Bills Lecture 5 (Pending Title)

Question 1

Synchotron Radiation

Answer 1

A method of making X-Rays

Particle accelerators accelerate electrons to near the speed of light. They give of X-rays when they slow down.

Question 2

X-Rays

Answer 2

Wavelength of 0.01 to 10 nm / 0.1 to 100 Å

Question 3

Parameters that describe an electromagnetic wave

Answer 3

Amplitude, Wavelength, and Phase (where a locus is on the wave)

Question 4

Destructive Interference

Answer 4

Zero amplitude, waves cancel out.

Question 5

Constructive Interference

Answer 5

Stronger Amplitude

Question 6

Intermediate between Destructive Interference and Constructive Interference

Answer 6

Reduced Amplitude

Question 7

Bragg’s Law

Answer 7

λ=2dsinθ
xy=yz=dsinθ
xyz=2dsinθ
Explains X-ray diffraction by treating crystals as consisting of layers or planes of semi-transparent mirrors. Some X-rays are reflected with an angle of reflection that equals the angle of incidence and some are transmitted through the plane and reflected by subsequent planes.

Question 8

Compton Effect

Answer 8

Reduction in frequency of scattered x-ray beam occurs since the collision of photons with the electron results in an energy loss.

Question 9

Elastic Scattering

Answer 9

Production of scattered radiation with the same wavelength as the incident beam.

Question 10

The Atomic Scattering Factor “f”

Answer 10

(sinθ)/λ
Amplitude of x-rays scattered by an atom in any direction. Given in arbitrary electron units (e.u.), such that the forward scattering at θ=0 is equal to the number of extra nuclear atoms Z.

Question 11

The Structure Factor “F”

Answer 11

F=Σf
I ∝ |F|^2 : Intensity is proportional to the magnitude of F squared.
This is the amplitude of X-rays scattered by one unit cell, in e.u.’s in the direction of the hkl reflection. Into a reflection hkl, each atom scatters its amplitude f. The amplitudes from all the atoms add together with phase differences dependent on hkl and the positions of the atoms.

Question 12

Vector Representation of a Wave

Answer 12

Argand Diagram
A + iB = c cosα + ic sinα is equivalent to c*exp[iα]
Where c = Amplitude and exp[iα] = Phase

Question 13

The Electron Density Equation

Answer 13

ρ(x,y,z) = 1/Vc * Σ F(h,k,l)*exp[-2πi(hx+ky+lz)]
Where:
ρ(x,y,z) = Electron density at any point x,y,z
Vc = Volume of the unit cell

Question 14

The Patterson Function

Answer 14

Puvw = 1/Vc Σ |F|^2 cos2π(hu + kv + lw)
The height of a Patterson peak depends on the number of electrons in the atoms between which the vector occurs and is proportional to their atomic numbers (Z).

Vectors between heavy atoms appear with heights equivalent to the square of their atomic numbers (Z) and so can stand out against the heavy-light, light-light atom vectors.

No phase information required, only intensity measurements and knowledge of symmetry

The Patterson function is centrosymmetric even if original structure is not.

Question 15

Types of Patterson Vector

Answer 15

Cross Vectors: Between atoms in the same asymmetric unit

Harker vectors: Between symmetry related atoms

Question 16

Harker Sections

Answer 16

Portions of the Patterson map that contain a large proportion of the readily interpretable information because they contain many such Harker peaks.
e.g. Using Patterson Function:
x-1/2, 1/2, -2z
Above shows y = 1/2, this is a Harker section, its location in a structure will be known, 0.5 .