UNIT 4

CRYSTAL STRUCTURE, X-RAYS AND ELECTRODYNAMICS

1 Question: Define the term Space Lattice?

Solution:

In a solid crystalline material, the atoms or molecules are arranged regularly and periodically in all three dimensions. The atomic arrangement in a crystal is called crystal structure.

To explain crystal symmetries easily, it is convenient to represent an atom or a group of atoms that repeats in three dimensions in the crystal as a unit. If each such atom or unit of atoms in a crystal is replaced by a point in space, then the resultant points in space are called space lattice.

Each point in a space lattice is called a lattice point. Each atom or unit of atoms is called basis or pattern.

A space lattice represents the geometrical pattern of crystal in which the surroundings of each lattice point is the same. If the surroundings of each lattice point is same or if the atom or all the atoms at lattice points are identical, then such a lattice is called Bravais lattice.

On the other hand, if the atom or the atoms at lattice points are not same, then it is said to be a non-Bravais lattice.

A space lattice or crystal lattice is defined as three dimensional infinite arrays of points in space in which every point has surroundings identical to that of every other point in the array.

2 Question: Define Basis in Crystal Structure?

Solution:

The atomic arrangement in a crystal is called crystal structure. The crystal structure is formed by associating every lattice point with an atom or an assembly of atoms or molecules or ions, which are identical in composition, arrangement and orientation, called the basis. i.e. an atom, or a group of atoms or molecules identical in composition is called the basis or the pattern.

The basis provides the number of atoms per lattice point, their types, mutual orientations and distances of separation between the atoms.

If the basis is substituted for the lattice points, then the resulting structure is called crystal structure as shown in Figure.

Lattice + basis = crystal structure.

The basis shown in Figure contains two different atoms. In copper and sodium crystals the basis is single atoms; In NaCl, the basis is diatomic and in CaF2 the basis is triatomic. A lattice is an imaginary assumption while the crystal structure is a real concept.

1 Figure:  Lattice sites, Basis and crystal structure

3 Question: What is Unit Cell and also discuss about Lattice Parameters?

Solution:

Unit cell

Unit cell is small repeating entity of the atomic structure. It is the basic building block of the crystal structure. It defines the entire crystal structure with the atom positions within.

Unit cells for most of the crystals are parallelepipeds or cubes having three sets of parallel faces. A unit cell is the basic structural unit or building block of the crystal.

2 Figure: unit cell and unit cell parameter

A unit cell is defined as the smallest parallelepiped volume in the crystal, which on repetition along the crystallographic axes gives the actual crystal structure or the smallest geometric figure, which on repetition in three-dimensional space, gives the actual crystal structure is called a unit cell.

Choice of a unit cell is not unique but it can be constructed in a number of ways as shown in figure. Following figure shows different ways of representing unit cells in a two-dimensional lattice. A unit cell can be represented as ABCD or A′B′C′D′ or A′′B′′C′′D′′, etc.

3 Figure: Construction of unit cell in different way

First we define crystallographic axes. These axes are obtained by the intersection of the three non-coplanar faces of the unit cell. The angles between these faces or crystallographic axes are known as interfacial or interaxial angles. The angles between the axes Y and Z is α, between Z and X is β and between X and Y is γ. The translational vectors or primitives a, b, c of a unit cell along X, Y, Z axes and interaxial angles α, β, γ are called  Cell parameters. These cell parameters are shown in (Figure 2). The cell parameters determine the actual size and shape of the unit cell.

4 Question: What do you meant by Primitive and non-primitive unit cells?

Solution:

The unit cell formed by primitives is called a simple or a primitive unit cell. A primitive unit cell contains only one lattice point. If a unit cell contains more than one lattice point, then it is called a non-primitive unit cell or a multiple unit cell. Most of the unit cells of various crystal lattices contain two or more lattice points and hence it is not necessary that unit cell should be a primitive unit cell.

