ref.:elements of x-ray diffraction / b.d. cullity and s.r. …elements of x-ray diffraction / b.d....
TRANSCRIPT
Dr. Somsak [email protected]
Tel 8 2 Tel: 5852, 5777Course materials:
http://www.sc.mahidol.ac.th/scpy/courses/scpy642_09.html
Ref.:Elements of x-ray diffraction / B.D. Cullity and S.R. Stock, Prentice Hall, 2001
Interference and Diffraction of Light
Interaction and Effects X‐rays
Scattering
Plane wave approximationPlane wave approximation
Incident plane wave
rki khpe ivvvv
⇒.ii khpe i =⇒
Outgoing plane wave
ffrki khpe f
vvvv
=⇒.
g g p
Related topics : Huygen Principle
Diffraction
0coscos =−=− θθ PKPKPRQKDifference in Phase between 1’ and 1a’ 0coscos == θθ PKPKPRQK
θθ sinsin ddLNML ′+′=+Difference in Phase between 1’ and 2’
Difference in Phase between 1 and 1a
θλ sin2dn ′=Complete Constructive interference Bragg’s Law
Bragg’s LawTwo geometrical facts: 1. The incident beam, the normal to the diffraction plane, the
d ff d b l ldiffracted beam always coplanar.2. The angle between the diffracted beam and the transmitted
beam is always 2θ This angle is normally used as diffraction beam is always 2θ. This angle is normally used as diffraction angle. To use it in Bragg’s law divide it by two.
1sin2
<=′
θλd
n
d ′< 2λ
θλ sin2n
d ′=
θλ sin2d=
Laue’s Equations
λαα ha =− )cos(cos λαα ha =)cos(cos 0
λββ kb =− )cos(cos λββ kb =)cos(cos 0
λγγ lc =)cos(cos λγγ lc =− )cos(cos 0
Direct and Reciprocal lattice
Lattice vectors a, b, c
Vector to a lattice point: d = ua + vb + wc
Lattice planes (hkl)
Lattice vectors a*, b*, c*
V t t i l l tti i tVector to a reciprocal lattice point: d* = ha* + kb* + lc*
Each such vector is normal to the real space plane (hkl)
Length of each vector d* = 1/d-spacing (distance between hkl planes)(distance between hkl planes)
Reciprocal Lattice and Diffraction
+= AvuAδPath difference
)SOA (S
(-S).OA.OAS0 +=+= OnOm
)S--OA.(S 0=
λπδφ /2=Phase difference
.OA.-).OAS-(S
- 0 Sπλ
π2
2==
321 bbb lkh ′+′+′=S
)(2
)()(2
rlqkph
rqplkh
′+′+′=++′+′+′=
ππφ
-
aaa.bbb- 321321
For diffraction to occur, S must start or end on points of the reciprocal lattice.
Direct and Reciprocal Lattice
Diffraction Direction
Limiting sphere : Rotating Ewald sphere about the origin
Translation to practice: Change incident angleChange incident angle
Diffraction Direction
θλ sin2d=
2
222
2
)(1
a
lkh
d
++=
ad
)(sin 2222
2 lkh ++λθ )(
4sin
2lkh
a++=θ
22i
λθ2110
2
2sin
a
λθ =Cubic
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+=
2
2
2
22
2
22
4sin
c
l
a
kh
a
λθ
Tetragonal
⎠⎝4 caa
Diffraction directions (θ) determined solely by the shape and size of the unit cell
Distance between planes
222 )( cubic
lkh
a
++
2
2
22
2
Tetragonal
)(
l
c
kh
a
lkh
+
++
2
2
2
2
2
2
222
cOrthohombi
g
cba
lkh
++
+
222 cOrthohombi
lkh++
Intensities of Diffracted Beams
Order of ComplexityX‐rays or Photons interacts through Coulomb interaction
with charged particles i e electronswith charged particles, i.e. electrons.
