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:--

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