4 Figure: Primitive and non-primitive unit cells

5 Question: What is the Difference between Crystalline solids and amorphous solids?

Solution:

6 Question: Derive Maxwell’s equations in vacuum and in non-conducting medium?

Solution:

Maxwell’s four equations are given by

∇·E = ρ/ε0        (1)

∇×E = −∂B/∂t        (2)

∇×H = J + ∂D/∂t       (3)

∇·B = 0        (4)

Free space or non-conducting medium. We know that non conducting medium means no current so conductivity is zero i.e. σ=0

So current density J=σE will also become zero as σ=0 Also free space means no charges which leads to ρ=0. These points mentioned below.

(a)    No condition current i.e σ=0, thus J=0 ( J=σE)

(b)   No charges (i.e ρ=0)

For the case of no charges or currents, that is, j = 0 and ρ = 0, and a homogeneous medium. Using these the Maxwell equation can rewritten as

∇.D=0 or ∇.E=0      as  ρ=0         (12)

∇  x E= -dB/dt  or    ∇ x E= -μ dH/dt    because  B = μH       (13)

∇ x H=d D/dt  or  ∇ x H = ε dE/dt         (J=0) and D = εE  (14)

∇.B=0                                      (15)

Now taking curl of second Maxwell’s equation (13) ,we get

∇ x(∇ x E)=- μ d/dt (∇ x H)

Applying standard vector identity, that is [∇ *(∇*E)=∇(∇.E)-∇2E] on left hand side of above equation, we get

∇ (∇ .E)-∇2E= -μ d/dt (∇ x H)                                                           (16)

Substituting equations (13) and (14) in equations (16) ,we get

-∇2E= – με d/dt (dE/dt)

Or                               ∇2E=με d 2 E/dT2       (17)

Equation (17) is the required wave equation in terms of electric field intensity, E for free space . This is the law that E must obey.

Wave Equation In Terms Of Magnetic Field Intensity, H

Take curl of fourth Maxwell’s equation(14) ,we get

∇x(∇xH)=ε d/dt(∇xE)

Applying standard vector identity that is

[∇*(∇*H)=∇ (∇.H)-∇2H]

On left side of above equation , we get

∇(∇.H)-∇2H= ε d/dt(∇xE)                                            (18)

Substituting equations (14) and (13) in equation(18) ,we get

-∇2H= – μεd/dt(dH/dt)

Or                      

∇2H=με d2H/dt2        (19)

Equations (19)  is the required wave equation in terms of magnetic field intensity, H and this is the law that H must obey

For vacuum μ=μ0 and ε=ε0, equations (17) and (19) will become

∇2 E=μ0ε0 d2E/dt2        (20)

And     ∇2H=    μ0ε0 d2H/dt2        (21)

This leads to an expression for the velocity of propagation

From equation both equations (20) and (21) have the form of the general wave equation for a wave

(x,t) traveling in the x direction with speed v. Equating the speed with the coefficients, we derive the speed of electric and magnetic waves, which is a constant that we symbolize with “c”

It is useful to note that in vacuum

c2=1/ε0μ0        

Where μ0 is the permeability of free space

7 Question: Write a Note on Maxwell’s equations?

Solution:

Maxwell’s four equations are given by

∇·E = ρ/ε0        (1)

∇×E = −∂B/∂t        (2)

∇×H = J + ∂D/∂t       (3)

∇·B = 0        (4)

These equations illustrate the unique coexistence in nature of the electric field and the magnetic field. The first two equations give the value of the given flux through a closed surface, and the second two equations give the value of a line integral around a loop. In this notation,

∇=(∂/∂x, ∂/∂y, ∂/∂z)

E is the electric vector

B is the magnetic induction

ρ is the electric charge density

j is the electric current density

ε0 is the permittivity of free space

c is the speed of light.

In addition to Maxwell equations, the following identities are useful:

J =  σE         (5)

D = εE         (6)

B = μH         (7)

Here,

D is the electric displacement

H is the magnetic vector

σ is the specific conductivity

ε is the dielectric constant (or permittivity)

μ is the magnetic permeability

8 Question: What are Bravais lattices?

Solution:

If the surroundings of each lattice point is same or if the atom or all the atoms at lattice points are identical, then such a lattice is called Bravais lattice.

A space lattice represents the geometrical pattern of crystal in which the surroundings of each lattice point is the same. If the surroundings of each lattice point is same or if the atom or all the atoms at lattice points are identical, then such a lattice is called Bravais lattice.