↓Many electrons build up an atomMany electrons build up an atom.
↓Many atoms orderly arrangement build up a unit cellMany atoms, orderly arrangement build up a unit cell.
↓Many unit cells constitutes materialMany unit cells constitutes material.
∴XRD f t l t t f it ll d t i tiXRD: x‐ray for crystal structure for unit cell determination.
Come basically from interactions of x‐ray with many electronselectrons
Scattering by an electron
Thomson found that intensity I of the beam scattered by a single electron of charge e coulombs and mass k kg at a distance r meters from the electron
ααμ 22
42
0 sinsin ⎟⎞
⎜⎛=⎟⎟
⎞⎜⎜⎛
⎟⎞
⎜⎛=
KI
eII αα
π 20220 sinsin4
⎟⎠
⎜⎝
=⎟⎟⎠
⎜⎜⎝
⎟⎠
⎜⎝
=r
Irm
II
α = angle between the scattering direction and direction of acceleration of the electron
Scattering by an electron
ααπμ 2
202
22
42
00 sinsin
4⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
r
KI
rm
eII
π4 ⎠⎝⎠⎝⎠⎝ rrm
θ2cos2KII =K
II θ2cos2r
II OzPz =
III
2rII OyPy =
22
)2cos( θ+=
+=
IIK
III
OzOy
PzPyP
22
2
)2cos22
( θ+=II
r
Kr
OO
yX-ray direction : OXUnpolarized beam
222 EEE + 2122 EEE
1 III 31
2
2)
2
2cos1(
22
θ+=
r
KI
r
O
2z
2y
2 EEE += 2212
z2y EEE ==
021
00 III zy == 4434421factoron polarizati
Compton Effect on Electron
θλλλ 2o
sin04860)A( ==Δ θλλλ 21 sin0486.0)A( =−=Δ
0 Å t θ 0 d0 Å at θ = 0 degree 0.05 Å at θ = 180 degree
Random backgro nd radiationRandom background radiation
Scattering by an atom
Atom composes of many electrons Scattering by atom ≈ scatterings from many electrons inside an atomScattering by atom ≈ scatterings from many electrons inside an atom
atoman by scattered wave theof amplitudef
electron oneby scattered wave theof amplitude=f
Scattering by a Unit Cell
Now consider ray 1´and 3´
)(/
)(13 λλδh
x
AC
ABRBS ===′′
y 3
)(/
)(13 haAC
δ)2( π
λδφ =
C id ´ d ´
λθδ ===′′ sin2 0012 hdMCN a
hxππλδ
φ 2)2(13
13 == ′′′′
Consider ray 1´ and 2´
h
aACdh ==00 huπφ 213 =′′
Scattering by a Unit Cell
)2i ( φAE
)(2 lwkvhu ++= πφ)2sin( 111 φπν −= tAE
)2sin( 222 φπν −= tAE )2sin( 222 φπνtAE
Scattering by a Unit Cell
xixeix sincos +=
φφφ sincos AiAAei +=
2 22AAeAeAe iii == − φφφ
2222 )sin(cos)sincos(sincos AAAiAAiA =+=−+ φφφφφφ
)(2 lwkvhuii feAe ++= πφ
K+++= ++++++ )(23
)(22
)(21
333222111 lwkvhuilwkvhuilwkvhui efefefF πππ
∑ ++=N
)(2 nnn lwkvhuinhkl efF π∑
1
Scattering by a Unit Cell
∑ ++=N
1
)(2 nnn lwkvhuinhkl efF π
electron oneby scattered wave theof amplitude
cellunit a of atoms theallby scattered wave theof amplitude=F
[ ]∑ +++++=N
1
)(2sin)(2cos nnnnnnn lwkvhuilwkvhufF ππ1
ibaF +=∑
N
∑N
∑ ++=1
)(2cos nnnn lwkvhufa π ∑ ++=1
)(2sin nnnn lwkvhufb π
222))(( baibaibaF +=−+= ))(( baibaibaF ++
[ ]222221111
2)(2cos)(2cos L++++++= lwkvhuflwkvhufF ππ
[ ]222221111 )(2sin)(2sin L+++++++ lwkvhuflwkvhuf ππ
Some Useful Relations
153 −=== iii eee πππ 1eee
1642 === iii eee πππ 1=== eeenine )1(=πe )1(−=
