Page 14 - Modern physical chemistry

1.3 Diffraction

In the simple diffraction setup of figure 1.2, monochromatic (sinusoidal) radiation
strikes a plane containing equispaced parallel scattering lines at angle 0 and is observed
at angle ¢J above the plane. Maxima in the radiation are found at certain angles, which
we number, taking 0 as the order when ¢J equals 0, 1 as the order for the first larger angle,
2 for the second larger one, and so on.
Each maximum occurs where neighboring rays reinforce each other as much as pos-
sible. That is, where a crest of one ray reaches the observer at the same time as a crest
of each of its neighbors, a trough at the same time as a trough of each of its neighbors.
But at other angles, the diffracted ray labeled 2 cancels part of the one labeled 1, and so
on. Thus, the intensity drops very fast on each side of a maximum.
Since the rays labeled 1 and 2 start in phase at their source, they are reflected in phase,
with the crests matching at the observation point, when the extra path length along 1 equals
an integral number of wavelengths. But from figure 1.2, this extra length equals the pro-
jection of the distance between scattering lines on the incident ray 1 minus its projection
on the reflected ray 2:
nA = d(cosO -COS41). [1.1]

Note that n is the order of the reflection, A the wavelength of the radiation, d the dis-
tance between lines, 0 the glancing angle at which the rays strike the grating, and ¢J the
angle at which the diffracted rays leave the grating.
A person can detennine d directly by counting the number of lines per millimeter
under a microscope, or indirectly by measuring how radiation of known wavelength is
diffracted. With the short wavelength radiation needed to study crystals and with avail-
able gratings, angles 0 and ¢J are quite small. Nevertheless, they can be very accurately
measured, so A can be found to about six significant figures.

Example 1.1

X rays of wavelength 0.710 A strike a grating at a glancing angle 0 equal to 5' 0". If the
grating contains 200 lines cm- evenly spaced, at what angle ¢J would the first-order dif-
1
fraction peak be found?
This diffraction is governed by formula (Ll) with

n=l, A = 0.710 x 10- 8 cm, d= 1 ,
200 cm- 1

(5minX3.1416 radians)
0= = 0.001454 radians.
( 60 min deg- ¥ 180deg)
1
Because the angles are small, higher terms in the expansions
0 2
cosO = 1 - - + ... ,
2
41 2
cos41 = 1 - - + ... ,
2
which are valid when the angles are in radians, can be neglected.
Then