Solid State Physics || UNIT 01|| Crystal Diffraction & Reciprocal lattice || 💐 EWALD Construction 💐||Geometrical Interpretation of the Bragg's Condition ||my notes||#physicsextreem

Bragg condition in terms of reciprocal lattice:---

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')

Waale Construction in the reciprocal lattice giving the possible directions of the detracted rays for the incident direction AO. The length of OB can also be seen to be (2Sin@)/(lambda) . Equating these expression for the length of vector OB ,

Or 

Which is the Bragg condition

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.

Solid State Physics ||Unit 01 || Crystal Diffraction & Reciorocal Lattice ||💥Interference Condition & The Reciprocal Lattice 💥 || mynotes || #physicsextreem

The condition for an Xray beam to be diffracted by a crystal may be expressed in an elegant form with the help of the reciprocal lattice transformation . We have seen that Xray diffraction is equivelent to reflection by the sets of parallel lattice planes in the crystal. Since in a crystal there are many sets of interpenetrating planes with various orientation and spacings . This requires the consideration of several sets of parallel planes which is very difficult . P.P Ewald developed a simple method for this purpose. We know that orientation or slope of a plane is determined by its normal as well. Further if the length of the normal is made proportional to 1/d(hkl), its length and direction uniquely describe the set of parallel planes. Now ti determine the various sets of parallel planes we might think in terms of such one dimensional normals instead of 2 dimesional plane. Now it can be shown that the terminal points of all such possible normals corresponding to all sets of parallel planes form a lattice array . Fig(1) makes this easy for a monoclinic crystal . Here only the unit cell of the crystal are looking along its unique axis ( taken prependicular to the planes of paper and designated as b ) and the four planes (100),(101),(102),(001) are shown in an edge view.

Fig (1):---

( each point (*) reoresents completely a parallel set of planes )

 Since alk these points are parallel to b ,their normals lie in the plane of paper . To locate the points we have priceeded as followes:--

(1) :-- The normal to each plane from a common origin is drawn.

(2):-- A point on the normal at a distance from the origin equal to 1/d(hkl) has been placed.

Indeed the collection of such points form a lattice array . This array is called the reciprocal lattice, because distance in this lattice arr reciprocal to these in tge crystal.

Let us now define a reciprocal lattice vector

. It is a vector whose magnitude is 1/d(hkl) and whose direction is parallel to the normal to the (hkl) planes.

The parallelopiped spanned by three non-coplanar reciprocal lattice vectors:-

is the unit cell in the reciprocal lattice.

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The Vector algebric analysis

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We shall now set up farmulae for finding the reciprocal lattice of a lattice algebrically.

Assume the primitive unit cell of volume V of the crystal lattice (fig2)

The volume V of the unit cell is equal to the area of the base (shaded), whose sides are b,c times the height of the cell, which is d(100)

Accordingly:-

An area is represented by the vector product of its sides so that it can be written as :--

here n is unit vector  in the direction of the normal to the plane (bc) . Further---

comparing eqs 1 and2, and expressing the volume in vector form ,we obtain:----

These three vectors are chosen as the reciprocal translational vectorsb, for defining the 3 dim reciprocal lattice translational vector as:----

Reciprocal lattice translation vectors represent a simple relationship to the crystal translational vectors:---

or in other words:---

relation 4  can be derived by forming the scalar product of both sides of 1st relation of 4 with b,c etc. Similarly eq7 can be derived by forming the scalar product of both sides of the first eq of 4 with a etc.

If a lattice is constructed using the reciprocal lattice vector (04), it follows that successive points in the direction represent successive submultiples h of the spacing of (100), in the b* direction , successive submultiples k of the spacing of (010) and in the c* direction , successive submultiples l of the spacing of (001)

That this is indeed so, is evident from 2.

Since:--

Therefore to reach  any reciprocal lattice point hkl one gives h units along a*, k units along b*, and l units along c*.

