственный Университетим. М.В. ЛомоносоваФакультет Наук о Материалах

Исаева А.А.

Научные руководителид.х.н., проф. Б.А. Поповкинк.х.н., асс. ФНМ А.И. Баранов

КАФЕДРА НЕОРГАНИЧЕСКОЙ ХИМИИЛаборатория Направленного Неорганического Синтеза

Ë Î Ì Î Í Î ÑÎ Â -2 0 0 3

Diffraction by incommensurately modulated phases

Joke Hadermann

At the end of this lecture you should be able to:

UNDERSTAND and INDEX

the reciprocal lattice of an

incommensurately modulated material.

One atom type A

a b

Basic cell, one plane

One atom type A

010

100

a b

[001]

Basic cell EDP

One atom type A

010

100

a b

[001]

Basic cell EDP

a=3 Å b=5 Å

One atom type A

010

100

a b

[001]

Basic cell EDP

1/3Å 1/5Å

a=3 Å b=5 Å

Alternation A and B atoms

a b

double cell model


a b

010

100

[001]

double cell, EDP=


a b

010

100

*bm

Gg2

Reflections at

[001]

double cell, g vectors=


a b

010

100

*bm

Gg2

Reflections at

[001]

double cell, g vectors=

q

*2

1bq

qmclbkahg ***

Gq0Gg

qmclbkahg ***

*** cbaq

Basic structure

reflections

All reflections

hkl0

hklm

qmGg

010

100

*2

1bq

0001

0100

1000

1001

[001]

q

double indexed with q double indexed with q

*bm

Gg2

010

100

q

*458.0. bmGg

*458.0 bq

0.458: q indicated

010

100

q

0001

0101 -

0100

1000

*458.0. bmGg

*458.0 bq

0.458 indexed with q

Projections from 3+d reciprocal space &

“simple” supercell in 3+d space

(Example in 1+1

reciprocal space)

a1*

a2*

q

Projections from 3+d reciprocal space &

“simple” supercell in 3+d space

(Example in 1+1

reciprocal space)

a1*

a2*

q

e2

a2*=e2+q

Basis vectors

*a*a1

*b*a2

*c*a3

qe*a 44

Basis vectors of the reciprocal lattice

*c*b*aq

0100

1000

0100

1000

0100

1000

0100

1000

all four with q

q=0.5b* q=0.458b*

q=0.25b* q=0.33b*

010

100

]0,3

1,

3

1[mGg

[001]

*0*3

1*

3

1cbaq

0001

0100

1000

0002

q

110, with q

010

100

q

0001

0101 -

0100

1000

0.458 indexed with q

hk0m: no conditions

{R|v} is an element of the space group of the basic structure

is a phase shift and is ±1

Space group of the

basic structure

components of q symmetry-operators

for the phase

Superspace groups: position and phase

(r,t) ( Rr + v, t + )

Example

Pnma(01/2)s00

Superspace groups

Incommensurately modulated materials

Reële ruimte Reciproke ruimte R*

a

200

000

a*

b*

a


3a

a* q=

1/3a*

a 3a

a*

q=1/3a* (3+Δ)a

q=(1/3-δ)a*


Exercise

Index the following experimental

diffraction patterns and derive the

possible superspace group(s) from the

reflection conditions.

The example is incommensurately

modulated LaSrCuO4-x.

First, index the commensurate parent

structure LaSrCuO4, this will help you with

indexing the next, incommensurate one.

1/6.50 Å 1/1.85 Å

1/6.50 Å

1/2.61 Å

1/2.61 Å

(Simulated ED patterns)

Given data:

cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.

1/1.85 Å

1/1.85 Å 1/2.61 Å

1/2.61 Å

1/1.85 Å

1/6.50 Å 1/1.85 Å

1/6.50 Å

1/2.61 Å

1/2.61 Å

Given data:


1/1.85 Å

001

002

200

The reflection in the red box is:

1/6.50 Å

2.61 Å

2.61 Å

Given data:


1.85 Å

001

002

200

The reflection in the red box is:

1/1.85 Å 1/2.61 Å

1/2.61 Å

1/1.85 Å

1/6.50 Å

002 1/1.85 Å

002

1/2.61 Å

1/2.61 Å

Given data:


1/1.85 Å

010

020

110

002 020

002

2.61 Å

2.61 Å

Given data:


1.85 Å

010

020

110

002 020

002

2.61 Å

2.61 Å

Given data:


1.85 Å

002 020

002

2.61 Å

2.61 Å

Given data:


1/1.85 Å

020

1/1.85 Å

002 020

002

2.61 Å

2.61 Å

Given data:


1/1.85 Å

100

200

020

020

002 020

002

2.61 Å

2.61 Å

Given data:


1.85 Å

100

200

020

020

002 020

002

2.61 Å

2.61 Å

Given data:


200

110

220

½ ½ 0

020

002 200

002

110 200

110

020

Determine the reflection condition for h0l.

