Chemistry Reference
In-Depth Information
Figure 4.26. The incident ( S o is the unit vector of the incident wave-normal) and
diffracted (“reflected”, S is the unit vector of the reflected wave-normal) X-ray beams
on two parallels crystal lattice planes ( hkl ) separated by a distance d hkl . The vector
b = S S 0 is normal to the lattice plane ( hkl ) with
| b |
=
| s s 0 |
=2sin θ .
relations are expressed in vector notations as below:
a ·
b = a ·
c = b ·
a = b ·
c = c ·
a = c ·
b =0
The magnitudes of the reciprocal vectors are the reciprocal of the spacing of
the corresponding planes in the real lattice, i.e., a · a = b · b = c · c =1.
It means that | a | is the reciprocal of the spacing of the a planes of the
real crystal lattice; | b | is the reciprocal of the spacing of the b planes of
the real crystal lattice; and | c | is the reciprocal of the spacing of the c
planes of the real crystal lattice. Thus, as shown in Figure 4.27, if PO
is the vector of length 1 in the direction of the incident wave-normal
( S o ), and PQ the vector ( S ) of the diffracted wave in the direction
of a maximum diffraction, OQ must be the vector ( b ) normal to one of
the lattice planes ( hkl ) with a magnitude of (2 )sin θ . When the Bragg
condition 2d hkl sin θ = λ is fulfilled, (2 )sin θ is equal to 1/ d hkl .OQis
therefore in the direction of a vector in the reciprocal lattice, and is equal to
the vector in magnitude (1/ d hkl ). If the vector PO is drawn so that O lies
at the origin of the reciprocal lattice, Q must lie at another point ( h , k , l )
of this lattice. In other words, if a sphere is drawn with P as the center and
1 as the radius, both the origin O and another point Q of the reciprocal
lattice must lie on the sphere. Otherwise there will be no diffraction.
For certain orientations of the crystal relative to the direction of the inci-
dent X-rays, there will be no reciprocal lattice points lying on the surface of
the diffraction sphere, and therefore no diffraction from the crystal planes.
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