Today’s Outline - February 27, 2020segre/phys570/20S/lecture_14.pdf · C. Segre (IIT) PHYS 570 -...

Today’s Outline - February 27, 2020

• Lattices & space groups

• Ewald sphere

• XRayView demonstration

• Crystal Truncation Rods

• Diffuse Scattering

• Modulated structures

• Lattice vibrations

Reading assignment: Chapter 5.2

Homework Assignment #04:Chapter 4: 2, 4, 6, 7, 10due Tuesday, March 10, 2020

C. Segre (IIT) PHYS 570 - Spring 2020 Feburary 27, 2020 1 / 22

• Ewald sphere

BCC structure factor

In the body centered cubic structure, there are 2 atoms in theconventional, cubic unit cell. These are located at

~r1 = 0, ~r2 =1

2(~a1 +~a2 +~a3)

the unit cell structure factor is thus

F bcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k+l))

= f (~G )×

{2 h + k + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2 +~a3)

F bcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k+l))

= f (~G )×

{2 h + k + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2 +~a3)

F bcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k+l))

= f (~G )×

{2 h + k + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2 +~a3)

F bcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k+l))

= f (~G )×

{2 h + k + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2 +~a3)

F bcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k+l))

= f (~G )×

{2 h + k + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2 +~a3)

F bcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k+l))

= f (~G )×

{2 h + k + l = 2n

0 otherwise

FCC structure factor

In the face centered cubic structure, there are 4 atoms in the conventional,cubic unit cell. These are located at

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3), ~r4 =

2(~a1 +~a3)

F fcchkl = f (~G )

e i~G ·~rj

= f (~G )(

1 + e iπ(h+k) + e iπ(k+l) + e iπ(h+l))

= f (~G )×

{4 h + k , k + l , h + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3), ~r4 =

2(~a1 +~a3)

F fcchkl = f (~G )

e i~G ·~rj

= f (~G )(

= f (~G )×

{4 h + k , k + l , h + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3), ~r4 =

2(~a1 +~a3)

F fcchkl = f (~G )

e i~G ·~rj

= f (~G )(

= f (~G )×

{4 h + k , k + l , h + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3), ~r4 =

2(~a1 +~a3)

F fcchkl = f (~G )

e i~G ·~rj

= f (~G )(

= f (~G )×

{4 h + k , k + l , h + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3), ~r4 =

2(~a1 +~a3)

F fcchkl = f (~G )

e i~G ·~rj

= f (~G )(

= f (~G )×

{4 h + k , k + l , h + l = 2n

0 otherwise

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3), ~r4 =

2(~a1 +~a3)

F fcchkl = f (~G )

e i~G ·~rj

= f (~G )(

= f (~G )×

{4 h + k , k + l , h + l = 2n

0 otherwise

Diamond structure

This is a face centered cubic structure with two atoms in the basis whichleads to 8 atoms in the conventional unit cell. These are located at

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3) ~r4 =

2(~a1 +~a3)

~r5 =1

4(~a1 +~a2 +~a3), ~r6 =

4(3~a1 + 3~a2 +~a3)

~r7 =1

4(~a1 + 3~a2 + 3~a3), ~r8 =

4(3~a1 +~a2 + 3~a3)

F diamondhkl = f (~G )

(1 + e iπ(h+k) + e iπ(k+l)

+ e iπ(h+l) + e iπ(h+k+l)/2 + e iπ(3h+3k+l)/2

+ e iπ(h+3k+3l)/2 + e iπ(3h+k+3l)/2)

This is non-zero when h,k,l all even and h +k + l = 4n or h,k ,l all odd

Diamond structure

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3) ~r4 =

2(~a1 +~a3)

~r5 =1

4(~a1 +~a2 +~a3), ~r6 =

4(3~a1 + 3~a2 +~a3)

~r7 =1

4(~a1 + 3~a2 + 3~a3), ~r8 =

4(3~a1 +~a2 + 3~a3)

(1 + e iπ(h+k) + e iπ(k+l)

+ e iπ(h+3k+3l)/2 + e iπ(3h+k+3l)/2)

Diamond structure

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3) ~r4 =

2(~a1 +~a3)

~r5 =1

4(~a1 +~a2 +~a3), ~r6 =

4(3~a1 + 3~a2 +~a3)

~r7 =1

4(~a1 + 3~a2 + 3~a3), ~r8 =

4(3~a1 +~a2 + 3~a3)

