Chapter 3 – Crystal Planes, X-ray Diffraction and Defects in Solids – Applied Physics

CHAPTER 3

Crystal Planes, X-ray Diffraction and Defects in Solids

3.1 Crystal planes, directions and Miller indices

Crystal planes are defined as some imaginary planes inside a crystal in which large concentration of atoms are present. Inside the crystal, there exists certain directions along which large concentration of atoms exists. These directions are called crystal directions. Figure 3.1 shows a two-dimensional lattice with different orientations of crystal planes.

 

Figure 3.1 A two-dimensional lattice with crystal planes

 

Crystal planes and directions can be represented by a set of three small integers called Miller indices [because Miller derived a method of representing crystal planes]. These integers are represented in general as h, k and l. If these integers are enclosed in round brackets as (hkl ), then it represents a plane. On the other hand, if they are enclosed in square brackets as [hkl], then it represents crystal direction perpendicular to the above-said plane. Next, we will see the way of obtaining Miller indices for a plane.

 

Figure 3.2 Miller indices for a plane ABC

 

  1. As shown in Fig. 3.2, take a lattice point as origin ‘0’ of crystallographic axes X, Y and Z in a space lattice. The unit cell translational distances or lattice constants along X, Y and Z directions are a, b and c, respectively. Let a crystal plane ABC intersect these three axes at 2a, 3b and c. In general, the intercepts can be represented as pa, qb, and rc.
  2. Divide these intercepts with lattice points translational distances along the axes to obtain intercepts of the plane in terms of multiples of unit cell translational lengths.

     

     

  3. Take the reciprocals of these multiples, they are in general

     

  4. Clear these fractions [by multiplying with LCM] to smallest integers having the same ratio as the fractions, enclose these integers in brackets.

     

     

Figure 3.3 Represent some important crystal planes and directions in a cubic crystal

 

 

Miller indices of the plane ABC is (326). In general, indices of a plane are represented as (hkl ) = (qr pr pq) or

 

Miller indices may be defined as a set of three small integers obtained by clearing the reciprocals of the three intercepts [in terms of multiples of unit cell edges] made by a plane on crystallographic axes.

Now, we will see the important features of Miller indices:

  1. Miller indices represent a set of equidistant parallel planes.
  2. If the Miller indices of a plane represent some multiples of Miller indices of another plane, then these planes are parallel. For example (844) and (422) or (211) are parallel planes.
  3. If (hkl ) are the Miller indices of a plane, then the plane divides the lattice constant ‘a’ along X-axis into h equal parts, ‘b ’ along Y-axis into k equal parts and ‘c ’ along Z-axis into l equal parts.
  4. If a plane is parallel to one of the crystallographic axes, then the plane intersects that axis at infinity and the Miller indices along that direction is zero.
  5. If a plane cuts an axis on the negative side of the origin, then the corresponding index is negative and is indicated by placing a minus sign above the index. For example, if the plane cuts on negative Y-axis, then Miller indices of the plane is (hl).
  6. When Miller indices are enclosed in curly brackets, {hkl}, they refer to planes which in the crystal are equivalent even though their Miller indices may differ. For example in a cubic lattice, all cube faces are equivalent, they are (100), (010), (001), (00), (00), (00); these planes are represented as {100}.Similarly, a full set of equivalent directions in a crystal is represented by a symbol <hkl>. For example, the eight body diagonals of a cube [111], [], [11], [11], [11], [1], [1], [1] are designated as <111>.

3.2 Distance of separation between successive hkl planes

The separation between successive parallel planes in rectangular axes crystal system can be extracted easily. Let us consider a rectangular [cartesian] coordinate system with origin ‘0’at one of the lattice points. Let (hkl) be the Miller indices of a plane ABC, which makes intercepts OA, OB and OC on X, Y and Z axes, respectively as shown Fig 3.4. A normal to this plane from the origin passes through a point N in the plane ABC, such that ON = d1. This normal makes α′, β′, and γ′ angles with X, Y and Z-axes, respectively. Since the plane segments ‘a’ into ‘h’ equal parts, b into k equal parts and c into l equal parts, then the intercepts OA, OB and OC are such that:

 

 

where a, b, c are the unit cell edge lengths along X, Y and Z-axes, respectively.

From Fig. 3.4

 

 

Let the coordinates of N be x, y and z along X, Y and Z axes, then:

 

Figure 3.4 Orthorhombic crystal

 

 

Also from Fig. 3.4:

 

 

Substitute Equation (3.4) in (3.3) gives:

 

 

Substitute Equation (3.2) in (3.5) gives:

 

 

Again substitute Equation (3.1) in (3.6)

 

 

Let be the Miller indices of the next plane A′ B′ C′, this plane makes intercepts OA′, OB′ and OC′ on X, Y and Z axes, respectively. A normal from the origin to this plane passes through a point N′, so that ON′ = d2. As the extension of d1 is d2, it makes same angles α′, β′ and γ′ with X, Y and Z-axes, respectively. Since the plane segments ‘a’ into h/2 equal parts, b into k/2 equal parts and c into l/2 equal parts, then the intercepts OA′, OB′ and OC′ are such that:

 

 

From Fig. 3.4,

 

 

Let the coordinates of N′ are x′, y′ and z′ along X, Y and Z-axes, respectively.

 

 

Also from Fig 3.4:

 

 

Substitute Equation (3.4′) in (3.3′) gives:

 

 

Substitute Equation (3.2′) in (3.5′) gives:

 

 

Again substitute Equation (3.1′) in (3.6′) gives:

 

 

Let the separation between the planes ABC and A′B′C′ is ‘d’.

 

 

Using Equation (3.8), we can determine the interplanar separation in orthorhombic crystals. For tetragonal crystal a = bc, substitute these values in Equation (3.8), we have:

 

 

For cubic crystals: a = b = c, substitute these values in Equation (3.8), we have:

 

 

The calculation of interplanar spacing for other crystal systems is complicated, so we will not discuss them.

3.3 Imperfections in crystals

In a sound crystal (or in an ideal crystal), the atoms are arranged regularly and periodically in three dimensions. But the grown crystals [or real crystals] may contain imperfections or defects. These defects are mainly divided into point, line, surface and volume defects. Point and line defects are discussed here, while surface defects are dealt with in Appendix B.