inin ππ − inin ee ππ =ii 2−ππ xee ii cos2=+ ππ
Structure‐Factor Calculations
a) A unit cell with one atom at the origin) g
ffeF i == )0(2π 22 fF =
b) A based‐centered cell two atoms; one at origin the other at 021
21
22
)1( )(
)2/2/(2)0(2
khi
khii
f
fefeF+
++=π
ππ
)1( )( khief ++= π
unmixedandfor2 khfF = mixed andfor 0= khF22 4
unmixed and for 2
fF
khfF
= 02 =F
Structure‐Factor Calculations
d) A face‐centered cell; four atoms; 21
21
21
21
21
21 0 0 0 000
)1( )()()(
)2/2/(2)2/2/(2)2/2/(2)0(2
lhilkikhi
lhilkikhii
f
fefefefeF+++
+++
+++
+++=πππ
ππππ
)1( )()()( lhilkikhi eeef +++ +++= πππ
indicesunmixedfor4 fF = indicesmixedfor0=F22 16
indicesunmixedfor 4
fF
fF
= 0
indices mixedfor 02 =F
F
Structure‐Factor Calculations
B i L tti t R fl ti ibl R fl ti Bravais Lattice type Reflection possibly present
Reflection necessarily absent
Simple all noneSimple all none
Base‐centered h and k unmixed h and k mixed
Body‐centered (h+k+l) even (h+k+l) odd
Face‐centered (h k and l) (h k and l) mixedFace centered (h, k, and l) unmixed
(h, k, and l) mixed
Step for calculating structure factor:
W it d th t iti1. Write down the atom positions.
2. Write down the equation for F as a product of the value of the
common translation terms and the basis atoms of the cell.common translation terms and the basis atoms of the cell.
3. As necessary, simplify the factor further.
Structure Factor: NaCl
r+/r⁻=0.69NaCl 5 640 Å
000000N 111111
NaCl 5.640 ÅLattice type: fcc
00
0
0000
00
Cl
000Na
21
21
21
21
21
21
21
21
21
21
21
21
Image credit: http://www.chemistry.wustl.edu/~edudev/LabTutorials/Water/PublicWaterSupply/images/nacl.jpghttp://wikis.lib.ncsu.edu/index.php/Halite-NaCl
Structure factor of NaCl
∑ ++=N
1
)(2 nnn lwkvhuinhkl efF π
00
0
0000
00
Cl
000Na
21
21
21
21
21
21
21
21
21
21
21
21
)2/(2)2/(2)2/(2)2/2/2/(2
)2/2/(2Na
)2/2/(2Na
)2/2/(2Na
)0(2Na
hikililkhi
lhilkikhii
efefefef
efefefefFππππ
ππππ
++++
+++=++
+++
000000Cl 222222
)(Cl
)(Cl
)(Cl
)(Cl efefefef ++++
[ ][ ]hkllkh
lhilkikhi eeefF πππ +++= +++
)(
)()()(Na 1
[ ]ihikillkhi eeeef ππππ ++++ ++ )(Cl
[ ])()()(Na 1 lhilkikhi eeefF +++ +++= πππ
( )[ ])()()(Cl 1 lkilhikhilkhi eeeef −−−−−−++ ++++ ππππ
[ ])()()(1 lhilkikhi eeefF +++ +++= πππ[ ]( )[ ])()()(
Cl
Na
1
1lhilkikhilkhi eeeef
eeefF+++++ ++++
+++=ππππ
[ ][ ])(ClNa
)()()(1 lkhilkilhikhi effeeeF +++++ ++++= ππππ
Structure factor of NaCl
[ ][ ])(ClNa
)()()(1 lkhilkilhikhi effefeeF +++++ ++++= ππππ
indicesmixedfor0=F
For mixed indices: e.g. 001, 011
0
indices mixedfor 02 =F
F
[ ])(ClNa4 effF lkhi+= ++π
For unmixed indices: e.g. 111, 200
[ ][ ][ ]2ClNa
2
ClNa
ClNa
16
even is if 4
ffF
l)k(hff
ff
+=
+++=
[ ]ClNa16 ffF +
odd; is if )(4 ClNa l)k(hffF ++−=
)(16 2ClNa
2 ffF −=
Calculation of Structure Factor
Step for calculating structure factor:1 Write down the atom positions1. Write down the atom positions.
