X-ray crystallography uses X-rays to determine the atomic and molecular structure of crystals. X-rays have wavelengths small enough (~0.1nm) to probe the distances between atoms in crystals. When X-rays hit a crystal, they cause the electrons of individual atoms to scatter the X-rays in all directions. The scattered X-rays interfere with one another, producing a diffraction pattern that can be used to reveal the crystal structure. Crystals are required because they produce repeated patterns that amplify the X-ray scattering, making it possible to determine atomic positions.

1. X-ray diffraction

2. 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?”

3. 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.

4. 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).

5. 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.

6. 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?”

7. 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.

8. Why do we use diffraction?

9. 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.

10. Scattering of X-rays by electrons of a single atom

11. 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

12. Scattering of X-rays by two adjacent atoms
*sin21
sin2
d
nd
or
dn
d* is the resolution of the X-ray data.

13. 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
, jfs

14. Representation of plane waves

15. 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

16. 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

17. 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

18. 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)].

19. 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.

20. 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)

21. 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.

22. 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.

23. 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
- - - - -

24. X-ray intensity data quality
h
N
i
ih
h
N
i
hih
merge
I
II
R
1
1

25. The reconstruction of electron density
rrSrrSF
r
FTdi
cellV
2exp.
rrF *
FT *
rFr 1
FTrrFTFT 1

26. The phase problem

27. 2. Patterson Function
1. Guess the structure 3. Molecular Replacement
4. Multiple Isomorphous Replacement
HPPH FFF
5. Multiple Anomalous Dispersion

28. Resolution and Model Building

29. Refinement
h
h
obs
calcobs
F
FF
R

30. Structure validation