(1) Point defects: As the name indicates, these defects are at some points in the crystal. So, these are also called zero-dimensional defects. The point defects are divided into three categories: (a) lattice site defects; this includes vacancies [Schottky defect] and interstitialcies [Frenkel defect], (b) compositional defects; this includes substitutional impurity and interstitial impurity and (c) electronic defects. These defects are discussed below.

(a) Lattice site defects: In this type of defects, some atoms may not be present in their regular atomic sites. They are:

(i) Vacancies: As shown in Fig. 3.5, at a lattice point, one or two or three atoms are missed, and this is referred to as single or double or triple vacancies, respectively. The vacancies are formed due to the imperfect packing during crystallization or due to thermal vibrations at high temperatures.

 

Figure 3.5 (a) Perfect crystal; (b) Vacancy defect

 

Schottky defect: In ionic crystals, if a cation vacancy exists, then in the very nearby place an anion vacancy also exists. i.e., usually an anion and cation pair is moved to the surface of the crystal, so that charge neutrality is maintained in the vacancy region as shown in Fig. 3.6. This is known as Schottky defect. Crystals such as NaCl, KCl, KBr, etc. show Schottky defect.

 

Figure 3.6 Schottky defect

 

(ii) Interstitial defect: If an atom is moved to an interstitial space in the crystal, then the defect is known as interstitial defect.

 

Figure 3.7 Frenkel defect

 

Frenkel defect: In ionic crystals, if a cation [positive ion] moves to an interstitial space, then a vacancy is formed in its atomic position. Here, charge neutrality is maintained in the defective region as shown in Fig. 3.7. This type of defect is known as Frenkel defect. Crystals such as CaF2, AgBr, AgI, etc. show Frenkel defect.

(b) Compositional defect: The presence of impurity atoms in the crystal leads to compositional defects. Impurity atoms are present at the sites of regular parent atoms or in the interstitial spaces. These defects are described below.

(i) Substitutional defect: As shown in Fig. 3.8(a), during crystallization few foreign atoms occupy the regular parental atoms sites. For example, in extrinsic semiconductors either third or fifth group atoms occupy the sites of silicon or germanium atoms.

 

Figure 3.8 (a) Substitutional defect; (b) Interstitial defect

 

(ii) Interstitial impurity defect: The spaces between the parental atoms in a crystal are known as interstitial spaces. Small-sized [lower atomic number] atoms, such as hydrogen, etc. may fit into these interstitial spaces. These atoms are known as interstitial atoms and the defect formed due to the presence of interstitial atoms is known as interstitial defect. This is shown in Fig. 3.8(b). If ‘r’ is the radius of a parent atom, then a octahedral and a tetrahedral space can accommodate an interstitial atom of radius 0.414r and 0.225r, respectively.

(c) Electronic defects: Non-uniformity of charge or energy distribution in the crystal is referred to as electronic defect. The presence of impurity atoms such as substitutional and interstitial atoms and vacancies can vary the uniform distribution of electronic charge in the crystal. So, the presence of these defects also leads to electronic defects. In semiconductors, temperature variation changes charge concentration, so the variation of temperature [i.e., thermal energy] leads to electronic defects.

Point defects are formed by thermal fluctuations, by severe deformation [i.e., by hammering or rolling] and by bombarding with high energetic particles.

(2) Line defects: If a crystal plane ends somewhere in the crystal, then along The edge of that incomplete plane produces defect in the crystal called line defect. The line defect is of two types: they are (i) edge dislocation and (ii) screw dislocation. These are described below.

(i) Edge dislocation: Figure 3.9(a) shows three-dimensional view and front face of a perfect crystal. The vertical crystal planes are parallel to side faces of a crystal is shown in the figure. One of the crystal planes does not pass from top to bottom face of the crystal, but ends some where in the crystal as shown in Fig. 3.9(b). In this crystal, just above The edge of incomplete plane, the atoms are in a state of compression so that the bond distances are less than normal values and below the edge of incomplete plane, the atoms are far apart, so the bond distances are larger than normal values. This situation extends all along the edge of this incomplete plane producing edge dislocation. The extra plane indicated in Fig. 3.9(b) can be either above or below the slip plane shown as dotted line X, Y in Fig. 3.9(c). If the incomplete extra plane is above the slip plane, then the edge dislocation is positive and is represented by the symbol on the other hand if it is below the slip plane, then the edge dislocation is negative and is represented by the symbol . If one plane of atoms glides over another separated by an integral multiple of interatomic distance is called slip, and the slip plane is the plane in which slip has taken place. Thus, the crystal consists of slipped and normal regions.

 

Figure 3.9 (a) Three-dimensional view of perfect crystal; Front view of perfect crystal; (b) Three-dimensional view of edge dislocation crystal; Front view of edge dislocation crystal; (c) Positive and negative edge dislocations

 

 

The magnitude and direction of the displacement of crystal planes due to edge dislocation can be represented by a vector called Burger's vector, which is perpendicular to the dislocation line. This indicates how much and in what direction the lattice above the slip plane is shifted with respect to the lattice below the slip plane. Figure 3.9(c) shows a method of determining Burger's vector for edge dislocation. To find the magnitude and direction of Burger's vector, one starts arbitrarily from a lattice point A, drawing atom-to-atom vectors round the dislocation in clockwise direction to form a closed circuit. Here, the number of vectors in horizontal direction at the top and bottom and vertical vectors at the left and right are equal, but the circuit is not closed unless we put the vector , as shown in the circuit. This is the Burger's vector for the above-said edge dislocation.

(ii) Screw dislocation: The crystal planes spiral about a line in the crystal, called dislocation line. The screw dislocation is shown in Fig. 3.10. Due to the spiralling of crystal planes, the atoms at one end of the plane are displaced by one atomic distance with respect to the other end of the plane in perpendicular direction to the plane. As shown in Fig. 3.10, the plane ABCD is the slipped area. The upper portion of the crystal has been sheared by one atomic distance compared to the right side region of the crystal. Slip has not taken place to the right side of AD, so AD is the dislocation line. Burger's circuit is completed around the dislocation. The Burger's vector is parallel to the dislocation line. By knowing the Burger's vector and dislocation line, the dislocation is completely described.