2. Write down the equation for F as a product of the value of the common translation of the value of the common translation terms and the basis atoms of the cell.
3. As necessary, simplify the factor further.3 y p y
4 Na at 000 + face‐centering translations,l f l4 Cl at ½½½ + face‐centering translations
⎤⎡ [ ]indices mixed
indices unmixed
0
4 )(ClNa
lkhieffF +++⎥⎦
⎤⎢⎣
⎡= π
Structure factor of Diamond
4 C at 000 + face-centering translations
[ ] indices unmixed1
4 2/)( lkhifF ++⎥⎤
⎢⎡ π
g4 C at ¼¼¼ + face-centering translations
[ ]indices mixed
10
2/)(c
lkhiefF +++⎥⎦
⎤⎢⎣
⎡= π
2 CfF = 2 CfF = 0=F
h,k,l unmixed & oddh,k,l unmixed & odd multiple of two
Cf C
, , u ed & odd p
h,k,l unmixed & even multiple of two
image credit: http://people.uis.edu/kdung1/Gemini/Diamond.jpg
multiple of two
Hexagonal LatticeSi l H l (SH) S h f l i tSimple Hexagonal (SH) - Spheres of equal size are most densely packed (with the least amount of empty space) in a plane when each sphere touches six other spheres arranged in the form of a regular hexagon. When these hexagonally closest packed planes (the plane through the g y p p ( p gcenters of all spheres) are stacked directly on top of one another, a simple hexagonal array results; this is not, however, a three-dimensional closest packed arrangement. The unit cell, outlined in black, is composed of one atom at each corner of a primitive unit cell (Z = 1)
SHof one atom at each corner of a primitive unit cell (Z 1), the edges of which are: a = b = c = 2r, where cell edges a and b lie in the hexagonal plane with angle a-b = gamma = 120 degrees, and edge c is the vertical stacking distance.
Hexagonal Closest Packing (HCP) - To form a three-dimensional closest packed structure, the hexagonal closest packed planes must be stacked such that atoms in successive planes nestle in the triangular "grooves" of thesuccessive planes nestle in the triangular grooves of the preceeding plane. Note that there are six of these "grooves" surrounding each atom in the hexagonal plane, but only three of them can be covered by atoms in the adjacent plane. The first plane is labeled "A" and the
HCPsecond plane is labeled "B", and the perpendicular interplanar spacing between plane A and plane B is 1.633r (compared to 2.000r for simple hexagonal). If the third plane is again in the "A" orientation and succeeding planes
t k d i th ti tt ABABA (AB) thare stacked in the repeating pattern ABABA... = (AB), the
resulting closest packed structure is HCP.
http://www.cartage.org.lb/en/themes/Sciences/Physics/SolidStatePhysics/AtomicBonding/ReciprocalLattice/CrystalLattice/hcp_aba.gif
Structure factor of Hexagonal
Atoms of the same kind!
