### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# SPEC TOP INORG MAT MATRL 286G

UCSB

GPA 3.94

### View Full Document

## 48

## 0

## Popular in Course

## Popular in Material Science and Engineering

This 7 page Class Notes was uploaded by Miss Alberto Prohaska on Thursday October 22, 2015. The Class Notes belongs to MATRL 286G at University of California Santa Barbara taught by R. Seshadri in Fall. Since its upload, it has received 48 views. For similar materials see /class/226927/matrl-286g-university-of-california-santa-barbara in Material Science and Engineering at University of California Santa Barbara.

## Similar to MATRL 286G at UCSB

## Reviews for SPEC TOP INORG MAT

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/22/15

Materials 286CUCSB Class VI Structure factors continued the phase problem Patterson techniques and direct methods Ram Seshadri seshadrimrlucsbedu Structure factors The structure factor for a system with many atoms each with its own form factor fj and sitting within the unit cell at a site with the crystallographic coordinates 147127 wj is described by N Fhkl Z fj exp27rihuj k Uj 107 j1 The square of the structure factor is an indication of the intensity of any spotpeakintensity in the diffraction pattern corresponding to the Bragg conditions being satis ed for the particular hkl There are other contributors to the intensity such as the Lorentz polarization factors and the Debye Waller factor that we will discuss at a later stage Some applications CsCl The CsCl structure is simple cubic ngm with the Cs atom at the corners of a cube at 000 and the Cl atom at G The structure factor for any hkl re ection is 1 1 1 Fhkl fcs exp27rih0 k0 0 fol exp27rih k5 which simpli es to Fhkl fcs fciexplm h k 1 When h k 1 is even explwih k 1 1 and when h k 1 is odd expm h k 1 71 This means for CsCl F 7 fcs fc1 if h k 1 is even W T fcs 7 fol if h k 1 is odd For an hcp metal The atoms are at 000 and g and both atoms have a form factor f 1 2 1 Fhkl f fexp2m hg kg We consider speci c cases F001ffexpl7rilff0 F002 ffexp27ri ff Qf 2m 1 F100ffexpi f fi 3 2 2 1 1 3 x3 F101 7 f fexp27 g 5 7 f5 if The corresponding intensities are obtained from 1th Fhkl Fikl Thus 001 0 002 41 27 1100 1 27 101 3f2 The Argand diagram can be used to graphically represent FM f gt F002 gt gt F100 1 2713 F101 J 5713 Diffraction from a centrosymmetric crystal In a centrosymmetric crystal uvw and 714712710 represent the same point Note that negative numbers are also written with overlines iu E In other words for any atom with form factor f at the position u v w there is an identical atom at iii Together the structure factor due to these is FM f exp2m hu w lw f exp2m ha kt m f exp27rihu kv 10 f exp727rihu kv 10 Now expi gt exp7i gt cos gt i sin gt cos gt 7 i sin gt 2 cos gt This means that for a centrosymmetric crystal Fhkl 2f cos27rhu kv 10 Friedel39s Law Friedel s law states that even if a crystal does not possess a center of symmetry it s diffraction pattern does 1W Fhkl ngl fexp27rihu kv 10 x f exp72m hu kv 10 f exp27rihu kv 10 x f exp2m hu Ev 10 By the same token m Fm F f exp2m hu Ev Tw x f exp72m hu Ev Tw f exp2m hu Ev Tw x f exp27rihu kv 10 In other words 7 as 7 FhkliFmanthkliFm and hid m Friedel s law is Violated in the case when diffraction is anomalous The phase problem Simply stated what is measured in an experiment is Ihkl but what is required is FM There is no information of phase in Ihkl Inverting the structure factor equation Since the atomic form factor is a re ection of the electron density7 one could write a the expression for the structure factor in continuous form rather than for discrete points Fhkl pzgz exp2m hz kg 12 cell where pzgz represents the electron density at point in the unit cell and the integration is performed for all points in the cell Assume that Fhkl can be measured Then pzgz can be obtained through the Fourier transform Of F hkl 1 pzgz V FW expli2m hm l kg l 2 where V is the volume of the unit cell One could write this as the summation 1 pzgz V Z Fhkl expli2m hm l kg l 2 hkl But we remember that the vector Fhkl can be written in terms of its amplitude and phase Fhkl thkll expli hklll where gthkl is the phase associated with the point in reciprocal space associated with the coordi nates hkl This means pyz Z thkll expli gthkl exppmm kg 12 hkl The structure factor has units of electrons and the density pzgz has units of electrons A s The summation is over all h k and 1 values