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- 1. 1 INDEX Unit Contents Page I 1.Crystallography 1-19 2.X-ray Diffraction & Defects in Crystals 20-27 II 3.Principles of Quantum Mechanics 28-38 4.Elements of Statistical Mechanics& Electron theory of Solids 39-61 III 5.Dielectric Properties 62-75 6.Magnetic Properties & Superconducting Properties 76-97 IV 7.Optics 98-119 8.Lasers & Fiber Optics 120-148 V 9.Semiconductor Physics 149-189 10.Nanotechnology 190-215
- 2. 2 UNIT-I CHAPTER-1 CRYSTALLOGRAPHY 1.1.1. Introduction to Bonding in Solids Atom consists of a positively charge nucleus surrounded by a negatively charged electron cloud. When the two atoms are brought closer, there will be both attractive and repulsive forces acting on them. Therefore, the two atoms take relative positions when there is a balance between these two forces. This is known as equilibrium position. The bonding forces which bind the atoms in a molecule and in the molecules in a solid are mainly four types of forces i.e. gravitational, electrical, and nuclear and weak forces. Chemical bond is defined as inter- atomic or inter-molecular or inter-ionic force of attraction which holds the atoms together. When the two atoms attain the equilibrium position, then the value of energy needed to move an atom completely away from its equilibrium position is termed as binding energy or cohesive energy. On the basis of the nature of forces which bind solid lattice together, the solids can be divided two mainly groups namely. Primary bonds Secondary bonds Primary bonds: Primary bonds are the strongest bonds which hold atoms together. The three types of primary bonds are, Ionic bond Covalent bond Metallic bond Secondary bonds: Secondary bonds are much weaker than primary bonds. Two types of Secondary bonds are, Hydrogen bond Vander Waals bond 1.2. Primary bonds: 1. Ionic bond: This bond is the simplest type of chemical bond called ‗Hetero polar Bond‘. Ionic bonds are mostly insulating in character. These are not pure elements, but they are compounds. Eg:-NaCl, KCl, KBr etc. The electronic structure of atoms is relatively stable when the outer shells contain eight electrons (two in the case of 1st shell). Sometimes it is possible by transferring the electrons from one atom to another, electron shells are filled the donor atom will take a positive charge and the acceptance will have a negative charge. When Na and Cl atoms are placed together, these are transfer of electrons from Na to Cl atoms resulting in a strong electrostatic attraction between the positive Na ion and negative Cl ions. Let us examine Na and Cl electronic configuration how they tends to form an ionic bond. Atomic number of Na is 11 Electronic configuration of Na is: 1s2 2s2 2p 6 3s1 It has one electron in the outer most orbits, this electron is called as valence electron Na atom try to lose this valence electron in order to attain insert gas configuration. The minimum energy required to detach the electron from Na atom is called as ionization energy. Na + ionization energy → Na+ + e- Ionization energy required = 5.1 eV Atomic number of Cl is 17 Electronic configuration is 1s2 2s2 2p 6 3s2 3s5 It has 7 electrons in its outermost orbit. It needs one electron to insert gas configuration. It readily accepts an electron and release some energy called electron affinity. Cl + e- →Cl- + Electron Affinity Electron Affinity = 3.6ev In order to produce two ions, NaCl needs energy equal to5.1ev-3.6eV = 1.5ev Na + Cl+1.5eV → Na+ + Cl- Na+ and Cl- attract each other and form a bond. This bond is called ―Ionic Bond‖.