21
32
31000 X
)2/3/23/(2)0(2 lkhii fefeF +++= ππ
)1( )2/3/)2((2 lkhief
ff+++= π
)1( 2 igefF π+=+= )2cos22(22
igfF π
Put [(h+2k)/3)+l/2]=g
)2(
)1()1(222
2222
igig
igig
eef
eefFππ
ππ
−
−
++=
++=[ ]−+=
+=
cos4
)1cos2(22
)2cos22(
22
22
f
gf
igfF
π
π
⎟⎠⎞
⎜⎝⎛ +
+=
=
23
2cos4
cos4
22 lkhf
gf
π
π
⎠⎝ 23
= 0 when (h+2k) is a multiple of 3 and l is odd
Structure factor of Hexagonal
[ ]+=
)12(22
)2cos22(22
22
f
igfF π
[ ]
⎞⎛
=
−+=
2
cos4
)1cos2(2222
22
lkh
gf
gf
π
π
⎟⎠⎞
⎜⎝⎛ +
+=
23
2cos4 22 lkh
f π
= 0 when (h+2k) is a multiple of 3 and l is odd 0 when (h+2k) is a multiple of 3 and l is odd
⎞⎛
when (h+2k) is a multiple of 3 and l is even h+2k l |F|2
nlkh
=⎟⎠⎞
⎜⎝⎛ +
+23
2
2
, where n is an integer 3m odd 03m even 4f 2
2
2
1cos
1cos
n
n
=
±=
π
π 3m even 4f 3m±1 odd 3f 2
224 fF = 3m±1 even f 2
Structure factor of ZnS
ZnS pdf file
Example
1. List the first few values of h2+k2+l2 of the following crystal structure lattice structure lattice
• Simple cubic
• BCC
• FCC
• Diamond
Scc 1,2,3,4,5,6,8,9,10,11,12 ,…B 2 4 6 8 10 12 14Bcc 2,4,6,8,10,12,14,…FCC 3,4,8,11,12,16,19,20 ,…Diamond 3,8,11,16,19 ,…
ExerciseBelow is XRD pattern of polycrystalline Al. the spectrum is obtained with θ‐2θ diffractrometer technique using CuKα (1.5402 Å) If Al belongs to a cubic lattice could we draw a conclusion of Å). If Al belongs to a cubic lattice, could we draw a conclusion of which sublattice in cubic to which aluminium belong?
Image credit: http://www.ntbase.net/imgAll/pr_al_xrd.gif (FCC)
l f ( l d d b d)1. Polarization factor (already described)
2. Structural factor (already described)2. Structural factor (already described)
3.Multiplicity factor
4.Lorentz factor
5 Absorption factor5.Absorption factor
6.Temperature factor
Factors affect relative intensity
1. Polarization factor2. Structural factor
4. Lorentz factor5. Absorption factorf
3. Multiplicity factor5. Absorption factor6. Temperature factor
)θA()(LP)(2 θFI { {)θA(.)(LP.p.)(
factor absorptiononpolarizati-Lorentzfactorty multiplici
2∝ θhklhkl FI321
)θA()(LPp)e p(
2N
)(2∝ ∑ ++ θπ nnn lwkvhuiMf )θA()(LPp)exp( 1
)(
factor etemperatur
⋅⋅⋅−∝ ∑ θnnnnn eMf
43421
Polarization Factor
+= III
2 )2cos( θ+=
+=
IIK
III PzPyP
2
2
)2(
)2cos(
θ
θ+=
IIK
IIr
OO
OzOy
2
22
21
)2cos22
(
θ
θ
+
+=
K
rOO
4434421
2
2)
2
2cos1(
θ+=
r
KIO
X ray direction : OX 2factoron polarizati
X‐ray direction : OXUnpolarized beam
Structure Form Factor
xixeix sincos +=
φφφ sincos AiAAei +=
2 22AAeAeAe iii == − φφφ
2222 )sin(cos)sincos(sincos AAAiAAiA =+=−+ φφφφφφ
)(2 lwkvhuii feAe ++= πφ
K+++= ++++++ )(23
)(22
)(21
333222111 lwkvhuilwkvhuilwkvhui efefefF πππ
∑ ++=N
)(2 nnn lwkvhuinhkl efF π∑
1
Multiplicity FactorF d i th th l h i th For powder specimens; there are more than one planes having the same d‐spacing. These planes contribute to the x‐ray intensity. Multiplicity factor; p = the number of permutations of position Multiplicity factor; p the number of permutations of position and sign of ±h, ±k, ±l for planes having the same d and F2.