The summation must be carried out on as ne a grid of m g and 2 points as possible to obtain a smooth electron density distribution in the unit cell Where the electron density is concentrated7 atoms are found Solving a crystal structure Assume you have a crystal about which you know nothing This crystal diffracts X rays and you collect as complete a 3D set of diffraction data as possible The question is whether the diffraction data has all the information necessary to completely determine the structure of the crystal Can the diffraction pattern be used to locate all atoms in the unit cell7 and their identities established The answer is yes7 at least in most cases This despite the fact that phase information is not available in the diffraction pattern Two methods are commonly used to get around the phase problem The rst is the use of so called Patterson synthesis and the second7 more widely used class of tools are referred to as Direct Methods Materials 286CUCSB Class V The form and structure factors intensities and the phase problem and systematic absences Ram Seshadri seshadri mrucsbedu The atomic form factor This discussion closely follows Elements of Modern X my Physics by Jens Als Nielsen and Des Morrow John Wiley 85 Sons Ltd 2001 and makes use of gures from their book httpntservfyskudkXBook Consider the scattering of X rays from two electrons one at the origin and the other separated by a distance 77 Let an incident X ray of wavevector E be scattered elastically to some k7 after it leaves the second electron The scattering wavevector cf is de ned as 4 2 I ll7 8m9j The diffraction condition is provided by the phase difference gz k7 77 cf 77 The scattering amplitude is given by AU f0 foeiw f01 6W The scattered intensity is the square of the scattering amplitude 16 AACf 2f ll COM 77 mensty un ts 01 f N 0 1 1 0 05 1 15 qun391ts 01 27W Scattering from 2 electrons separated by F when 1 is in the same direction as F What is f0 7 It re ects the ability of the electron to scatter In the case of atoms7 f0 is replaced by f7 the atomic scattering factor l t d f tt b t atomic scattering factor f amp 1 u e 0 sea ermg y a 0m amplitude of scattering by a single electron Consider an atom has with a spherically symmetric distribution of electrons This distri bution is represented by some The scattering from the entire atom can be written as an integral over all the space within which the electrons are enclose 19gt mm where cmquot is the usual phase factor The limiting conditions are q a 0 when f Z where Z is the atomic number7 and q a 00 when f 0 At 1 07 all the scattered radiation is in phase When 1 start to become large7 the phase differences between the scattering will increase and destructive interference will tend to drive the scattering to 0 The form factors of floppy77 atoms and ions tend to die out faster that the form factors of compact77 atoms Tabulations of the form factor The calculated 1 form factors for the different elements and their important ions can be found tabulated using nine terms For example7 for Si 11 b1 12 b2 13 b3 14 b4 C l l 62915 l 24386 l 30353 l 32333 l 19891 l 06785 l 15410 l 816937 l 11407 l 1Such calculations by Di Ti Cromer form some of the most cited papers of all time The following function makes use of these 9 constants to evaluate fs where s q47T sin 9 A 4 M Z c j1 Go to httpwww structurellnlgovXraycompscatfachtm to make plots of fs for different elements Form factors for neutrons When nuclei which are very very small scatter neutrons with wavelengths of the order of 1 A s is effectively 0 and the scattering the so called scattering length remains constant throughout the scattering diagram The Debye formula Based on an extension ofthe two electron scattering problem see Als Nielsen one can arrive at the very general Debye formula for scattering of X rays by molecules7 crystals etc For N atoms7 each with its form factor f the scattering intensity is given by lt N are X f J j1 gt lf1l2 lf2l2lf1vl2 orient av sin 7 sin 7 2f1f2M2flfNM quN sin 7 sin 7 2f2f37q 23 2f2fN q 2N qrgg 1er sin qu sin qu N 2fN71f1M 2fN71fN QTN711 QTN71N The different 77 represent the distances between atom 239 and atom j So given a system where all the atom positions are known7 the scattering can be calculated This applies for glasses7 crystals7 nanoparticles The structure factor See Hammond and the handout Friedel39s law See Hammond and the handout

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.