 

Figure 3.10 Screw dislocation and Burger's vector

3.4 Energy for the formation of a vacancy and number of vacancies — at equilibrium concentration

Energy supply to a crystal moves some of the atoms present at regular atomic sites in the interior of the crystal to the surface, so that vacancies are formed inside the crystal. If we supply energy to an ionic crystal, then either cation–anion pairs are moved to the surface [Schottky defect] or cations are moved to interstitial spaces [Frenkel defect], so that vacancies are formed inside the ionic crystal. We shall find the relation between number of vacancies and energy of formation of a vacancy in all the above cases.

(i) In metallic crystals: Let a crystal contains N number of atoms. The energy required to move an atom at a regular atomic site in the interior of the crystal to the surface is Ev i.e., the energy required to create a vacancy. To create ‘n’ number of isolated vacant sites, The energy required is nEv. At some thermal equilibrium temperature ‘T’, let ‘n’ number of vacancies present in the crystal. The number of ways these ‘n’ vacancies are created is given by (P).

 

 

The vacancies created inside the crystal produces disorder in the crystal. The disorder can be measured in terms of entropy. The increase in entropy (S) due to the increase of vacancies is:

 

 

where KB is Boltzmann constant, substituting Equation (3.11) in (3.2), we have:

 

 

The creation of vacancies produces not only the change in entropy but also change in free energy (F) of the crystal.

 

 

where U = nEv = internal energy of crystal at temperature T K. Equations (3.12) and (3.14) are taken from thermodynamics. Substituting Equation (3.13) in (3.14) gives:

 

 

The logarithmic term in the above equation can be simplified using Stirling's approximation,

 

 

Equation (3.15) becomes:

 

 

In thermal equilibrium at constant volume, the free energy is minimum with respect to changes in ‘n’.

 

 

Taking exponential on both sides of Equation (3.17), we have:

 

 

 

The above equation indicates that by decreasing equilibrium temperature, the concentration of vacancies decreases.

(b) In Ionic crystals: Here, we see Schottky and Frenkel defects separately.

(i) Schottky defect: In ionic crystals, equal number of cations [positive ions] and anions [negative ions] vacancies are formed i.e., usually cation–anion-paired vacancies are formed, so that charge neutrality is maintained in the crystal. The energy required to move a cation and an anion from interior of the crystal to the surface is EP. At some thermal equilibrium temperature (T), let ‘n’ pairs of cation–anion vacancies present in a crystal containg ‘N ’ pairs of ions. The number of ways these n-pairs of vacancies are created is given by (P).

 

 

The vacancies created inside the crystal produces a disorder in the crystal. The disorder can be measured in terms of entropy. The increase in entropy (S), due to the creation of n pairs of vacancies is

 

 

where KB is Boltzmann constant, substituting Equation (3.19) in (3.20) we have:

 

 

The vacancies produce not only change in entropy but also change in free energy (F) of the crystal.

 

 

where U = n EP = internal energy of the crystal at temperature T. Equations (3.20) and (3.22) are taken from thermodynamics; substituting Equation (3.21) in (3.22) gives:

 

 

 

The logarithmic term in the above equation can be simplified using Stirling's approximation:

 

 

... Equation (3.23) becomes:

 

 

In thermal equilibrium at constant volume, the free energy is minimum with repect to the changes in ‘n’.

 

 

Taking exponential on both sides of Equation (3.25), we get:

 

 

 

(ii) Frenkel defects: Let the ionic crystal contains N number of atoms and the number of interstitial spaces are slightly less than the number of atoms. Let Ni be the number of interstitial spaces in a perfect crystal. The amount of energy required to displace an atom from regular atomic site to an interstitial position is Ei. At some thermal equilibrium temperature, let there be ‘n’ number of cation site vacancies and same number of interstitial atoms. The number of ways the ‘n ’ Frenkel defects can be formed is:

 

 

The increase in entropy (S) due to the creation of Frenkel defects is given by:

 

 

Substituting Equation (3.27) in (3.28), we get:

 

 

These defects produce not only change in entropy but also change in free energy (F) given by:

 

 

Equations (3.28) and (3.30) are taken from thermodynamics. Substituting Equation (3.29) in (3.30), we have:

 

 

The logarithmic term in the above equation can be simplified by applying Stirling's approximation

 

 

 

Substituting Equation (3.32) in (3.31), we have:

 

 

 

At thermal equilibrium, the change in free energy is minimum w.r.t ‘n’, so we have:

 

 

 

Taking exponential on both sides, we get:

 

 

The above equation shows that n is proportional to (NNi)1/2

3.5 Diffraction of X-rays by crystal planes and Bragg's law

The visible light rays when pass through a sharp edge of an object can form some bright regions inside the geometrical shadow of the object. This is due to the bending nature of light, called diffraction. Diffraction of visible light rays can also be produced using plane-ruled grating. This grating consists of about 6000 lines/cm; so that the spacing between any two consecutive lines in the grating is of the order of the wavelength of visible light used to produce diffraction. The wavelength of X-rays is of the order of an angstrom, so X-rays are unable to produce diffraction with plane optical grating. To produce diffraction with X-rays, the spacing between the consecutive lines of grating should be of the order of few angstroms. Practically, it is not possible to construct such a grating. In the year 1912, a German physicist Laue suggested that the three-dimensional arrangement of atoms in a crystal can serve as a three-dimensional grating. Inside the crystal, the spacing between the crystal planes can work as the transparent regions as between lines in a ruled grating. Laue's associates Friedrich and Knipping succeeded in diffracting X-rays by passing through a thin crystal.

In 1913, W.L. Bragg and his son W.H. Bragg gave a simple interpretation of the diffraction pattern. According to Bragg, the diffraction spots produced are due to the reflection of some of the incident X-rays by various sets of parallel crystal planes. These planes are called Bragg's planes. The Bragg's interpretation is explained in the following topic.

Bragg's law: W.L. Bragg and W.H. Bragg considered the X-ray diffraction as the process of reflection of X-rays by crystal planes as shown in Fig. 3.11. A monochromatic X-ray beam of wavelength λ is incident at an angle θ to a family of Bragg planes. Let the interplanar spacing of crystal planes is ‘d’. The dots in the planes represent positions of atoms in the crystal. Every atom in the crystal is a source of scatterer of X-rays incident on it. A part of the incident X-ray beam AB, incident on an atom at B in plane l, is scattered along the direction BC. Similarly, a part of incident X-ray DE [in parallel to AB] falls on atom at E in plane 2 and is scattered in the direction EF and it is parallel to BC. Let the beams AB and DE make an angle θ with the Bragg's planes. This angle θ is called the angle of diffraction or glancing angle.