- 3. 3 Characteristics of ionic bonded Materials Ionic bonds are strong. They have high melting points and boiling points. They are transparent to visible light. They have closed packed structure. They are highly soluble in polar solvents like H20, liquid NH3 and insoluble in non-polarsolvent. Non directional, because the charge distribution is special in nature. 2. Covalent Bond: Covalent bonds are called homo polar bounds. In this neighboring atoms share their valence electrons in the formation of a strong covalent bound. Eg: Cl2, O2, etc. Consider the Cl atom, which has seven electrons in the valence shell. Spins of 6 electrons are paired, where as the spin of 7 the electron is unpaired. The unpaired electron is always looking for another unpaired electron. Which comes closer to it. In such case another Cl atom comes near the first Cl atom. The two unpaired electrons of the two atoms get paired. . Characteristics of covalent bonded materials These bounds are usually hard and brittle. Binding energy is high. M.P and B.P are high but low compared to ionic crystal. Covalent bonds are insoluble in H2o. These materials are soluble in non-polar solvents like C6H6. They are transparent to longer wavelengths, but opaque to shorter wavelength. 3. Metallic Bond: Metallic bonds are similar to covalent bonds. In this outer electrons have high mobility. In metallic bonding each atom of metal give one or two value electrons to the crystal. These hold the atoms together and are not bound to individual atoms, but move freely throughout the whole metal. In metal, the ionization energies are low, so metallic atoms gives their valence electrons to crystal. The valence electrons will form electron cloud or electron gas, which is occupied throughout the metal space. The positive ions are held together by the electrostatic forces due to the free electrons. The electrostatic interaction between the positive ion and the surrounding electron could hold the metal together. So it is called the metallic bond. Eg; - Al, Ag, Na, Cu etc.
- 4. 4 Fig: Atomic arrangement in metallic crystals Characteristics of Metallic bonded materials They have high thermal conductivity due to the presence of free electrons. They have high electrical conductivity. Metallic bonds are non directional. The metallic bound is comparatively weaker than the ionic and covalent bounds. 1.3. Secondary bonds 4. Hydrogen bond: Covalently bonded atoms often produce an electronic dipole configuration. The hydrogen atom has the positive end of dipole. If the bonds arise as the result of electrostatic attraction between the atoms. It is known as hydrogen bound. The hydrogen form bounding with electro negative. Ni, O2, Fl2 usually of another molecule. Let us consider the example of H2O molecule. The greater electro-negative of O2. The electrons tend to stay closer to O2 atom than the hydrogen atoms. The O2 atom acts as negative end of the dipole. While H+ atom acts as the positive end. The positive end attracts the negative end of another. H2O molecule and thus bonding the molecules together. Characteristics of Hydrogen bonded materials The bonding is relating strong as compared to other dipole-dipole interactions. The hydrogen bounds are directional. They have low melting points. They are transparent to light. They are soluble in both polar and non-polar solvents. They are good insulators of electricity. They have low densities. 5. Vander Waal’s Bond: Weak and temporary (fluctuating) dipole bounds between hydrogen are known as vandarwaals bounding. They are non-directional. If the symmetrical distribution of electrons around the nucleus is distributed, the center of positive and negative charges may not coincide at that movement giving rise to weak fluctuating dipole. Fig: Van der Waal’s bond Characteristics of van der waal‘s bonded materials The molecular crystals have small binding energy. Molecule structure can be both crystalline and non-crystalline. They are usually transparent to light. They have low melting point. They are soluble in both polar and non-polar solvents. They are good insulators.