On the other hand, if the atom or the atoms at lattice points are not same, then it is said to be a non-Bravais lattice.

For representing the type of distribution of lattice points in space, seven different co-ordinate systems are required. These co-ordinate systems are called crystal systems.

The crystal systems are named on the basis of geometrical shape and symmetry. The seven crystal systems are:

(1) Cubic

(2) Tetragonal

(3) Orthorhombic

(4) Monoclinic

(5) Triclinic

(6) Trigonal (or Rhombohedral)

(7) Hexagonal.

In 1948, Bravais showed that 14 lattices are sufficient to describe all crystals. These 14 lattices are known as Bravais lattices and are classified into 7 crystal systems based on cell parameters or lattice points present per unit cell.

9 Question: Describe seven crystal systems with diagrams?

Solution:

Bravais lattices are categorized as

10 Question: Describe FCC crystal structure?

Solution:

Face-centered lattice (F) 

Number of atoms

In case of face-centered lattice, in addition to the 8 atoms at 8 corners each contributing 1/8thto the unit cell, six atoms will be present at the center of six faces of the cell each contributing 1/2nd to the unit cell. Therefore number of atoms or lattice points in a face centered unit cell becomes 8 x 1/8 + 6 x 1/2 = 4.

Figure: Face-centered lattice (F)

Coordination number

In Face Centred Cubic system 12 neighbour atoms are in touch with the selected atom as shown in figure. So the coordination number of FCC is twelve.

Figure: Face Centred Cubic Crystal

Atomic Radius

For FCC         r =

11Question: Obtain the relations between the edge of the unit cell and atomic radius for the SC, BCC and FCC lattices?

Solution:

Atoms are not hard spheres with distinct boundaries, so their sizes are not determined directly. However, the positions of the atoms in a solid can be determined by x-ray diffraction, and the sizes of the atoms are inferred from those distances.

In this method, the radius of an atom is determined from the unit cell edge length (a), which is determined from the location of the atoms, and the assumption that the atoms touch as shown in below.

Simple or Primitive Cubic Crystal

2r = a

Body Centred Cubic Crystal

Face Centred Cubic Crystal

Relationship of Atomic Radii and Unit Cell Edge Lengths

Setting the edge length equal to 2r in the sc unit cell, and applying the Pythagorean theorem to the triangles shown in Figures we obtain the relationships between the atomic radius (r) of the atom and the edge length of the unit cell (a) given in Equation

For SC           r = a/2

For BCC        r =

For FCC         r =

Describe BCC structure?

12 Question: Find the packing fraction for SC, BCC and FCC crystals?

Solution:

Packing Efficiency

Packing efficiency is the fraction of the unit cell that is occupied by particles. Spheres cannot be packed without creating some void space, but the amount of void space depends upon how well they are packed. Packing efficiency (PE) is that fraction of the unit cell volume that is actually occupied by particles, not void space. Thus, the packing efficiency of a unit cell is

Packing Fraction =

PE =  x 100%

a = the length of a side of the unit cell, so a3 is the volume of the unit cell.

r = the radius of the atoms that occupy the unit cell, so (4/3)πr3 is the volume of a single atom in the unit cell.

N = the number of atoms in the unit cell, so N(4/3)πr3 is the volume occupied by all of the atoms in the unit cell.

The unit cell size (a) is directly proportional to the atom size (r), so one can be defined in terms of the other.

Consequently, the packing efficiency depends only upon the cell type, not its size. As shown in the next sections, the relationship between a and r depends only upon the unit cell type.

Packing in Simple Cubes

Figure

In the simple cubic unit cell, the spheres touch along the cell edge. Consequently, the edge length is two atomic radii

(a = 2r).

The properties of the simple cubic unit cell:

r = 0.5a

N = 1 atom/unit cell

PE =  x 100%

Packing efficiency = 52%

Coordination number = 6

Packing in Body-Centered Cubes

Figure

For the body-centered cubic unit cell, the spheres make contact along the body diagonal (bd), which is four atomic radii (each yellow arrow is one atomic radius), i.e.,  bd = 4r.