 

Figure 3.11 Bragg's law

 

If the path difference between the rays ABC and DEF is equal to λ, 2λ, 3λ…etc. or , i.e., integral multiples of wavelength, where n = 1, 2, 3,…etc. are called first-order, second-order, third-order…etc. maxima, respectively. As path difference is equal to , then the rays reflected from consecutive planes are in phase; so, constructive interference takes place among the reflected rays BC and EF, hence the resulting diffracted ray is intense. On the other hand, if the path difference between the rays ABC and DEF is …etc., then the scattered rays BC and EF are out of phase so that destructive interference takes place and hence the resulting ray intensity is minimum. To find the path difference between these rays, drop perpendiculars from B on DE and EF. The intersecting points of perpendiculars are P and Q as shown in Fig. 3.11. The path difference between the rays is PE + QE. From the figure, we know that BE is perpendicular to plane 1 and BP is perpendicular to AB. So, as the angle between ray AB and plane 1 is θ, then PBE = QBE = θ. In the triangle PBE, or PE = d sin θ. Similarly, EQ = d sin θ.

... For constructive interference, PE + EQ = or d sin θ + d sin θ =

i.e., 2d sin θ = nλ

The above equation is called Bragg's law.

3.6 Powder method

X-ray powder method is usually carried for polycrystalline materials. The powder photograph is obtained in the following way. The given polycrystalline material is ground to fine powder and this powder can be taken either in a capillary tube made up of non-diffracting material or is just struck on a hair with small quantity of binding material and fixed at the centre of cylindrical Debye-Scherrer camera as shown in Fig. 3.12(a).

 

Figure 3.12 (a) Debye-Scherrer cylindrical camera; (b) Film mounted in camera; (c) Film on stretchout

 

A stripe of X-ray photographic film is arranged along the inner periphery of the camera. A beam of monochromatic X-rays is passed through the collimator to obtain a narrow fine beam of X-rays. This beam falls on the polycrystalline specimen and gets diffracted. The specimen contains very large number of small crystallites oriented in random directions. So, all possible diffraction planes will be available for Bragg reflection to take place. Such reflections will take place from many sets of parallel planes lying at different angles to the incident X-ray beam. Also, each set of planes gives not only first-order reflections but also of higher orders as well. Since all orientations are equally likely, the reflected rays will form a cone whose axis lies along the direction of the incident beam and whose semi-vertical angle is equal to twice the glancing angle (θ), for that particular set of planes. For each set of planes and for each order, there will be such a cone of reflected X-rays. There intersections with a photographic film sets with its plane normal to the incident beam, form a series of concentric circular rings. In this case, a part of the reflected cone is recorded on the film and it is a pair of arcs, the resulting pattern is shown in Fig. 3.12(c). Diameter of these rings or corresponding arcs is recorded on the film, and using this the glancing angle and interplanar spacing of the crystalline substance can be determined. Figure 3.12(b) shows the film mounted in the camera and the X-ray powder pattern obtained. The film on spread-out is shown in Fig 3.12(c). The distance between any two corresponding arcs on the film is indicated by the symbol S.

In case of cylindrical camera, the diffraction angle θ is proportional to S. Then,

      where R represents the radius of the camera.

If S1, S2, S3…etc. are the distances between symmetrical lines on the stretched film, then,

 

 

Using these values of θn in Bragg's equation = 2 dhkl sin θn

where    n = 1, 2, 3,…etc = order of diffraction

            dhkl = interplanar spacing

            θn = angle of diffraction for nth order

The interplanar spacing dhkl can be calculated.

3.7 Laue method

In Laue method, a narrow beam of white X-rays [usually in the wavelength range, 0.2 to 2.0 Å] is obtained by passing X-rays through a collimator ‘C’. This beam is allowed to fall on a stationary single crystal ‘S’ as shown in Fig. 3.13(a). The crystal act as a 3-dimensional diffraction grating to the incident beam. The processes of reflection of X-rays by crystal planes is considered as X-ray diffraction. The diffraction phenomenon satisfies Bragg's law, = 2d sin θ. where n = 1, 2, 3,…represent the order of diffraction, λ = wavelength of diffracted X-rays from a system of crystal planes with interplanar spacing ‘d’ and θ = glancing angle i.e., the angle made by X-rays with a crystal plane. As the crystal is not rotated, so, the angle ‘θ’ is fixed for a set of planes having separation ‘d’. Different sets of crystal planes satisfy Bragg's law with different wavelengths of X-rays and produce diffraction. The diffracted X-rays from a set of planes produce constructive interference, if they are in phase and form an intense beam, and this produces dark spots on photographic film. If the diffracted rays are out of phase, they produce destructive interference so that photographic film is unaffected.

 

Figure 3.13 (a) X-ray diffraction by crystal plane; (b) Lane pattern for NaCl crystal

Laue photograph is obtained either by allowing the transmitted diffracted rays or by back-reflected diffracted rays on photographic film as shown in Fig. 3.13(b).

As we observe the diffracted film, the diffracted spots lie on certain curves. These curves are either ellipses or hyperbolas on transmission Laue photograph and hyperbolas on back-reflection Laue photograph. The way of arrangement of spots on a film is a characteristic property of the crystal. Laue method is useful to decide the crystal symmetry and orientation of the internal arrangement of atoms/molecules in the crystal. Cell parameters of a crystal cannot be determined using Laue method. For transmission Laue method, the crystal should be thin.

Laue method can be used to study imperfections or strains in the crystal. The presence of above defects forms streaks instead of spots in the Laue photograph.

Formulae

Solved Problems

1. A beam of X-rays of wavelength 0.071 nm is diffracted by (110) plane of rock salt with lattice constant of 0.28 nm. Find the glancing angle for the second-order diffraction.

 

(Set-1–Sept. 2007), (Set-2, Set-3–Sept. 2006), (Set-2–May 2006), (Set-3–May 2004), (Set-4–May 2003)

Sol: Given data are:

Wavelength (λ) of X-rays = 0.071 nm

Lattice constant (a) = 0.28 nm

Plane (hkl) = (110)

Order of diffraction = 2

Glancing angle θ = ?

