BT631-15-X-Ray_Crystallography_Diffracton

X-ray diffraction

Why do we use X-ray?

What is of significance is the comparison between the size of the obstacle (i.e. a significant

linear dimension of the obstacle) and the wavelength of the wave motion with which the

obstacle interacts.

Boat = 3 m

Larger wave = 20-30 m

Smaller wavelength = 1 m

“Why can we not use light waves to study the structure of a crystal?”

The nature of the interaction is dependent of the relative magnitudes of the wavelength of the

wave motion and the dimensions of the obstacle.

If the wavelength is much greater than a significant linear dimension of the obstacle, then the

obstacle has only a small effect on the behavior of the wave.

If the wavelength of the wave motion is smaller than or comparable to a significant linear

dimension of the obstacle, then the behavior of the wave is affected to large extent, and events

such as reflection may occur.

Thus, if we wish to study an object by means of its effect upon a wave motion, then me must

choose a wave form whose wavelength is smaller than or comparable to a significant linear

dimension in the obstacle.

Only in this event will the object cause a pronounced effect on the wave motion and it is from

this effect that we derive information about the obstacle.

A crystal contains two significant linear dimensions. One of these is

the macroscopic length of the crystal itself, which we may take as

typically 1 mm.

We now place the crystal in the path of a light beam of wavelength 600 nm, corresponding to

yellow light. The wavelength of the light is approximately 103 times smaller than the

macroscopic crystal size, but the order of 103 times greater than the intermolecular spacing.

The other significant linear dimensions is the spacing between the

molecules in the crystal, which we expect to be in the range of the

molecular dimensions, of the order of 1 nm (10 Å).

The light wave will be affected by the macroscopic crystal size, but will not be noticeably

perturbed by the intermolecular spacing. The light therefore sees only the crystal as a whole,

and not the molecular scale structure. So, by looking a the effect of the crystal on a beam of

yellow light, we can detect only the macroscopic features, and can gain no information on

molecular structure. Light is therefore not a suitable wave motion for investigating he detailed

molecular architecture of crystals.

Now, if we put the crystal in a beam of X-rays, the situation is changed. The wavelength of

the X-ray is of the order of 0.1 nm (1 Å), quite comparable to the molecular spacing within

the crystal. Hence we may predict that the molecular scale structure will have a large effect on

the X-rays.

“Why can we not use X-ray microscope to study the structure of a crystal?”

XM-1 held the world record in spatial

resolution with Fresnel zone plates down to

15 nm and is able to combine high spatial

resolution with a sub-100 ps time resolution to

study e.g. ultrafast spin dynamics.

In July 2012, a group at DESY

claimed a record spatial resolution of

10 nm, by using the hard X-ray

scanning microscope at PETRA III.

Why do we use diffraction?

Why do we need a crystal for X-ray diffraction?

Furthermore, the order of the three-dimensional array of molecules implies that the effect of

any one molecule on the X-rays is repeated in a regular manner throughout the crystal.

This makes the effect of an individual molecule much amplified, so that relatively large

overall effects are observable.

The effect of the molecular structure and organization of a crystal is to cause diffraction of the

X-ray beam, and from the nature of the diffraction, the molecular structure of the crystal may

determined.

Scattering of X-rays by electrons of a single atom

Path difference

Δp = (λr.s0 – λr.s1) = (s1 – s0).r λ = S.rλ

Scattering of X-rays by electrons of an atom

Phase difference

Δφ=(2π/λ) x path difference = 2πS.r

Scattering of X-rays by two adjacent atoms

*sin21

sin2

dnd

or

dn

d* is the resolution of the X-ray data.

Scattering of X-rays from a molecule

j

atoms

j

js if SrFs 2exp.1

0

,

The scattering function or molecular

diffraction envelop of the entire

molecule.

sF

atomic scattering factors0

, jf s

Representation of plane waves

Addition of plane waves

sincossincos iFiiBA FF

sincos iei

Euler’s Formula for the complex exponential

function

ii eFe ..FF

It implies that

Thus, the addition of n waves F=F1+F2+F3… Fn

becomes

n

j

n

j

n

j

jj

i

j

i

j iFeFe jj

1 1 1

exp...FF

Scattering of X-rays from a molecular crystal

1

0

1

0

1

0

2exp.2exp.2exp.w

w

v

v

u

u

cellcryst wiviui ScSbSaFF SS

The Fourier Transform decomposes any function into a sum of sinusoidal basis functions.

Each of these basis functions is a complex exponential of a different frequency. The Fourier

Transform therefore gives us a unique way of viewing any function - as the sum of simple

sinusoids.

Fourier Transforms

Fourier Transforms

Convolution Theorem:

Convolution: take one function, f(r), and put it down at every point of a second function, g(r);

f(r)*g(r). Here * is the convolution operator.

Convolution Theorem states: FT [f(r) * g(r)] = F(S) • G(S); that is, the Fourier transform of

one function convoluted with another is the same as the Fourier transform of the first

multiplied by the Fourier transform of the second (here we will set a convention; real space

functions and their coordinate symbols are expressed in lower case, e.g. f, g, and r, and their

Fourier transforms and frequency space coordinates are expressed in upper case, e.g. F, G, and

S). The converse is also true: F(S) * G(S) = FT [f(r)•g(r)].

A crystal is a convolution of one function (a motif) with another (a lattice)

Lattice * Motif = Crystal

2-dimensional crystal formed by convoluting a motif with a 2-dimensional lattice.

A 3-dimensional crystal would result from convolution of a motif with a 3-dimensional lattice.

FT (lattice) -- another lattice with spacings which are reciprocal of those of the original lattice

Note that the coordinates of Reciprocal space are inverse distance; for example, if a and b are

measured in Å, the dimensions of the reciprocal lattice, 1/a=a* and 1/b=b* will be Å-1.

Fourier Transform of a Crystal (Diffraction Pattern)

FT (motif) -- a continuous function -- no sharp discontinuities

FT(crystal) = FT[(motif) * (lattice)] = FT(motif) . FT(lattice) = FT(motif) . (Reciprocal

Lattice).

Thus, the continuous Fourier transform of the motif is sampled at the points of the Reciprocal

lattice; the Fourier transform of the crystal is only non-zero at the points of the Reciprocal

lattice.

Diffraction can be described as the transformation of the electron density from real space R

into reciprocal space R*. The reciprocal space diffraction pattern is then given by the

amplitudes of the Fourier transform (FT) of the crystal structure.

X-ray intensity Data Collection

Index Intensity Error

0 0 6 56268.9 3346.5

0 0 9 1961 96.7

0 0 12 175988 7349

0 0 15 1919.5 115.8

0 0 18 11464 516.1

0 0 21 178022 7456

0 0 24 22022.4 952.3

0 0 27 36798.8 1406

0 0 36 62468.2 2375.5

- - - - -

X-ray intensity data quality

h

N

i

ih

h

N

i

hih

merge

I

II

R

1

1

The reconstruction of electron density

rrSrrSFr

FTdi

cellV

2exp.

rrF* FT *

rFr1FTrrFTFT 1

The phase problem

2. Patterson Function

1. Guess the structure 3. Molecular Replacement

4. Multiple Isomorphous Replacement

HPPH FFF

5. Multiple Anomalous Dispersion

Resolution and Model Building

Refinement

h

h

obs

calcobs

F

FF

R

Structure validation