The Bragg equation has a simple geometrical significance in the reciprocal lattice. Let we draw a vector AO of length 1/lambda in the direction of the incident X-ray beam and terminating at the origin of the reciprocal lattice. The tail of this vector does not necessarily have to rest at a reciprocal lattice point. Now we construct a sphere of radius 1/(lambda) about the point A as centre . Supposed this sphere intersects some point B of the reciprocal lattice, the indices of which being (h'k'l') of the lattice ,as such it must be normal to the (h'k'l') plane of the direct lattice , AE is this plane and therefore angle EAO=@ (thi tha){ the Bragg angle)also vector OB must be of length n/d, n being the largest integral factor common to the three integers (h'k'l')
Discussion:---- it is evident from this geometrical construction that Bragg condition will be satisfied for a given wavelength whenever the surface of a sphere of radius 1/(lambda) drawn about A intersects with a point of the reciprocal lattice. Now from figure 1 it is clear that for vector AO as the incident beam direction , the vector OB represents a normal to the reflecting plane , the vector AB is a vector in the direction of the diffracted beam and the angle between the vector AO and a plane normal to OB (i.e Vector AE) is the aporopriate Bragg angle in each case. This statement can be understood more easily by translating vector AB parallel to itself until its tail lies upon point O, when it will be seen that vector AO, OB and AB represents the familiar relation of the incident beam direction,scattering normal and diffracted beam direction.We also observe that when Bragg condition is satsfied, the vector AO and AB form an isosceles triangle with a reciorocal lattice vector OB. Again the desposition of vectors is such that AB vector must be the vector sum of AO vector and OB vector.
If the sphere passes through no points that would indicate that the particular wavelength in question would not be diffracted by that crystal in that orientation and the readjustment of the orientation is required. In fact the construction can be used to examine the different ways in which a crystal can be made to satisfy the Bragg reflection condition for its various planes. The construction also represent the experimental conclusion that
"Xray diffraction can not occur if wavelength is greater than (2a). This is because a sphere of radiud less than 1/2a can not pass through any point of the lattice and the above construction is not possible.If longer the vector AO ( shorter the wavelength ) , the greater is the likelihood of the spheres intersecting a point and hence of diffraction. This is known as " Ewald Construction ".
Vector form of the Bragg equation
The Bragg equation has a more elegant form in the reciprocal lattice. For the proof of this , let imagine all the vectors of fig (1) to be magnified by a constant scale factor of 2π and to relabel them as shown in fig (2):---
It is simply 2π times the vector AO. The disposition of the vectors , however is of the same type as in fig 2 and therefore for diffraction it is necessary that the magnitude of A'B' must be equal to the magnitude of A'O'. Thus the Bragg condition imply that:--
This is the vector form of the Bragg equation . Here G is 2π times a reciprocal lattice vector and K is a vector of magnitude 2π/(wavelength) along the direction of the incident Xray beam, known as wave-vector.
If we call the scattered wave vector as K', then in this type of scattering :- K'=K+G
2 and 3 show that the scattering changes only the direction of K and that the scattered wave differs from the incident wave by a reciprocal lattice vector G.
This is therfore an example of elastic scattering.
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