Bragg's law is 2d sinθ = nλ

because rock salt is FCC

 

 

Substitute in Bragg's equation

 

 

 

 

2. A beam of X-rays is incident on a NaCl crystal with lattice plane spacing 0.282 nm. Calculate the wavelength of X-rays if the first-order Bragg reflection takes place at a glancing angle of 8°35′. Also calculate the maximum order of diffraction possible.

 

(Set-4–Sept. 2007), (Set-3–May 2007), (Set-2–May 2004), (Set-3–May 2003)

Sol: Given data are:

NaCl crystal is FCC

Lattice plane spacing (d) = 0.282 nm

Wavelength of rays (λ) = ?

Order of diffraction (n) = 1

Glancing angle θ = 8°35′

Bragg's equation is = 2d sin θ

          1λ = 2 × 0.282 × 10–9 sin (8°35′)

          = 0.0842 nm

Maximum order of diffraction (nmax ) = ?

2d sinθ = nλ

if θ = 90° then n = nmax

... 2d = nmax λ

 

 

3. The fraction of vacant sites in a metal is 1 × 10−10 at 500 ° C. What will be the fraction of vacancy sites at 1000 ° C?

 

(Set-4–Sept. 2006), (Set-1–May 2004), (Set-2–May 2003)

Sol: The number of vacancies at temperature (TK) in a metal is represented by:

 

 

The given data are:

 

 

 

 

Taking logarithms on both sides of the above Equations (1) and (2), we get:

 

 

 

Dividing Equation (4) by (3),

 

 

Take exponential on both sides,

 

 

The fraction of vacancy sites at 1000°C is 8.466 × 10–7.

4. Calculate the ratios d100:d110:d111 for a simple cubic structure.

 

(Set-2–Nov. 2004), (Set-2–Nov. 2003)

Sol: Let ‘a’ be the lattice constant of cubic structure, then,

 

 

 

5. The Bragg's angle in the first order for (220) reflection from nickel (FCC) is 38.2°. When X-rays of wavelength 1.54 Å are employed in a diffraction experiment. Determine the lattice parameter of nickel.

 

(Set-2–May 2008)

Sol: Order of diffraction, n = 1

Diffraction angle, θ = 38.2°

Wavelength of light, λ = 1.54 Å

Plane of reflection = (220)

Lattice parameter, a = ?

Bragg's law is 2d sin θ =

 

 

 

6. Monochromatic X-rays of λ = 1.5 A.U are incident on a crystal face having an interplanar spacing of 1.6 A.U. Find the highest order for which Bragg's reflection maximum can be seen.

 

(Set-4–May 2006)

Sol: Given data are

Wavelength of light (λ) = 1.5 Å

Interplanar spacing (d) = 1.6 Å

Glancing angle (θm) = 90°

Order of diffraction (n) = ?

Bragg's law

= 2d sin θ

 

 

... The maximum order of diffraction is 2

7. The distance between (110) planes in a body centred cubic structure is 0.203 nm. What is the size of the unit cell? What is the radius of the atom?

 

(Set-3–Sept. 2007), (Set-3–May 2006)

Sol: The given data are

The distance between (110) planes of BCC structure (d110) = 0.203 nm = 0.203 × 10–9 m

Length of unit cell (a) = ?

Volume of unit cell (a3) = ?

Radius of the atom (r) = ?

 

 

Volume of unit cell a3 = 0.02364 × 10–27 m3

 

 

8. Monochromatic X-rays of λ = 1.5 A.U. are incident on a crystal face having an interplaner spacing of 1.6 A.U. Find the highest order for which Bragg's reflection maximum can be seen.

 

(Set-1–Sept. 2006)

Sol: Given data are wavelength of X-rays, λ = 1.5 Å

Interplanar spacing, d = 1.6 Å

For highest order of diffraction, θ = 90°

Highest order of diffraction, n = ?

Formula    2d sin θ = n λ

 

 

... Highest order of diffraction is 2.

9. Calculate the glancing angle at (110) plane of a cubic crystal having axial length 0.26 nm corresponding to the second order diffraction maximum for the X-rays of wavelength 0.65 nm.

 

(Set-1–May 2007)

Sol: The given data are

Edge length of cubic system, a = 0.26 nm

Wavelength of X-rays λ = 0.065 nm

Glancing angle, for plane (110), θ = ?

Order of diffraction, n = 2

Separation between (110) planes of a cube,

 

 

Bragg's law

 

 

10. The Bragg's angle for reflection from the (111) plane in a FCC crystal is 19.2° for an X-ray wavelength of 1.54 A.U. Compute the cube edge of the unit cell.

 

(Set-2, Set-4–May 2007)

Sol: The given data are

Bragg's angle, θ = 19.2°

Wavelength of X-rays, λ = 1.54 Å

Order of diffraction, n = 1

Cube edge, a = ?

Bragg's law

2d sinθ =

2d sin 19.2° = 1 × 1.54

 

 

 

11. The Bragg's angle in the first order for (220) reflection from nickel (FCC) is 38.2°. When X-rays of wavelength 1.54 Å are employed in a diffraction experiment. Determine the lattice parameter of nickel.

 

(Set-2–May 2008)

Sol: Order of diffraction, n = 1

Diffraction angle, θ = 38.2°

Wavelength of light, λ = 1.54 Å

Plane of reflection = (220)

Lattice parameter, a = ?

Bragg's law is 2d sinθ =

 

 

 

12. Copper has FCC structure with lattice constant 0.36 nm. Calculate the interplanar spacing for (111) and (321) planes.

Sol: Given data is:

lattice constant (a) = 0.36 nm = 0.36 × 10–9 m

Interplanar spacing (d) for (111) plane is:

 

 

Interplanar spacing for (321) plane

 

 

13. The first-order diffraction occurs when a X-ray beam of wavelength 0.675 Å incident at a glancing angle of 5 °25′ on a crystal. What is the glancing angle for third-order diffraction to occur?

Sol: Wavelength of X-rays (λ) = 0.675 Å

Glancing angle for first order (n = 1) diffraction (θ1) = 5°25′

Find the glancing angle for third order (n = 3) diffraction (θ3) = ?

Bragg's equation is 2d sin θ = nλ

For first order, 2d sin θ1 = 1λ

2d sin 5°25′ = 0.675 × 10–10 m

 

 

For third-order diffraction,

2d sin θ3= 3λ

 

 

14. What is the angle at which the third-order reflection of X-rays of 0.79 Å wavelength can occur in a calcite crystal of 3.04 × 10–8 cm spacing.

