Page 17 - Modern physical chemistry

6 Structure in Solids

C
FIGURE 1.4 Reflection of X rays from parallel atomic planes in a crystal.

Since the neighboring rays, FA and GB, originate at the same distant source, they are
nearly parallel. And, equiphase surfaces intersect the two rays at right angles, as line AD
does. Since the detector is so far away, the reflected rays, AH and BI, are also nearly par-
allel. A maximum in intensity is observed where these arrive in phase. Then the wavelets
at points A and E, on a perpendicular to the two rays, are in phase. Such occurs when
length DB + BE is an integral number of wavelengths A. .
Each side of acute angle DAB is perpendicular to a side of the acute angle between
FA and the first parallel plane. Therefore, the two angles are equal. Similarly, angle EAB
equals the reflection angle O. In addition, hypotenuse AB of the two small right triangles
equals the interplanar spacing d. So from the definition of the sine, we have

DB = dsinO [1.2]
and
BE = dsinO. [1.3]

Setting the sum of these lengths equal to an integral number of wavelengths A. yields
the Bragg equation
nA. = 2d sin 0 where n = an integer. [1.4]
When n is 1, the reflection is said to be first order, when n is 2, second order, and so on.

Example 1.2

A crystal reflects X rays of wavelength 1.539 A most strongly when it is oriented so
the angle between the incident rays and the reflecting planes is 37° 15' . If the order of
the reflection is 1, what is the interplanar spacing?
Put the given data into formula (1.4),

(1 X 1.539 A) = 2d sin( 37° 15'),

and solve for d:
d = 1.539 A = 1.271 A.
2(0.6053)