The body diagonal is related to the face diagonal (fd) and the cell edge (a) by the Pythagorean theorem. The relationship between the cell edge and the face diagonal was derived in the discussion of the fcc unit cell. The properties of the body-centered cubic unit cell:

For BCC        r =

r = 0.433 a

N = 2 atoms/unit cell

PE =  x 100%

Packing efficiency = 68%

Coordination number = 8

Packing in Face-Centered Cubes

Figure

For FCC         r =

r= 0.354a

N = 4 atoms/unit cell

PE =  x 100%

Packing efficiency = 74%

Coordination number = 12

As shown in the previous sections, the packing efficiency of a unit cell depends only on the cell type; it is independent of edge length and atomic radius because both cancel in the calculation. As might be expected, the packing efficiency increases as the number of atoms in the unit cell and the number of atoms that are packed around each atom (coordination number) increases.

13 Question: An atomic solid crystallizes in a body center cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60.3 pm. If the atomic weight of A is 48, Find the density of the solid?

Solution:

According to given condition,

0.13 a = 60.3 ⟹ a = 463.8

So Density =

=

=  1.75 g/cc

14 Question: Deduce the expression for the interplanar distance in terms of Miller indices for a cubic structure.

Solution:

Distance of separation between successive hkl planes

The separation between successive parallel planes in rectangular axes crystal system can be extracted easily.

Consider a plane ABC which belongs to a family of planes. h, k, l are the Miller indices of this plane, which represents the set of planes. The perpendicular ON from the origin O to the plane represents the inter-planar spacing d = ON of this family of planes. Let ON make an angle α’, β’, γ’ with the x, y, z axes respectively.

The intercepts of the plane on the three axes are OA= a/h; OB= a/k; OC=a/l Where ‘a’ is the length of the cube edge. Then from figure, we have

Cos = = =

Cos = = =

Cos = = =

From the figure ON =

d =

But,  

Substituting the values of cos’, cos’, cos’ in equation (2), We get,

For the orthogonal coordinates Substituting the values of cos’, cos’, cos’ in equation (2), We get,

( )2 + ( )2 +( )2 = 1

()2 + ()2 +( )2 = 1

() () = 1

d =

d =

15 Question: Sketch the following planes of a cubic unit cell: (111) and (210)

Solution:

1. The (111) surface

Intercepts: a , a , a

Fractional intercepts: 1 , 1 , 1

Miller Indices: (111)

Ii.            The (210) surface

Intercepts: ½a , a ,

Fractional intercepts: ½ , 1 ,

Miller Indices: (210)

16 Question: Explain Bragg’s Law and its importance?

Solution:

Consider a set of parallel planes called Bragg’s planes. Each atom is acting as a scattering center. The intensity of the reflected beam at certain angles will be maximum when the path difference between two reflected waves from two adjacent planes is an integral multiple of λ.

Let ‘d’ be the distance between two adjacent planes, 'λ’ be the wavelength of the incident x-ray, ‘θ’ be the glancing angle.

The path difference between the rays reflected at A & B is given by

= CB + BD= d sinθ + d sinθ = 2dsinθ

If the two consecutive planes scatter waves in phase with each other, then the path difference must be an integral multiple of wavelength. 

= nλ

For the reflected light intensity to be maximum, the path difference

nλ= 2dsinθ

Where ‘n’ is the order of scattering.

This is called Bragg’s law

Therefore, Bragg’s law states that X-rays diffracted from different parallel planes of a crystal interfere constructively when the path difference is integral multiples of wavelength of X-rays.

From Bragg’s law nλ = 2d sin θ, since maximum possible value for sin θ is 1,

nλ/2d ≤ 1 or λ ≤ 2d.

This sets the limitation on the wavelength, i.e. in order to get the diffraction pattern by a crystal, the wavelength of X-rays should not exceed twice the inter-planar spacing.

Importance of Bragg’s law:

1. Bragg’s law is the essential condition to be satisfied by crystal planes in order to get diffraction pattern from a crystal.

2. It is used to calculate inter-planar spacing. Knowing the values of inter-planar spacing, lattice parameters can be determined.

17 Question: Discuss x-ray diffraction method?

Or

Discuss Laue’s Method?