Sol: Wavelength of X-rays, λ = 0.79 Å = 0.79 × 10–8 cm

Interplanar spacing, d = 3.04 × 10–8 cm

Order of diffraction, n = 3

Angle of diffraction, θ = ?

2d sinθ =

 

Multiple Choice Questions

  1. Crystal directions are defined as: ( )
    1. certain directions inside the crystal along which large concentration of atoms exists
    2. certain directions inside the crystal along which low concentration of atoms exists
    3. certain directions inside the crystal along which no atoms are present
    4. none
  2. Crystal planes and directions can be represented by a set of __________ small integers. ( )
    1. 2
    2. 3
    3. 4
    4. 6
  3. To represent crystal direction, the Miller indices should be enclosed in ( )
    1. square brackets
    2. round brackets
    3. curly brackets
    4. none
  4. If the Miller indices of two planes are (211) and (422), then they are: ( )
    1. parallel
    2. perpendicular
    3. they are at an angle of 45°
    4. none
  5. If the Miller indices of a plane along Y and Z-direction is zero, then: ( )
    1. the plane is perpendicular to X-axis
    2. the plane is parallel to Y-axis
    3. the plane is parallel to X-axis
    4. the plane is parallel to Z-axis
  6. If the Miller indices of a plane is (hl), then the plane: ( )
    1. intersects negative X-axis
    2. intersects negative Z-axis
    3. intersects negative Y-axis
    4. intersects positive Y-axis
  7. If {hkl} are the Miller indices in cubic system, they represent: ( )
    1. (100) and (00) planes
    2. (010) and (00) planes
    3. (001) and (00) planes
    4. all
  8. The Miller indices <hkl> in cubic system represent the following directions: ( )
    1. [11], [11], [11]
    2. [1], [1], [1]
    3. [111], []
    4. all
  9. If (hkl ) represents the Miller indices of planes in cubic crystal of lattice constant ‘a’, the separation between the parallel planes is: ( )
  10. Crystal defects are: ( )
    1. point and line defects
    2. surface defects
    3. volume defects
    4. all
  11. Point defects are: ( )
    1. lattice site defects
    2. compositional defects
    3. electronic defects
    4. all
  12. Electrical charge neutrality is maintained in: ( )
    1. Schottky defect
    2. Frenkel defect
    3. both a and b
    4. none
  13. Schotty defect may exist in: ( )
    1. NaCl crystal
    2. KCl crystal
    3. KBr crystal
    4. all
  14. Substitutional defect and interstitial impurity defect belong to: ( )
    1. compositional defect
    2. Schottky defect
    3. Frenkel defect
    4. lattice site defects
  15. Non-uniformity of charge or energy distribution in the crystal is referred to as: ( )
    1. point defect
    2. electronic defect
    3. Schottky defect
    4. Frenkel defect
  16. Point defects in crystals are formed by: ( )
    1. thermal fluctuations
    2. Large deformation
    3. bombarding with high energetic particles
    4. all
  17. Edge dislocation and screw dislocation belong to: ( )
    1. electronic defects
    2. compositional defects
    3. line defects
    4. point defects
  18. Just above the edge of an incomplete crystal plane in a crystal, the bond distances are __________. ( )
    1. equal to normal values
    2. lesser than normal values
    3. greater than normal values
    4. none
  19. If the incomplete plane is below the slip plane, then the edge dislocation is: ( )
    1. positive
    2. negative
    3. both a and b
    4. none
  20. In edge dislocation, the Burger's vector is __________ to the dislocation line. ( )
    1. parallel
    2. at an angle of 45°
    3. perpendicular
    4. at an angle of 60°
  21. If Ev is the energy required to form a vacancy in the crystal containing ‘N’ atoms at temperature ‘T’, the number of vacancies in the crystal is [KB = Boltzmann constant] ( )
  22. If Ep is the energy required to move an anion–cation pair from interior to the surface of an ionic crystal containing N pairs of ions, the formation of n pairs of vacancies at temperature ‘T’ is given by [KB = Boltzmann constant] ( )
  23. Let Ei is the energy required to move a cation [positive ion] to the interstitial space of an ionic crystal containing ‘N’ number of atoms and Ni be the number of interstitial spaces. The number of ways n cations are moved to interstitial spaces, at temperature T is given by [KB = Boltzmann constant] ( )
  24. If a monochromatic X-ray of wavelength ‘λ’ incident at an angle ‘θ’ on a parallel set of crystal planes of separation ‘d’, then the Bragg's law for constructive interference is [n = 1, 2, 3,…= order of diffraction] ( )
    1. 2d sin θ =
    2. d sin θ =
    3. 2λ sin θ = nd
    4. λ sin θ = nd
  25. Miller indices of the plane parallel to X and Y axes are: ( )
    1. (001)
    2. (010)
    3. (100)
    4. (111)
  26. The crystal planes are defined as some imaginary planes inside a crystal in which __________ of atoms are present. ( )
    1. large concentration
    2. low concentration
    3. medium concentration
    4. none
  27. Crystal planes and directions can be represented by a set of three small integers called: ( )
    1. plane indices
    2. Miller indices
    3. direction indices
    4. none
  28. If the Miller indices are enclosed in round brackets, then it represents a crystal ( )
    1. plane
    2. direction
    3. set of directions
    4. system of planes
  29. Miller indices may be defined as a set of three integers obtained by clearing the reciprocals of the __________ made by a plane on crystallographic axes. ( )
    1. intercepts
    2. relations
    3. both a and b
    4. none
  30. Miller indices represent a set of equidistant __________ planes. ( )
    1. perpendicular
    2. intersecting
    3. parallel
    4. none
  31. If (hkl) is the Miller indices of a plane, then the plane divides the lattice constant ‘a’ along ‘X’ axis into ______. ( )
    1. h equal parts
    2. k equal parts
    3. l equal parts
    4. all
  32. Point defects in crystals are also called as __________ defects. ( )
    1. three-dimensional
    2. two-dimensional
    3. one-dimensional
    4. zero-dimensional
  33. By moving an anion and a cation from interior of an ionic crystal to the surface of the crystal leads to __________ defect. ( )
    1. Frenkel
    2. Schottky
    3. both a and b
    4. none
  34. Vacancies and interstitial defects belong to __________ defects. ( )
    1. lattice site
    2. Schottky
    3. Frenkel
    4. none
  35. The crystal defect formed by moving a cation to interstitial spaces in an ionic crystal is known as _______ defect. ( )
    1. Schottky
    2. point defect
    3. Frenkel
    4. none
  36. Examples for Frenkel defect ( )
    1. CaF2
    2. AgBr
    3. AgI
    4. all
  37. Presence of impure atoms in the crystal leads to __________ defects. ( )
    1. Schottky
    2. Frenkel
    3. compositional
    4. none
  38. If ‘r’ is the radius of a parent atom of a crystal, then octahedral and tetrahedral spaces can accommodate an interstitial atom of radius __________ and __________, respectively. ( )
    1. 0.414r, 0.225r
    2. 0.225r, 0.414r
    3. 0.0225r, 0.0414r
    4. 0.0414r, 0.0225r
  39. Extrinsic semiconductors contain __________ crystal defect. ( )
    1. interstitial
    2. substitutional
    3. Frenkel
    4. Schottky
  40. Just below the edge of an incomplete crystal plane in a crystal, the bond distances are ______ normal values. ( )
    1. same as
    2. more than
    3. less than
    4. none
  41. If a crystal plane ends some where inside the crystal, then the defect along the edge of the incomplete plane is called __________. ( )
    1. edge dislocation
    2. screw dislocation
    3. Schottky defect
    4. interstitial defect
  42. If the incomplete plane is above the slip plane in the crystal, then the edge dislocation is: ( )
    1. Schottky
    2. Frenkel
    3. negative
    4. positive
  43. The magnitude and direction of the displacement of crystal planes due to edge dislocation can be represented by a vector called: ( )
    1. Burger's vector
    2. Laue's vector
    3. both a and b
    4. none
  44. In screw dislocation, the atoms at one end of a plane are displaced by __________ distance with respect to the other end of the plane, perpendicular to plane. ( )
    1. 3 atomic
    2. 2 atomic
    3. 1 atomic
    4. none
  45. In screw dislocation, Burger's vector is __________ to dislocation line. ( )
    1. perpendicular
    2. parallel
    3. both a and b
    4. none
  46. By decreasing the equilibrium temperature of a crystal, the concentration of vacancies __________. ( )
    1. decreases
    2. increases
    3. remains the same
    4. none
  47. To produce diffraction with X-rays, the spacing between the consecutive lines of grating should be of the order of __________ angstroms. ( )
    1. thousands of
    2. hundreds of
    3. few
    4. none
  48. In 1912, Laue suggested that a crystal can serve as a __________ for X-ray diffraction. ( )
    1. three-dimensional grating
    2. two-dimensional grating
    3. one-dimensional grating
    4. none of the above
  49. __________ and __________ succeeded in diffracting X-rays by passing through a thin crystal. ( )
    1. Friedrich and Knipping
    2. Bragg and Knipping
    3. Friedrich and Laue
    4. none of the above
  50. If the path difference between the X-rays reflected by successive crystal planes is …, then the intensity of diffracted ray: ( )
    1. will not change
    2. is minimum
    3. is maximum
    4. none
  51. If the path difference between the X-rays reflected by successive crystal planes is , where n = 1, 2, 3,…, then the intensity of diffracted ray: ( )
    1. is minimum
    2. is maximum
    3. remains the same
    4. none
  52. X-ray powder method is usually carried for __________ materials. ( )
    1. polycrystalline
    2. powder
    3. single crystal
    4. amorphous
  53. Using powder diffraction, __________ of a crystal can be determined. ( )
    1. the interatomic spacing
    2. the interplanar spacing
    3. both a and b
    4. none
  54. In powder method, _____________ chromatic X-rays are used. ( )
    1. mono
    2. poly
    3. both a and b
    4. none
  55. In Laue method, __________ X-rays are used. ( )
    1. monochromatic
    2. white
    3. both a and b
    4. none
  56. In transmission Laue method, the diffracted spots lie on the curves of: ( )
    1. ellipses
    2. hyperbolas
    3. a or b
    4. none
  57. In back reflection Laue method, the diffracted spots lie on curves of: ( )
    1. hyperbola
    2. parabolas
    3. ellipses
    4. none
  58. Laue method is useful to decide the __________ and orientation of the internal arrangement of atoms/molecules in the crystal. ( )
    1. cell parameters
    2. crystal symmetry
    3. both a and b
    4. none
  59. The diffracted spots will be in the form of______________, if the crystal contains imperfections or strains. ( )
    1. streaks
    2. spots
    3. both a and b
    4. none