- 5. 5 1.4. Variation of Interatomic Forces with Interatomic Spacing: Electrical forces are responsible to binding the atoms gives different solid structures. Magnetic and gravitational forces are negligible in the formation of solids. NaCl crystal is more stable than the Na and Cl atoms. Similarly silicon is more stable than silicon atoms. This shows that silicon atoms attract each other, when they come closer. This is the force which is responsible in the forming are crystal. It means that the energy of crystal is lower than the energy of these energies is called as cohesive energy of the crystal. Sometimes it also called as building energy. Consider two atoms at a separate ion of ‗r‘ where both the attractive and repulsive forces are present. The net force between the atoms is given as, F(r) = 𝑨 𝒓𝑴 - 𝑩 𝒓𝑵 → (1) Where A, B, M, N are constants which are the characteristics of the atom. Fig: Variaration of interatomic force with interatomic spacing The first term in the above equation represents attractive force. Where M approximately equal to 2.i.e. M∝2 (according to inverse quire law). The second term represents repulsive force which is very strong at small distances and the value is 9.N∝9 ∴ N>M The variation of the attractive and repulsive forces with distance of separation r0 Where the separation between atoms = 𝒓𝟎 The resultant force = 0 This separation 𝒓𝟎is called equilibrium distance, at equilibrium position, F(r=𝒓𝟎)=0 𝑨 𝒓𝟎 𝑴 - 𝑩 𝒓𝟎 𝑵 =0 𝑨 𝒓𝟎 𝑴 = 𝑩 𝒓𝟎 𝑵 𝒓𝟎 𝑴 𝒓𝟎 𝑵= 𝑩 𝑨 𝒓𝟎 𝑵−𝑴 = 𝑩 𝑨 ; 𝒓𝟎= 𝑩 𝑨 𝟏 𝑵−𝑴 1.5. Cohesive Energy: The potential energy is due to the interaction between the two atoms and it depends on the interaction spacing. The potential energy is, W(r) = 𝒅𝒖(𝒓) = 𝑭(𝒓) → (1) ∴ F(r) = 𝑨 𝒓𝑴 - 𝑩 𝒓𝑵 → (2) Substitute eq (2) in eq(1) W(r) = 𝑨 𝒓𝑴 − 𝑩 𝒓𝑵 dr