Accordingly the reciprocal lattice vector can be written in vector notation as :---

collection of points at the terminals of the set of vectors confirms the following conditions:--

1. The vector sigma(hkl) is normal to the crystal plane (hkl) 

2. The length of the vector sigma (hkl) is equal to / d(hkl)

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Proof of 1st property:--

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assume plane (hkl) intercepts a-axis at a/h, b-axis at b/k, c-axis at c/l.

(a/h-b/k), (-a/h+c/l), (b/k-c/l) are the vectors lying in the plane (hkl)

Consider the product as:---

Eq 10 is the. required proof.

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Proof of 2nd property

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In view of property 1st, n unit vector normal to plane (hkl) is parallel to sigma (hkl)

Thus:---

Or we can write as;---

eq 12 proves the 2nd  condition.

Thus the reciprocal lattice spanned by eq 4 agrees with that traced out by the geometrical method.

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Unit cell

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Volume of a unit cell of the reciprocal lattice is inversely proportional to the volume of a unit cell of the direct lattice.

Using eq 4, this becomes:--

 --------      ----------     -----------

Solid State Physics || Unit 01 || Crystal Diffraction & Reciprocal lattice ||The Von Laue Treatment ||Laue Diffraction equations 💥|| mynotes ||#physicsextreem

The Laue equations are derived from a simple static atomic model of a crystal structure. It describes effectively scattering of Xray from different atoms . The Bragg equation (2dSin@=n lambda) is derived as a direct consequence of the Laue equation. We consider the nature of the Xray diffraction pattern produced by identical scattering centre located at the lattice point of a space lattice.

We first look at the lattice points of a space lattice. Considering the scattering from any two lattice points P1,P2 in given figure, seperated by vector r.The unit incident wave normal is S0 and the unit scattered wave normal is S. We examine at a point 'a' long distance away ,the difference in phase of the radiation scattered by P1 and P2.

If P2A and P1B are the projections of vector r on the incident and scattered wave directions, the path difference between the two scattered wave is -----

where vector S=vector S0-vectorS

The vector S happens to represent the direction of the normal to a plane that reflects the incident direction into the scattering direction as shown in the following figure:---

If 2@ is the angle vector S makes with vector S0, then @ is the angle of incidence and from the figure we see that 

|S|=2.Sin@ as S and S0 are unit vectors.The phase fifference (phai) is equal to  (2.pai/lambda).path difference.

We have 

For this condition , the separate scattered amplitudes add up constructively and the intensity in the diffracted beam is maximum. If a,b,c are the primitive translation vectors , we have for the diffraction maxima:-------

Where h,k,l are integers . We have for direction cosine:---

This is Laue equation These eq have a direct geometrical interpretation .

The Laue equation state that in a diffraction direction the dcs are proportional to h/a, k/b,l/c.

Solid State Physics ||Unit 01||Crystal Diffraction & Reciprocal Lattice ||💥BRAGG'S LAW 💥||mynotes ||#physicsextreem

W.L Bragg found that position of the diffracted beams produced by a crystal can be determined by a simple model .This model assumes that xrays are reflected specularly from the various planes of atoms in the crystals. The diffracted beams are found only for special case in which the reflections from parallel planes of atoms  interfere constructively.

Xray is most useful for this purpose because its wavelength is nearly equal to (2d) {It is Bragg Condition}

We consider in the crystal a series of atomic planes which are considered to be partly reflecting for radiation of wavelength  (lambda) and which are spaced equal distance "d" apart. The radiation is incident in the plane of paper. The path difference for rays reflected from adjacent planes is 2d sin@ . 

Reinforcement of the radiation reflected from successive planes will occur when the path difference is an integral no "n" of wavelengths.

( refrence:---https://www.xtal.iqfr.csic.es/Cristalografia/parte_05_5-en.html )

This is Bragg's law.Here n being the order of diffrection .It is clear that there are only certain directions in which the reflections of a given wavelength from all parallel planes add up in phase to give a strong diffracted beam. We also reached the conclusion that a beam of monochromatic Xrays , incident on a crystal with an arbitrary angle @. 

(Ref :-- RL Singhal, C Kittel, Gupta-kumar)