Solution:

The Laue method was the first diffraction method ever used. This method is used to study the orientation of crystal and to verify crystal symmetry. A beam of white radiation, the continuous spectrum from an x-ray tube, is allowed to fall on a fixed single crystal. The Bragg angle θ is therefore fixed for every set of planes in the crystal, and each set picks out and diffracts that particular wavelength which satisfies the Bragg law for the particular values of d and involved. Each diffracted beam thus has a different wavelength.

There are two variations of the Laue method, depending on the relative positions of source, crystal, and film. In each, the film is flat and placed perpendicular to the incident beam. The film in the transmission Laue method (the original Laue method) is placed behind the crystal so as to record the beams diffracted in the forward direction. This method is so called because the diffracted beams are partially transmitted through the crystal. These curves are generally ellipses or hyperbolas for transmission patterns.

In the transmission Laue method, the film is placed behind the crystal to record beams which are transmitted through the crystal.

One side of the cone of Laue reflections is defined by the transmitted beam. The film intersects the cone, with the diffraction spots generally lying on an ellipse

In the back-reflection Laue method the film is placed between the crystal and the x-ray source, the incident beam passing through a hole in the film, and the beams diffracted in a backward direction are recorded. In either method, the diffracted beams form an array of spots on the film as shown in Figure. This array of spots is commonly called a pattern, more specifically, Laue pattern, but the term is not used in any strict sense and does not imply any periodic arrangement of the spots. On the contrary, the spots are seen to lie on certain curves, as shown in figure for transmitted pattern. These curves are generally hyperbolas for back-reflection patterns.

In the back-reflection method, the film is placed between the x-ray source and the crystal. The beams which are diffracted in a backward direction are recorded.

One side of the cone of Laue reflections is defined by the transmitted beam. The film intersects the cone, with the diffraction spots generally lying on an hyperbola.

The spots lying on any one curve are reflections from planes belonging to one zone. This is due to the fact that the Laue reflections from planes of a zone all lie on the surface of an imaginary cone whose axis is the zone axis. The positions of the spots on the film, for both the transmission and the back-reflection method, depend on the orientation of the crystal relative to the incident beam, and the spots themselves become distorted and smeared out if the crystal has been bent or twisted in any way.

The facts account for the two main uses of the Laue methods include: the determination of crystal orientation and the assessment of crystal perfection.

Laue method is useful to decide the crystal symmetry and orientation of the internal arrangement of atoms/molecules in the crystal.

The atomic arrangement in a crystal can be analysed by studying the positions and intensities of spots in Laue pattern.

As several wavelengths of X-rays can reflect in different orders from the same set of planes with the different order reflections superimposed on the same spot in the film, the intensity of the spots and hence the 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.

18 Question: What do you mean line and continuous spectrum of x-ray?

Solution:

Origin of Continuous X-ray Spectra

The line spectrum of an element originate when electrons are expelled from inner shells of its atoms and the electrons from higher levels fall into these vacancies in the inner shells. Each such transition causes energy loss equal to the energy difference of shells which appears as an X-ray photon.

Continuous X-ray spectrum is the result of inverse photoelectric effect. Continuous X-ray is produced by the phenomenon of “Bremsstrahlung” which is German word meaning “breaking radiation”. Electrons are emitted from the cathode in the X-rays are accelerated towards the target or plate and strike it, penetrate deep into the interior of its atoms and are attracted by the nuclei due to strong electrostatic interaction. The electron therefore deviates from its original path as shown in figure

The deviation of the electron from its straight line path is equivalent to its collision with the nucleus. The electrons are loss more of its kinetic energy due to the collision.

This energy appears as an X-ray photon.

Ei & Ef are initial kinetic energy and the final kinetic respectively.

 Ei - Ef h

The electron may suffer several collisions with various nuclei before coming to rest and each collision is accompanied by the emission of an X-ray photon. Hence, a number of photons of different frequencies are emitted. As there are large numbers of electrons in the beam and each electron collision in a different way, we get photons of almost all frequencies or wavelengths, thus producing a continuous X-ray spectrum.