(1 + e iπ(h+k) + e iπ(k+l)

+ e iπ(h+3k+3l)/2 + e iπ(3h+k+3l)/2)

Diamond structure

~r1 = 0, ~r2 =1

2(~a1 +~a2), ~r3 =

2(~a2 +~a3) ~r4 =

2(~a1 +~a3)

~r5 =1

4(~a1 +~a2 +~a3), ~r6 =

4(3~a1 + 3~a2 +~a3)

~r7 =1

4(~a1 + 3~a2 + 3~a3), ~r8 =

4(3~a1 +~a2 + 3~a3)

(1 + e iπ(h+k) + e iπ(k+l)

+ e iπ(h+3k+3l)/2 + e iπ(3h+k+3l)/2)

Heteroatomic structures

← bcc

sc →

← diamondfcc →

← bccsc →

← diamondfcc →

← bccsc →

← diamond

fcc →

← bccsc →

← diamondfcc →

The Ewald sphere

The Ewald sphere is a construct which per-mits the enumeration of reflections which ful-fill the Laue diffraction condition.

The sphere radius is set by the length of the ~kand ~k ′ vectors which characterize the incidentand scattered (where the detector is placed)x-rays and ∆~k being the bandwidth of theincident x-rays

As the detector moves, ~k ′ rotates but theEwald sphere remains constant.

The xrayview program can be used to gain a more intuitive understandingof the Ewald sphere.

http://www.phillipslab.org/software

The Ewald sphere

Ewald sphere & the reciprocal lattice

The reciprocal lattice isdefined by the unit vec-tors ~a∗1 and ~a∗2.

The key parameter is therelative orientation of theincident wave vector ~k

As the crystal is rotatedwith respect to the inci-dent beam, the reciprocallattice also rotates

When the Ewald sphere intersects a reciprocal lattice point there will be adiffraction peak in the direction of the scattered x-rays. The diffractionvector, ~Q, is thus a reciprocal lattice vector

~Ghlk = h~a∗1 + k~a∗2

When the Ewald sphere intersects a reciprocal lattice point there will be adiffraction peak in the direction of the scattered x-rays.

The diffractionvector, ~Q, is thus a reciprocal lattice vector

~Ghlk = h~a∗1 + k~a∗2

~Ghlk = h~a∗1 + k~a∗2C. Segre (IIT) PHYS 570 - Spring 2020 Feburary 27, 2020 7 / 22

Ewald construction

It is often more con-venient to visualize theEwald sphere by keep-ing the reciprocal latticefixed and “rotating” theincident beam to visual-ize the scattering geome-try.

In directions of ~k ′ (detec-tor position) where thereis no reciprocal latticepoint, there can be nodiffraction peak.

If the crystal is rotated slightly with respect to the incident beam, ~k , theremay be no Bragg reflections possible at all.

Ewald construction

Polychromatic radiation

If ∆~k is large enough,there may be more thanone reflection lying onthe Ewald sphere.

With an area detector,there may then be multi-ple reflections appearingfor a particular orienta-tion (very common withprotein crystals wherethe unit cell is verylarge).

In protein crystallography, the area detector is in a fixed location withrespect to the incident beam and the crystal is rotated on a spindle so thatas Laue conditions are met, spots are produced on the detector at thediffraction angle

Multiple scattering

If more than one recipro-cal lattice point is on theEwald sphere, scatteringcan occur internal to thecrystal.

The xrays are first scat-tered along ~kint thenalong the reciprocal lat-tice vector which con-nects the two points onthe Ewald sphere, ~G andto the detector at ~k ′.

This is the cause of monochromator glitches which sometimes removeintensity but can also add intensity to the reflection the detector is set tomeasure.

Multiple scattering

The xrays are first scat-tered along ~kint

thenalong the reciprocal lat-tice vector which con-nects the two points onthe Ewald sphere, ~G andto the detector at ~k ′.

Multiple scattering

The xrays are first scat-tered along ~kint thenalong the reciprocal lat-tice vector which con-nects the two points onthe Ewald sphere, ~G

andto the detector at ~k ′.

Multiple scattering

Laue diffraction

The Laue diffractiontechnique uses a widerange of radiation from~kmin to ~kmax

These define two Ewaldspheres and a volumebetween them such thatany reciprocal latticepoint which lies inthe volume will meetthe Laue condition forreflection.

This technique is useful for taking data on crystals which are changing ormay degrade in the beam with a single shot of x-rays on a 2D detector.