Answers

1. a 2. b 3. a
4. a 5. a 6. c
7. d 8. d 9. c
10. d 11. d 12. c
13. d 14. a 15. b
16. d 17. c 18. b
19. b 20. c 21. a
22. c 23. a 24. a
25. a 26. a 27. b
28. a 29. a 30. c
31. a 32. d 33. b
34. a 35. c 36. d
37. c 38. a 39. b
40. b 41. a 42. d
43. a 44. c 45. b
46. a 47. c 48. a
49. a 50. b 51. b
52. a 53. b 54. a
55. b 56. c 57. a
58. b 59. a

Review Questions

  1. Derive Bragg's law of X-ray diffraction.

     

    (Set-1–Sept. 2006), (Set-4–May 2006), (Set-3–May 2003), (Set-3–Nov. 2003)

  2. What are Miller indices? How are they obtained?

     

    (Set-1–May 2006), (Set-1, Set-2, Set-3, Set-4–June 2005), (Set-4–Nov. 2004), (Set-1–May 2003), (Set-4–Nov. 2003)

  3. Explain Schottky and Frenkel defects with the help of suitable figures.

     

    (Set-2–Sept. 2007), (Set-2–May 2007), (Set-4–Sept. 2006), (Set-1, Set-2, Set-3, Set-4–June 2005), (Set-4–Nov. 2004), (Set-1–May 2003), (Set-4–Nov. 2003)

  4. State and explain Bragg's law.

     

    (Set-1–Sept. 2007), (Set-2, Set-3–Sept. 2006), (Set-2–May 2006), (Set-3–May 2004), (Set-4–May 2003)

  5. Describe with a suitable diagram, the powder method for the determination of crystal structure.

     