According to classical theory, the continuous X-ray spectrum should consists of all frequencies between zero and infinity. But according to quantum theory, there is a minimum wavelength up to which is the continuous X-ray spectrum extends. To find the value of min, let the velocity of the incident electron decrease from vi to vf due to the interaction with the nucleus. Then the frequency of X-ray photon is given by

h = mvi2 - mvf2                       …………….(1)

When electron is brought to rest by electrostatic interaction, vf =0 and photon has maximum frequency or minimum wavelength.

hmax = mvi2        …………….(2)

If V is potential difference between cathode and target or plate the

mvi2 = eV        …………….(3)

From (2) and (3), we have

hmax = eV        …………….(4)

λmin =

From (4) we have

λmin =

λmin = Å        …………….(5)

The lowest wavelength limit of continuous spectra is inversely proportional to the accelerating potential of X-ray tube λmin

Continuous spectrum of X-rays consists of all radiation of all possible X-ray wavelengths within a certain range starting from min (i.e. the wavelength corresponding to minimum energy) onwards. The continuous spectrum is also known as general spectrum or white spectrum.

The plot of relative intensity of X-ray spectra versus wavelength in various potential differences applied to an X-ray tube between the cathode and target or plate is shown in figure. It is clear that by varying the X-ray tube accelerating potential, the min shifts toward left i.e. the X-ray achieve more energy; the intensity of X-rays also increases with voltage.

19 Question: Explain Mosley’s Law and its importance?

Solution:

This law advanced chemistry by immediately sorting the elements of the periodic table in a more logical order. Moseley could predict the existence of several then-unknown elements. A law that relates the frequency of the spectral lines of the characteristic X-radiation of a chemical element to its atomic number. This law was experimentally established by H. Moseley in 1913.

According to Moseley’s law, the square root of the frequency v of a spectral line of the characteristic radiation of an element is a linear function of its atomic number Z:

Where R is the Rydberg constant

Sn is the screening constant

n is the principal quantum number.

On a Mosley plot the dependence of on Z is a series of lines (such as the Kα lines, Lα lines, and Mα lines, which correspond to the values n = 1,2, and 3).

Mosley’s law was incontrovertible proof of the correctness of the arrangement of the elements in D. I. Mendeleev’s periodic system of the elements and the law helped to clarify the physical significance of Z.

According to Moseley’s law, the characteristic X-ray spectra do not display the periodic regularities that are inherent in optical spectra. This indicates that the inner electron shells of the atoms of all elements, which are manifested in the characteristic X-ray spectra, have an analogous structure.

Figure

Subsequent experiments revealed some deviations from a linear

Subsequent experiments revealed some deviations from a linear Moseley plot for the Kα lines, Lα lines, and Mα lines of characteristic X-radiation. The atomic number of the element Z is plotted along the axis of abscissas and the quantity (where c is the velocity of light) is plotted along the axis of ordinates.

Dependence for the transition groups of elements the deviations being due to the change in the order in which the outer electron shells are filled and also for heavy atoms, in which the deviations result from relativistic effects (and may provisionally be explained by the fact that the velocities of the inner electrons are comparable with the velocity of light).

The position of the spectral lines on a Moseley plot may vary somewhat, depending on a number of factors, such as the number of nucleons in the nucleus (isotope shift) and the state of the outer electron shells (chemical shift). The study of these shifts makes it possible to obtain detailed information about the atom.

Importance of Moseley's law:

• Using this law Moseley arranged K and Ar, Ni and CO in a proper way in Mendeleev's periodic table. 
• This law was held to the discovery of many new elements like Tc (43), Pr (61), Rh (45). 
• The atomic number of rare earth have been determined by this law.

20 Question: Moseley's Law for characteristic X-rays is √ν=a(Z−b)

In this formula

A both a and b are independent on the material.

B a is independent but b depends on the material.

C b is independent but a depends on the material.

D both a and b depend on the material.

Solution:

'A' is correct! The constants a and b are independent of the material. They depends on X-ray series.

21 Question: In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction (n = 1) occurred at an angle θ = 25.25°. Determine the spacing between the diffracting planes in copper.

Solution:
The distance between the planes is found by solving the Bragg equation,

nλ = 2d sin θ,

For d.

This gives:   d=nλ2sinθ

=1(0.1315nm)2sin(25.25∘)=0.154nm

Bottom of Form