hkl hhl 0kl 0kk hhh 0k0 00lhkl hhl 0kl 0kk hhh 0k0 00lCubic 48* 24 24* 12 8 6
Hexagonal &24* 12* 12* 12* 6 6 2
Rhombohedral24 12 12 12 6 6 2
Tetragonal 16* 8 8 8* 4 4 2Orthorhombic 8 4 4 4 2 2 2Orthorhombic 8 4 4 4 2 2 2Monoclinic 4 2 2Triclinic 2
Note that, in cubic crystal, for example, hhl stands for such indices as 112 (or 211), 0kl for such indices as 012 (or 210), 0kk for such indices as 011 (or 110), etc.* These are the usual multiplicity factors. In some crystals, planes having these indices comprise two forms with the same spacing but different structure, and the multiplicity factor for each form is half the values given above. In the cubic system, for example, there are some crystals in which permutations of the indices (hkl) produce planes which are not structurally equivalent; in such crystals (AuBe is an example), the plane (123), for example, belongs to one form and has a certain structure factor, while the plane (321) belongs to another
form and has a different structure factor. These are 48/2=24 planes in the first form and 24 planes in the second.
Temperature FactorEffects of thermal vibration1. unit cell expands → periodicity changes → position of 2θ
• can be used to determine thermal expansion coefficientp2. Intensities of the diffraction lines decrease3. Intensity of background scattering between line increase.
Image credit: http://www.ruppweb.org/write\AEG.gif
Temperature Factor (II)Meff −= 0
222 sinsin ⎞⎛⎞⎛⎟⎞
⎜⎛ θθu 22
22 sinsin
82 ⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛=
λθ
λθππ Bu
d
uM
22 i6 ⎞⎛⎤⎡ θTh2
2 sin
4)(
6⎟⎠⎞
⎜⎝⎛⎥⎦⎤
⎢⎣⎡ +=
λθφ
θx
xmk
ThM u=amplitude of thermal vibration
θ=diffraction angleh=Planck’s constant
20232
23426
2
2
)10)(1038.1(
)1063.6)(1002.6(66
×Θ××
=Θ −−
−
A
T
mk
Th k =Boltzmann’s constantT = absolute temperaturem = A/N = atomic weight/Avagadro#
2
41015.1
Θ×
=A
T Θ = Debye characteristic tempx= Θ/T
Temperature FactorMeff −= 0
2sin
⎟⎠⎞
⎜⎝⎛−
= λθ
B
fef = fef
Temperature-diffuse scattering
Lorentz Factor
21δ −=′′ CBAD
[ ])cos()cos(
coscos 12
21
θθθθθθ
Δ+−Δ−=−=
a
aa
a
aa
CBAD
θθθθ
δ
sin2
coscos 12
21
Δ=−=
−=′′
Ba θθ sin2 Δ=
BNa λθθor
sin2 =Δ
BNa θλθsin2
or
=Δ
Lorentz Factor
Distribution of plane normals for a particular cone of diffracted raysp y
2
cos
2
)90sin(2.2
BB
r
rr
N
N θθπ
θπθ Δ=
−Δ=
Δ o
22 rN π
θθθθ
θθ cossin4
1
2sin
cos
2sin
1cos
2sin
122
==⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
Lorentz Factor
Lorentz Polarization Factorθθθ
cossin
2cos12
2+
Lorentz Factor2
Lorentz Polarization Factorθθθ
cossin
2cos12
2+
Absorption Factor