002 200

002

110 200

110

h0l: h+l=2n

h = 2n

l = 2n

002 200

002

110 200

110

h0l: h+l=2n

Determine the reflection condition for h0l.

002 200

002

110 200 110

hhl: h+l=2n

h = 2n

l = 2n

h0l: h+l=2n

Determine the reflection condition for hhl.

002 200

002

110 200

110

hhl:

l = 2n

h0l: h+l=2n

Determine the reflection condition for hhl.

002 200

002

110 200

110

hhl: l = 2n h0l: h+l=2n hk0: h+k=2n

h = 2n

k = 2n

Determine the reflection condition for hk0.

002 200

002

110 200

110

hhl: l = 2n h0l: h+l=2n hk0: h+k=2n

Determine the reflection condition for hk0.

Solved reflection conditions

002 200

002

110

h0l:h+l=2n hhl:l=2n hk0:h+k=2n

Also (from rest of the zones) hkl: h+k+l=2n.

110 -

200

110

Determine space group

• International Tables: I--- • Most symmetrical I4/mmm

Incommensurate vs. basic cell

Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350

LaSrCuO3.52

Identify and index the subcell reflections.

Propose a modulation vector.

q=αa*

α<0.5

q=αa*

α>0.5

200

002

[010]

q=αa*

α<0.5

Index the satellite indicated in green.

- 001

0001

100

0001

Index the satellite indicated in green.

Index the next satellite indicated in green.

0002

2002 -

2001 -

-

2002 -

Which of the indicated vectors corresponds to the

modulation vector chosen on the previous slides?

Is the proposed

vector still valid?

yes

no

Is the proposed

vector still valid?

no

You need

q1

q2

q1 ànd q2

q1=αa*+βb*

q2= -αa*+βb*

α=β<0.25

You need

q1 ànd q2

q1=αa*+βb*

q2= -αa*+βb*

α=β<0.25

20011

20011 -

20011 -

Index the reflection indicated in red.

20011 -

Index the reflection indicated in red.

Index the others patterns consistently with this new choice.

002 [010] 200

002

[010]

200

[110] [001]

020 200

110

002 -

Index the reflection indicated in yellow.

20011

20011

20011 -

-

00010

00001

- 20011


20011 -

00010

00001

- 20011


00010

00001

- 20011

10110

10111

10111 -


00010

00001

- 20011

10111 -


00010

00001

- 20011

00002

00002

00001 -

-


00010

00001

- 20011

00002 -

00010

00001

- 20011

Derive the reflection conditions for hklmn.

hklmn:

h+k+l+m+n=2i

h+k+l=2i

m+n=2i

hhl:l=2n

hkl:h+k+l=2n

00010

00001

- 20011

hklmn:

h+k+l=2i

hhl:l=2n

hkl:h+k+l=2n

Derive the reflection conditions for hklmn.

00010

00001

- 20011

Derive the reflection conditions for hhlm0.

hhlm0: l=2i

h=2i

l,m=2i

hhl:l=2n

hkl:h+k+l=2n

00010

00001

- 20011

Derive the reflection conditions for hklm.

hhl:l=2n

hkl:h+k+l=2n

hhlm0:

l,m=2i

but

hhlm0:m=2i

is sufficient

00010

00001

- 20011

Derive the reflection conditions for hklm.

hklmn:h+k+l=2i

hhlm0:m=2i

(-hhl0n: n=2i)

hhl:l=2n

hkl:h+k+l=2n

00010

00001

- 20011

Determine the superspace group.

I4/mmm(α α0, -αα0)00mg

http://stokes.byu.edu/iso/ssg.php

Stokes et al., Acta Cryst. A67, 45-55 (2011)







You know cell parameters,

modulation vector and

superspace group.

At the end of this lecture you should be able to:

UNDERSTAND and INDEX

the reciprocal lattice of an

incommensurately modulated material.

Recap? http://www.slideshare.net/johader

http://www.slideshare.net/johader




Aperiodic crystal workshop 2013: TEM

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