Reglas de Seleccion Hcp

7/22/2019 Reglas de Seleccion Hcp

1/28

Suggested Reading

ages - n e rae c enry

Pages 59-61 in Engler and Randle

1


2/28

Structure Factor (Fhkl

)

2 ( )i j iN

i hu kv lw

hkl iF f e

Describes how atomic arrangement (uvw)influences the intensity of the scattered beam.

. ., ,expect in a diffraction pattern.

2


3/28


)

The amplitude of the resultant wave is given by aratio of am litudes:

amplitude of the wave scattered by all atoms of a UCF

amp tu e o t e wave scattere y one e ectron

to |Fhkl|2.

3


4/28

Some Useful Relations

ei = e3i = e5i = = -1

e2

i = e4

i = e6

i = = +1eni = (-1)n, where n is any integer

eni = e-ni, where n is any integer

eix + e-ix =2 cosx

Needed for structure factor calculations

4


5/28

Fhkl

for Simple Cubic

Atom coordinate(s) u,v,w:

2 ( )

1

i j i

Ni hu kv lw

hkl i

i

F f e

hklF fe f

structure factor equation for simple cubic crystals, the solution is alwaysnon-zero.

5

Thus, all reflections are allowed for simple cubic (primitive) structures.


6/28

Fhkl

for Body Centered Cubic


2 ( )

1

i j i

Ni hu kv lw

hkl i

i

F f e

, , .h k l

22 (0) 2 2 2

ii

hklF fe fe

1 i h k lF f ehkl

When h+k+l is even Fhkl = non-zeroreflection.

6When h+k+l is odd Fhkl = 0

no reflection.


7/28

Fhkl

for Face Centered Cubic


2 ( )

1

i j i

Ni hu kv lw

hkl i

i

F f e

,,0; ,0,;

0,,.

2 2 2h k h l k l

i i i

hklF e e e e

7

1

i h k i h l i k l

hkl

F f e e e


8/28

Fhkl

for Face Centered Cubic

1 i h k i h l i k lhklF f e e e

Substitute in a few values of hkl and you will find

When h,k,l are unmixed (i.e. all even or all odd), thenFhkl = 4f. [NOTE: zero is considered even]

hkl = or m xe n ces .e., a com na on o oand even).

8


9/28

Fhkl

for NaCl Structure

Atom coordinate(s) u,v,w: +

2 ( )

1

i j i

Ni hu kv lw

hkl i

i

F f e

0,0,0; ,,0; This means these

coordinates , , ;

0,,.(u,v,w)

Cl at ,, + FC transl. ,,; ,,

The re-assignment of coordinates is

, , ; , , 1,,1; 0,,0

,1,1. ,0,0

based upon the equipoint concept inthe international tables for

crystallography

9 Substitute these u,v,w values into Fhkl equation.


10/28

Fhkl

for NaCl Structure contd

For Na: 2 ( )1

i j i

Ni hu kv lw

hkl i

i

F f e

2 (0) ( ) ( ) ( )i i h k i h l i k l

( ) ( ) ( )1Na

i h k i h l i k l

Naf e e e

For Cl:

2 ( ) 2 ( ) 2 ( )( ) 2 2 2l k hi h k i h l i k li h k l

Clf e e e e

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2

( )

2 2i h k l i i ih k h l k lCl

i h k l i i i

Cl

f e e e e

f e e e e

l k h

l k hThese terms are all positive and even.

10

e er e exponen s o oreven depends solely on the remaining

h,

k, and

lin each exponent.


11/28

Fhkl

for NaCl Structure contd

2 ( )

1

i j i

Ni hu kv lw

hkl i

i

F f e

Therefore Fhkl:

( ) ( ) ( )

( ) ( ) ( ) ( )

1 i h k i h l i k l

hkl Na

i h k l i i i

F f e e e

l k h

which can be simplified to*:

( ) ( ) ( ) ( )1i h k l i h k i h l i k lhkl Na ClF f f e e e e

11


12/28

Fhkl

for NaCl Structure

When hkl are even Fhkl = 4(fNa +fCl)

hkl Na - Cl

Superlattice reflections

When hkl are mixed Fhkl = 0

o re ec ons

12


13/28

(200)100

80

90

NaClCuKradiation

(220)%)

60

70

ntensity(

50

30

40

(222)(420)

(422) (600)(442)

20

13

(311) (331)

(333)(511)

(440)

(531)

20 30 40 50 60 70 80 90 100 110 1202

()0


14/28

Fhkl

for L12 Crystal Structure

Atom coordinate(s) u,v,w: A B

2 ( )

1

i j i

Ni hu kv lw

hkl i

i

F f e

,,0; ,0,;

0,,.