1. Hull/Debye‐Scherer CameraA(θ)=f(θ,μr)A(θ) f(θ,μr)
⎟⎞
⎜⎛ compactρ
μ0
xeII = −
⎟⎟⎠
⎞⎜⎜⎝
⎛=
solid
compactsolidcompact ρ
ρμμ
ρμμ
*
)(mt coefficien absorptionlinear ; 1-
m==
Absorption Factor2. θ‐2θ Diffractrometera= volume fraction of the specimen containing particles having the
i i f diff i f h i id bcorrect orientation for diffraction of the incident beamb=fraction of incident energy which is diffracted by one unit volume
dxeablIdI BCABD
)(0
But
(ergs/sec) μ +−=
xBC
xABl
sin ,
sin ,
sin
1
But
βγγ===
dxeabI
dI x )sin/1sin/1(0
Therefore
βγμ
βγγ
+−= dxeabI
dI x θμ sin/20 −=dxedI D sin γ= dxedI D γsin
=
Absorption factor for thick film2. θ-2θ Diffractrometer
abI θi/20 dxeabI
dI xD
θμ
γsin/20
sin−=
In case of infinitely thick samples
μ20abI
dIIx
DD ∫∞=
==
In case of infinitely thick samples
How thick is infinite???
1000)at x(
)0at x( sin/2 ==== θμe
tdI
dI
D
D
μ20x=
The incident beam is reduced by a factor.
t t1
)(θA μθsin45.3
=t
by a factor.
constant2
)( ==μ
θA μExample: powder Ni ρ(powder)=0.6 ρ(bulk) = 0.6 × 8.9 g/cm3
261 1 t k θ 90° t 0 132μ=261 cm-1, taken θ~90°, t = 0.132 mm
Absorption factor for thin film2. θ-2θ Diffractrometer
abI θi/20 dxeabI
dI xD
θμ
γsin/20
sin−=
In case of samples of thickness t
=
∫tx
dII
In case of samples of thickness t
==∫
0x
DD dIIAbsorption factor A = ID(x=t)/ID(x=∞)
⎬⎫
⎨⎧−=
−θμ
sin
20 1
t
eabI
μ2 t
⎭⎬
⎩⎨μ1
2e
θμ
sin
2
1t
eA−
−=In t → ∞ limit ID=abI0/2μ
Absorption factor for thin film
μ2 t−
θsin1 eA −=
Diffraction and Reflection
Differences between diffraction and reflection1 Diffracted beam from a crystal is built up of rays scattered by 1. Diffracted beam from a crystal is built up of rays scattered by
all the atoms of the crystal which lie in the path of the incident beam. The reflection of visible light takes place in a c de t bea . e e ect o o v s b e g t ta es p ace athin surface layer only.
2. The diffraction of monochromatic x‐rays takes place only at those particular angles of incidence which satisfy Bragg’s law.The reflection of visible light takes place at any angle of incidenceincidence.
3. The reflection of visible light by a good mirror is almost 100 percent efficient. The intensity of a diffracted x‐ray beam is p y yextremely small compared to that of the incident beam.
N k h diff b diff i d fl i ! Thi Now we know the difference between diffraction and reflection! This course is just about diffraction even when we say reflection.
Direct and Reciprocal lattice
⎟⎠⎞
⎜⎝⎛
ו×
=ו
×=
cba
cba
aaa
aab *
321
321
⎞⎛
⎟⎠⎞
⎜⎝⎛
ו×
=ו
×=
b
cba
acb
aaa
aab *
321
132
⎟⎠⎞
⎜⎝⎛
ו×
=ו
×=
cba
bac
aaa
aab *
321
213