Laue diffraction

Diffraction resources

XRayView

http://www.phillipslab.org/downloads

Bilbao Crystallography Server

http://www.cryst.ehu.es/

Hypertext Book of Space Groups

http://img.chem.ucl.ac.uk/sgp/mainmenu.htm

GSAS-II

https://subversion.xray.aps.anl.gov/trac/pyGSAS

XRayView

GSAS-II

XRayView

GSAS-II

XRayView

GSAS-II

XRayView demonstration

Exercise 1 - Ewald sphere

Exercise 4 - Wavelength

Exercise 8 - Laue diffraction

Exercise 9 - Serial crystallography

Diffraction from a Truncated Surface

For an infinite sample, the diffractionspots are infinitesimally sharp.

With finite sample size, these spotsgrow in extent and become more dif-fuse.

If the sample is cleaved and left withflat surface, the diffraction will spreadinto rods perpendicular to the surface.

The scattering intensity can be ob-tained by treating the charge distri-bution as a convolution of an infinitesample with a step function in the z-direction.

CTR Scattering Factor

The scattering amplitude FCTR along a crystal truncation rod is given bysumming an infinite stack of atomic layers, each with scattering amplitudeA(~Q).

FCTR = A(~Q)∞∑j=0

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

this sum has been discussed previ-ously and gives

or, in terms of the momentumtransfer along the z-axis,Qz = 2πl/a3

since the intensity is the square of the scattering factor

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3

=A(~Q)

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

e iQza3j

=A(~Q)

1− e iQza3=

1− e i2πl

ICTR =∣∣∣FCTR

∣∣∣2 =

∣∣∣A(~Q)∣∣∣2

(1− e i2πl) (1− e−i2πl)=

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl)

Dependence on Q

When l is an integer (meeting the Laue condition), the scattering factor isinfinite but just off this value, the scattering factor can be computed byletting Qz = qz + 2π/a3, with qz small.

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (πl + qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2=

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Dependence on Q

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2=

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Dependence on Q

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2=

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Dependence on Q

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2=

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Dependence on Q

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Dependence on Q

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2=

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Dependence on Q

ICTR =

∣∣∣A(~Q)∣∣∣2

4 sin2 (Qza3/2)

∣∣∣A(~Q)∣∣∣2

4 sin2 (qza3/2)

∣∣∣A(~Q)∣∣∣2

4(qza3/2)2=

∣∣∣A(~Q)∣∣∣2

q2za23

0 2π 4π

Absorption Effect

Absorption effects can be included as well by adding a term for each layerpenetrated

e iQza3je−βj

=A(~Q)

1− e iQza3e−β

This removes the infinity andincreases the scattering profileof the crystal truncation rod

0 2π 4π

Absorption Effect

e iQza3j

e−βj

=A(~Q)

1− e iQza3e−β

0 2π 4π

Absorption Effect

e iQza3je−βj

=A(~Q)

1− e iQza3e−β

0 2π 4π

Absorption Effect

e iQza3je−βj

=A(~Q)

1− e iQza3e−β

0 2π 4π

Absorption Effect

e iQza3je−βj

=A(~Q)

1− e iQza3e−β

0 2π 4π

β=0.2

Absorption Effect

e iQza3je−βj

=A(~Q)

1− e iQza3e−β

0 2π 4π

β=0.2

Density Effect

The CTR profile is sensitive to the termination of the surface. This makesit an ideal probe of electron density of adsorbed species or single atomoverlayers.

F total = FCTR + F top layer

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

where z0 is the relative displace-ment of the top layer from thebulk lattice spacing a3This effect gets larger for largermomentum transfers

0 2π 4π

Density Effect

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

0 2π 4π

Density Effect

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

0 2π 4π

Density Effect

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

0 2π 4π

Density Effect

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

where z0 is the relative displace-ment of the top layer from thebulk lattice spacing a3

This effect gets larger for largermomentum transfers

0 2π 4π

Density Effect

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

where z0 is the relative displace-ment of the top layer from thebulk lattice spacing a3

This effect gets larger for largermomentum transfers

0 2π 4π

z0=+0.05

z0=-0.05

Density Effect

=A(~Q)

1− e i2πl

+ A(~Q)e−i2π(1+z0)l

0 2π 4π

z0=+0.05

z0=+0.10

Today’s Outline - February 27, 2020segre/phys570/20S/lecture_14.pdf · C. Segre (IIT) PHYS 570 -...

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