    (Set-1–Sept. 2007), (Set-2–Sept. 2006), (Set-2, Set-3–May 2006), (Set-3–May 2004), (Set-4–May 2003)

  6. Explain Bragg's law of X-ray diffraction.

     

    (Set-2, Set-4–Sept. 2007), (Set-2, Set-3, Set-4–May 2007), (Set-1–May 2006), (Set-4–Sept. 2006), (Set-3–Nov. 2004), (Set-2–May 2004)

  7. Describe Laue's method for determination of crystal structure.

     

    (Set-2–May 2008), (Set-2, Set-4–Sept. 2007), (Set-3–May 2007), (Set-4–Sept. 2006), (Set-2–May 2004), (Set-3–May 2003)

  8. Explain the significance of Miller indices.

     

    (Set-1–May 2004), (Set-2–May 2003)

  9. Derive an expression for the number of Schottky defects in equilibrium at a temperature T.

     

    (Set-4–Sept. 2006), (Set-1–May 2004), (Set-2–May 2003)

  10. Explain the various point defects in a crystal.

     

    (Set-1–Sept. 2007), (Set-1–Nov. 2004), (Set-1–Nov. 2003)

  11. Obtain the expression for the equilibrium concentration of vacancies in a solid at a given temperature.

     

    (Set-1–Sept. 2007), (Set-1–Nov. 2004), (Set-1–Nov. 2003)

  12. Deduce the expression for the interplanar distance in terms of Miller indices for a cubic structure.

     

    (Set-3–Sept. 2008), (Set-2–Nov. 2004), (Set-2–Nov. 2003)

  13. Sketch the following planes of a cubic unit cell: (001), (120) and (11).

     

    (Set-2–Sept. 2007), (Set-2–Nov. 2004), (Set-2–Nov. 2003)

  14. Define Miller indices. Sketch the following atomic planes in a simple cubic structure (010), (110) and (111).

     

    (Set-4–May 2004)

  15. How can the interplanar spacing of a set of Miller planes be calculated in terms of Lattice parameters?

     

    (Set-4–May 2004)

  16. What is Bragg's law? Explain.

     

    (Set-2–May 2008)

  17. What are Miller Indices? Draw (111) and (110) planes in a cubic lattice.

     

    (Set-2, Set-4–May 2007)

  18. Draw the (112) and (120) planes and the [112] and [120] directions of a simple cubic crystal.

     

    (Set-1–May 2007)

  19. Sketch the following planes of a cubic unit cell: (001), (120) and (11).

     

    (Set-4–Sept. 2006)

  20. What is Frenkel defect? Explain.

     

    (Set-1, Set-3–May 2007)

  21. Describe edge and screw dislocations. Draw Burger's circuit and slip planes for them.

     

    (Set-4–Sept. 2007), (Set-4–May 2006)

  22. Explain the significance to Burger's vector.

     

    (Set-2, Set-4–Sept. 2007), (Set-2–May 2007), (Set-3–Sept. 2006), (Set-4–May 2006)

  23. Describe Bragg's X-ray spectrometer and explain how Bragg's law can be verified.

     

    (Set-1–Sept. 2006), (Set-4–May 2006)

  24. Explain the influence of point defects in crystals and how do they aff ect the properties of materials.

     

    (Set-3–Sept. 2007)

  25. Obtain an expression for the energy required to create a vacancy in the crystal.

     

    (Set-3–Sept. 2007)

  26. Derive an expression for the interplanar spacing in the case of a cubic structure?

     

    (Set-1–May 2007)

  27. Derive an expression for the energy change due to creation of vacancies inside a solid.

     

    (Set-2–May 2006)

  28. Derive an expression for the concentration of Frenkel defects present in a crystal at any temperature.

     

    (Set-1, Set-3–May 2007)

  29. Sketch the planes (120), (23) and directions [100] and [211].

     

    (Set-4–Sept. 2008)

  30. Explain how the X-ray diffraction can be employed to determine the crystal structure. Give the ratio of interplanar distances of (100), (110) and (111) planes for a simple cubic structure.

     

    (Set-3–Sept. 2007), (Set-3–May 2006)

  31. Distinguish between Frenkel defects and Schottky defects.

     

    (Set-2–May 2006)

  32. Explain edge dislocation, screw dislocation and significance of Burger's vector.

     

    (Set-3–Sept. 2006)

  33. Write short notes on Burger's vector in dislocations.
  34. What are Miller indices? Derive an expression for the interplanar spacing between two adjacent planes of Miller indices (hkl) in a cubic lattice of edge length ‘a’.
  35. Explain and illustrate, with neat sketches, the edge and screw dislocations; show the Burger's vector in them.
  36. Write short notes on interstitial defects of crystals.
  37. What are point defects in crystals? Derive an expression for the concentration of Schottky defect in a crystal.
  38. Explain the principle, procedure and advantage of Debye–Scherrer method of X-ray diffraction.
  39. Mention the different kinds of crystal imperfections.
  40. Compare and contrast Frenkel and Schottky defects.
  41. Write short notes on screw dislocation.
  42. What are crystal imperfections? Explain.
  43. Distingu ish between edge and screw dislocations. What is Burger's vector?
  44. Discuss the Schottky defect in the case of ionic crystals.
  45. Explain the powder method of crystal structure analysis.
  46. What are Miller indices? How they are determined?
  47. Show that the number of Frenkel defects in equilibrium at a given temperature is proportional to (NNi )1/2, where N be the number of atoms and Ni be the number of interstitial atoms.
  48. Obtain the Miller indices of a plane which intercepts at a, b/2 and 3c in simple cubic unit cell. Draw a neat diagram showing the plane.
  49. What do you understand by Miller indices of a crystal plane? Show that in a cubic crystal the spacing between consecutive parallel planes of Miller indices (hkl) is given by:

     

     

  50. Define Schottky defect and derive an expression for the density of Schottky defects at a specified temperature.
  51. Calculate the first nearest neighbour atom distance in ZnS (i.e., from Zn to S atoms) system.
  52. Derive an expression for the interplanar distance in the case of cubic systems following Miller indices concept.
  53. Define a Frenkel defect and derive an expression for the density of such defects as a function of temperature.
  54. Write short notes on the Burger's vector in dislocation with appropriate diagrams.
  55. Derive the Bragg's law of X-ray diffraction and obtain the relation that connects the interplanar distance ‘d’ in orthogonal systems with lattice parameters a, b and c.