020 x ray diffraction VẬT LÝ CHẤT RẮN

Chapter Two: Reciprocal SpaceIn this chapter, we will introduce the idea of reciprocal space from the perspective of diffraction experiments.. X-ray scattering Bragg reflection momentum

Chapter Two: Reciprocal Space

In this chapter, we will introduce the idea of reciprocal space from the perspective of diffraction experiments X-ray and neutron diffraction were both very important for elucidating crystal structures They also suggested a fundamentally new way of looking at solids – in reciprocal space (often called momentum space)

X-ray scattering Bragg reflection momentum space reciprocal space Fourier transforms

Why do we use x-rays, neutrons and electrons to

investigate matter?

We need to use

particles with

wavelengths ~ 2 Å (remember – matter has wave-like

characteristics)

Neutrons 20 meV X-rays 6 keV

Electrons 0.35 eV

Wave length λ versus energy E

momentum p

Light: λ = c / ν h ν = E = c p

∴ λ = h / p = c h / E

Matter: non-relativistic, mass m

λ = h / p = h / m v (de Broglie wavelength)

E = p2 / 2m p = √ (2mE)

∴ λ = h / √(2mE)

What happens during a diffraction

experiment?

- X-rays (or neutrons, or electrons) of a single wavelength (and therefore, energy) are incident upon a crystal The incoming

rays are of the proper wavelength for diffraction (on the order

of the interatomic spacing), and thus we see diffraction peaks at certain values of θ, the scattered beam angle Each one of these peaks is from a plane of atoms within the crystal This is an

elastic process

Diffraction peaks

are observed as a

function of

scattered angle

Miller indices for planes of atoms

Experimental setup

Because there are many different

planes of atoms, we will see reflections

at many different angles The

particular crystal structure of a material

can be obtained by observing which

reflections cancel out because of the

arrangements of the atoms (we will

talk about this later) Why do we see

Bragg peaks at all?

Approach to Bragg scattering

For this to happen, the extra distance travelled by light ray B must be a multiple of the

wavelength

Bragg scattering occurs when reflections

from parallel beams interfere

constructively

W L Bragg (1913) came up with this simple theory for x-ray diffraction Bragg Condition for scattering from successive planes

2d sin θ = n λ : Condition for

constructive interference of x-rays – Bragg peaks

How are diffraction experiments

done?

X-ray diffraction

d = λ/(2 sin θ) = 0.154 x 10-9 / (2 x sin 22.2º) = 2.04 Å

Experimental setup (how do we

produce x-rays?)

X-ray production is via

Bremstrahlung radiation (think of

the inverse photoelectric effect –

instead of light hitting a target and

electrons being emitted, electrons hit

a target and photons are emitted)

The electrons are produced typically

from a tungsten source, are

accelerated towards a metal such as

copper, and when they hit the

surface, they slow down This

“braking radiation” is a broad band

of light which is emitted as the

electron slows down (charged

particles under acceleration emit

radiation)

Voltage to accelerate e

-Tungsten source produces e - by heating

Copper target gives off xrays

Broad band of x-rays produced

Characteristic radiation

On top of this Bremstralung radiation, there are a few very strong bursts of x-rays at very precise energies These are due to electrons hitting target atoms, and inducing inner shell electron transitions This process occurs when an inner shell electron is ejected, and to take it’s place, an electron from one of the higher energy shells makes a transition and gives off light (in the form

of x-rays) It is these x-rays that are used in diffraction.

This shows a K-shell (often called the 1s orbital) transition

Why are there 2 transitions

for this K process?

The first

diffraction

experiments :

Electron

diffraction

from Ni

Electron source Electron detector

For interference from the first plane alone, the

condition is different

Davisson – Germer experiment (1927)

of electron diffraction (matter

is wave-like!)

Note: the diffraction condition is different in this case (this is a case of

experimental geometry)

ki kf

| ki | = | kf | = k = 2π/λ (momentum of the x-ray, conserved due to elastic collision)

vector for planes

Now we need to talk about k vectors

Light path difference x

x = 2d sin θ

d

Phase shift of the lower scattered x-rays: ϕ = k x = 2 d k sin θ

(when we have Bragg reflection, the waves are in step, so the

phase difference is 2π Using this, we can get Bragg’s law:

2π = 2 d k sin θ = 2 d (2π/λ) sin θ 2d sin θ = nλ (n=1) In

general, though, they are not in step, and will have some

phase which will depend upon the scattered angle)

Diffraction conditions

kf

ki

f - ki : momentum transfer = 2 k sin θ

What we really need to figure out is how the reflected x-rays are out of step with one another (what is the

phase change from one plane to the next?)