2 2 2h k h l k li i i A B B BF f e f e f e f ehkl

14( ) ( ) ( )

A B

i h k i h l i k lF f f e e e

hkl


15/28

Fhkl

for L12 Crystal Structure

( ) ( ) ( )A B i h k i h l i k lF f f e e ehkl

(1 0 0) Fhkl =fA +fB(-1-1+1) =fAfB

(1 1 0) Fhkl =fA +fB(1-1-1) =fAfBA B

(1 1 1) Fhkl =fA +fB(1+1+1) =fA +3fB

(2 0 0) Fhkl =fA +fB(1+1+1) =fA +3fB

= + - + - =

(2 2 0) Fhkl =fA +fB(1+1+1) =fA +3fB

(2 2 1) Fhkl =fA +fB(1-1-1) =fAfB

(3 0 0) Fhkl =fA +fB(-1-1+1) =fAfB

(3 1 0) Fhkl =fA +fB(1-1-1) =fAfB

= =

15

(2 2 2) Fhkl =fA +fB(1+1+1) =fA +3fB


16/28

Intensity (%)

100

(111)

Example of XRD patternfrom a material with an

L12 crystal structure90

70

50(200)

30

(220)(311)

2

10

20(100)

(110)

(210) (211) (221)(300)

310

(222)

320 (321)

16

20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 1150


17/28

Fhkl

for MoSi2

Atom positions: Mo atoms at 0,0,0; ,,

c

a oms a , ,z; , ,z; , , +z; , , -z; z=

MoSi2 is actually body centered tetragonal witha = 3.20 and c = 7.86 z

c c ac

x y

x y

z

xy

z

b

x

y

z

a

b

b b

Viewed down z-axis

17

Viewed down x-axis Viewed down y-axis


18/28

Fhkl

for MoSi2

Substitute in atom positions:

( )2

1

i j iN

i hu kv lw

hkl i

i

F f e

Mo atoms at 0,0,0; ,,

Si atoms at 0,0, ; 0,0,z; ,,+z; ,,-z; z=1/3

2 2 22 ( ) 2 ( )2 (0) 2 2 2 2 2 6 2 2 63 3

i i ii ii

F f e f e f e f e f e f ehkl Mo Mo Si Si Si Si

2 ( ) 2 ( ) 3 33 31

i h i hi ii h k lF f e f e e e e

Mo Si

Now we can plug in different values for h k l to determine the structure factor.

or =

5(0 ) (0)(0) ( 0)3 33 3

1 0 1 02 ( ) 2 ( )1 0 0

21

0

1 1 1 1 1 1 0

i ii ii

hkl Mo Si

hkl

F f e f e e e eF

182You will soon learn that intensity is proportional to ; there is NO REFLECTION!

o

lhkF


19/28

Fhkl

for MoSi2 contd

. For h k l = 0 0 1

5(1) (1)

1 13 33 3

0 0 0 02 ( ) 2 ( )0 0 10

i ii ii

hkl Mo SiF f e e f e e e e

223(1 ) (2 ( ) )

(1 1) ( 1 1) 0

i i

Mo Si

Mo si

f e f COS e

f f

For h k l = 1 1 0

2 0 NO REFLECTION!hkl

F

1 1 0 1 1 0 1 1 00 2 (0) 2 (0)2 (0) (0) 2 2(1 ) ( )

(2) (4)

i i ii i

hkl Mo Si

i i i

Mo Si

Mo si

F f e e f e e e e

f e f e e e e

f f

2 POSITIVE! YOU WILL SEE A REFLECTION

hklF

19

to the rules for diffraction

h + k + l = even


20/28

100(103)

80

90

MoSi2CuKradiation

%)

60

70

(101)

nt

ensity(

50

30

40

(002)(213)

20(112) (200)

(202) (211) (116) (206)

2020 30 40 50 60 70 80 90 100 110 1202

()0

(006)(204) (222) (312)

(314)


21/28

Fhkl

for MoSi2 contd

21

- .PCPDFWIN v. 2.1


22/28


) for HCP

2N

i hu kv lw

1

hkl i

i

e

influences the intensity of the scattered beam.

i.e.,

It tells us which reflections (i.e., peaks, hkl) toexpect in a diffraction pattern from a given crystal

22

, , .


23/28

, , ,lattice point coordinates are:

0 0 0

1 2 1

3 3 2

Therefore, the structure factor becomes:

2h k l

3 3 21hkl iF f e


24/28

We sim lif this ex ression b lettin :2 13 2

h kg

which reduces the structure factor to:

2 ig

We can simplify this once more using:

from which we find:

cose e x

2 2 2 2h k l 3 2

hkl i


25/28

Selection rules for HCP

0 when 2 3 and oddh k n l 2

2

2when 2 3 1 and eveni

hklf h k n lF

2

4 when 2 3 and even

i

if h k n l

For your HW problem, you will need these things todo the structure factor calculation for Ru.

HINT: It might save you some time if you alreadyhad the ICDD card for Ru.


26/28

List of selection rules for different crystals

Crystal Type Bravais Lattice Reflections Present Reflections Absent

Simple Primitive, P Any h,k,l None

Body-centered Body centered, I h+k+l = even h+k+l = odd

Face-centered Face-centered, F h,k,l unmixed h,k,l mixed

NaCl FCC h,k,l unmixed h,k,l mixed

Zincblende FCC Same as FCC, but if all even

and h+k+l4N then absenth,k,l mixed and if all even

and h+k+l4N then absent

Base-centered Base-centered h,kboth even or both odd h,kmixed

Hexagonal close-packed Hexagonal h+2k=3Nwith l even

h+2k=3N1 with l oddh+2k=3N1 with l even

h+2k=3Nwith l odd

26


27/28

What about solid solution alloys?

If the allo s lack lon ran e order then ou

must average the atomic scattering factor.

falloy

=xAf

A+ x

Bf

B

where xn is an atomic fraction for the atomic

const tuent

27


28/28

Exercises

For CaF2 calculate the structure factor anddetermine the selection rules for allowed

reflections.

28

Reglas de Seleccion Hcp

Documents

Transcript of Reglas de Seleccion Hcp