Objectives

By the end of this section you should:• understand the concept of diffraction in crystals• be able to derive and use Bragg’s law• know how X-rays are produced• know the typical emission spectrum for X-rays,

the source of white radiation and the K and K lines

• know about Compton scattering

Diffraction - an optical grating

XY

1

2

a

Coherent incident light Diffracted light

Path difference XY between diffracted beams 1 and 2:

sin = XY/a

XY = a sin

For 1 and 2 to be in phase and give constructive interference, XY = , 2, 3, 4…..n

so a sin = n where n is the order of diffraction

Consequences: maximum value of for diffraction

sin = 1 a =

Realistically, sin <1 a >

So separation must be same order as, but greater than, wavelength of light.

Thus for diffraction from crystals:

Interatomic distances 0.1 - 2 Å

so = 0.1 - 2 Å

X-rays, electrons, neutrons suitable

Diffraction from crystals

XY

Z

d

Incident radiation “Reflected” radiation

Transmitted radiation

1

2

?

Beam 2 lags beam 1 by XYZ = 2d sin

so 2d sin = n Bragg’s Law

XY

Z

d

Incident radiation “Reflected” radiation

Transmitted radiation

1

2

We normally set n=1 and adjust Miller indices, to give

2dhkl sin =

2d sin = n

e.g. X-rays with wavelength 1.54Å are reflected from planes with d=1.2Å. Calculate the Bragg angle, , for constructive interference.

= 1.54 x 10-10 m, d = 1.2 x 10-10 m, =?

d2

nsin

nsind2

1

n=1 : = 39.9°

n=2 : X (n/2d)>1

1d

ha

kb

lc2

2

2

2

2

2

2

Use Bragg’s law and the d-spacing equation to solve a wide variety of problems

2d sin = n

or

2dhkl sin =

Example of equivalence of the two forms of Bragg’s law:

Calculate for =1.54 Å, cubic crystal, a=5Å

2d sin = n

(1 0 0) reflection, d=5 Å

n=1, =8.86o

n=2, =17.93o

n=3, =27.52o

n=4, =38.02o

n=5, =50.35o

n=6, =67.52o

no reflection for n7

(2 0 0) reflection, d=2.5 Å

n=1, =17.93o

n=2, =38.02o

n=3, =67.52o

no reflection for n4

X-rays with wavelength 1.54 Å are “reflected” from the (1 1 0) planes of a cubic crystal with unit cell a = 6 Å.

Calculate the Bragg angle, , for all orders of reflection, n.

Combining Bragg and d-spacing equation

056.06

0112

2

222

2 a

lkh

d

1

18d2 d = 4.24 Å

d = 4.24 Å

d2

nsin 1

n = 1 : = 10.46°

n = 2 : = 21.30°

n = 3 : = 33.01°

n = 4 : = 46.59°

n = 5 : = 65.23°

= (1 1 0)

= (2 2 0)

= (3 3 0)

= (4 4 0)

= (5 5 0)

2dhkl sin =

X-rays and solids

X-rays - electromagnetic waves

So X-ray photon has E = hX-ray wavelengths vary from .01 - 10Å; those used

in crystallography have frequencies 2-6 x 1018Hz

Q. To what wavelength range does this frequency range correspond?

c =

cmax = 1.5 Å

min = 0.5 Å

In the classical treatment, X-rays interact with electrons in an atom, causing them to oscillate with the X-ray beam.

The electron then acts as a source of an electric field with the same frequency

Electrons scatter X-rays with no frequency shift

Anode Cathode

Beryllium windowFilament

e.g. tungsten

Pyrex glass envelope (under vacuum)

Production of X-rays

Electrons stopped by target; kinetic energy converted to X-rays

continuous spectrum of “white” radiation, with cut-off at short (according to h=½mv2)

Wavelength not characteristic of target

Incident electrons displace inner shell electrons, intershell electron transitions from outer shell to inner shell vacancy.

“line” spectra

Wavelength characteristic of target

Two processes lead to two forms of X-ray emission:

I

E

Typical emission spectrum

Each element has a characteristic wavelength.

For copper, the are:

CuK1 = 1.540 Å

CuK2 = 1.544 Å

CuK = 1.39 Å

Many intershell transitions can occur - the common transitions encountered are:

2p (L) - 1s (K), known as the K line

3p (M) - 1s (K), known as the K line

(in fact K is a close doublet, associated with the two spin states of 2p electrons)

Copper K X-rays have a wavelength of 1.54 Å and are produced when an electron falls from the L shell to a vacant site in the K shell of a copper atom. Calculate the difference in the energy levels between the K and L shells of copper.

E = h

= c/ = (3 x108) / (1.54 x 10-10)

= 1.95 x 1018 Hz

E = h = 6.626 x 10-34 x 1.95 x 1018

= 1.29 x 10-15 J

~ 8 keV

Some radiation is also scattered, resulting in a loss of energy [and hence, E=h, shorter frequency and, c= , longer wavelength.

The change in frequency/wavelength depends on the angle of scattering.

This effect is known as Compton scattering

It is a quantum effect - remember classically there should be no frequency shift

)cos1(cm

h

e

Calculate the maximum wavelength shift predicted from the Compton scattering equation.

)cos1(cm

h

e

cm

h2

e

831

34

1031011.9

10626.62

= 4.85 x 10-12 m = 0.05Å

Filter

As well as characteristic emission spectra, elements have characteristic absorption wavelengths

e.g. copper

We want to choose an element which absorbs K [and high energy/low white radiation] but transmits K

e.g. Ni K absorption edge = 1.45 Å

As a general rule use an element whose Z is one or two less than that of the emitting atom

Monochromator

Choose a crystal (quartz, germanium etc.) with a strong reflection from one set of lattice planes, then orient the crystal at the Bragg angle for K1

= 1.540 Å = 2dhklsin

Example: A monochromator is made using the (111) planes of germanium, which is cubic, a=5.66Å. Calculate the angle at which it must be oriented to give CuK1 radiation

22

222

2 )66.5(

3

a

lkh

d

1

)27.32(

540.1sin

d2sin 11

d=3.27Å

=2d sin

= 13.62°

SummarySummary Crystals diffract radiation of a similar order of

wavelength to the interatomic spacings

We model this diffraction by considering the “reflection” of radiation from planes - Bragg’s Law

X-rays are produced by intershell transitions - e.g. L-K (K) and M-K (K)

The interaction of X-rays with matter produces a small wavelength shift (Compton scattering)

Filters can be used to eliminate K radiation; monochromators are used to select K1 radiation.