- 6. 6 W(r) = A 𝒓−𝑴 dr – B 𝒓−𝑵 dr W(r) = A 𝒓−𝑴+𝟏 −𝑴+𝟏 - 𝑩𝒓−𝑵+𝟏 −𝑵+𝟏 + C W(r) = 𝑨 𝑴−𝟏 𝟏 𝒓𝑴−𝟏 + 𝑩 𝑵−𝟏 𝟏 𝒓𝑵−𝟏 + ∴ W(r) = - 𝑨 𝑴−𝟏 𝟏 𝒓𝑴−𝟏 + 𝑩 𝑵−𝟏 𝟏 𝒓𝑵−𝟏 + C ∴ 𝑨 𝑴−𝟏 = a; 𝑩 𝑵−𝟏 = b; M-1 = m; N-1 = n ∴ W(r) = −𝒂 𝒓𝒎 + 𝒃 𝒓𝒏 + C Applying boundary conditions r →∝ , W→ 0, we get C=0 ∴W(r) = −𝒂 𝒓𝒎 + 𝒃 𝒓𝒏 → (3) In eq(3), the quality −𝒂 𝒓𝒎 represents attractive potential energy and 𝒃 𝒓𝒏 represents repulsive potential energy. At equilibrium distance (𝒓𝟎) the potential energy should be minimum mathematically it is represented as, 𝒅𝑾 𝒅𝒓 r=0=0 𝒅 𝒅𝒓 −𝒂 𝒓𝟎 𝒎 + 𝒃 𝒓𝟎 𝒏 = 0 -a 𝒅𝒓𝟎 −𝒎 𝒅𝒓 + b 𝒅𝒓𝟎 −𝒏 𝒅𝒓 = 0 -a (-m) 𝒓𝟎 −𝒎−𝟏 + b(-n) ) 𝒓𝟎 −𝒏−𝟏 =0 𝒂𝒎 𝒓𝟎 𝒎+𝟏 - 𝒃𝒏 𝒓𝟎 𝒏+𝟏 𝒂𝒎 𝒓𝟎 −𝒎−𝟏 𝒓𝟎 𝒏+𝟏 =𝒃𝒏 𝒓𝟎 𝒏−𝒎 = 𝒃𝒏 𝒂𝒎 𝒓𝟎= 𝒃𝒏 𝒂𝒎 𝟏 𝒏−𝒎 → (4) Second derivation:- 𝒅𝟐𝑾 𝒅𝒓𝟐 > 0 am 𝒅𝒓𝟎 −𝒎−𝟏 𝒅𝒓 - bn 𝒅𝒓𝟎 −𝒏−𝟏 𝒅𝒓 > 0 am (-m-1)𝒓𝟎 −𝒎−𝟏 - bn(-n-1)𝒓𝟎 −𝒏−𝟏 >0 −𝒂𝒎(𝒎+𝟏) 𝒓𝟎 𝒎+𝟐 + 𝒃𝒏(𝒏+𝟏) 𝒓𝟎 𝒏+𝟐 >0 𝒃𝒏(𝒏+𝟏) 𝒓𝟎 𝒏+𝟐 > 𝒂𝒎(𝒎+𝟏) 𝒓𝟎 𝒎+𝟐
- 7. 7 𝒏 + 𝟏 > 𝒂𝒎 𝒃𝒏 𝒎 + 𝟏 𝒓𝟎 𝒏+𝟐−𝒎−𝟐 (n+1) > 𝒂𝒎 𝒃𝒏 (m+1)𝒓𝟎 𝒏−𝒎 (n+1)> 𝒂𝒎 𝒃𝒏 (m+1) 𝒃𝒏 𝒂𝒎 [i.e from eqn (1)] n+1 > m+1 n>m Calculation of Cohesive Energy Potential energy for two atoms is given as, W(r) = −𝒂 𝒓𝒎 + 𝒃 𝒓𝒏 → (1) When the distance of separation is ′𝒓𝟎‘ the energy is minimum.(𝑾𝒎𝒊𝒏) 𝑾𝒎𝒊𝒏 = −𝒂 𝒓𝟎 𝒎 + 𝒃 𝒓𝟎 𝒏 → (2) For minimum the first derivation is zero. 𝒅𝑾 𝒅𝒓 𝒓=𝒓𝟎 =0 𝒅 𝒅𝒓 −𝒂 𝒓𝟎 𝒎 + 𝒃 𝒓𝟎 𝒏 = 0 -a 𝒅𝒓𝟎 −𝒎 𝒅𝒓 + b 𝒅𝒓𝟎 −𝒏 𝒅𝒓 = 0 𝒂𝒎 𝒓𝟎 𝒎+𝟏 − 𝒃𝒏 𝒓𝟎 𝒏+𝟏 = 0 𝒂𝒎 𝒓𝟎 𝒎+𝟏= 𝒃𝒏 𝒓𝟎 𝒏+𝟏 𝒓𝟎 𝒎−𝒏 = 𝒃𝒏 𝒂𝒎 𝒓𝒐 𝒏 = 𝒓𝒐 𝒏 𝒃𝒏 𝒂𝒎 → (3) Substitute eq (3) in eq(2) 𝐖(𝐫) 𝒓=𝒓𝟎 =𝑾𝒎𝒊𝒏= −𝒂 𝒓𝟎 𝒎 + 𝒃 𝒓𝟎 𝒎 × 𝒂𝒎 𝒃𝒏 ∴ 𝑼𝒎𝒊𝒏= −𝒂 𝒓𝟎 𝒎 𝟏 − 𝒎 𝒏 1.6. Calculation of Cohesive Energy for Ionic Solids: Electrons are transferred from electro positive atoms like Na,Ca,K,Mg to electronegative atoms like O, Fe, ,Cl. Let us consider, ions of charges z1e, z2e separated by a distance ‗r‘. ∴ Attractive Force = 𝒛𝟏𝒆𝒁𝟐𝒆 𝟒𝝅€𝟎𝒓𝟐 For complete crystal, coulomb‘s potential energy Wc = 𝑨𝒛𝟏𝒆𝒁𝟐𝒆 𝟒∏€𝟎𝒓𝟐 Where A modeling‘s constant. Due to overlap of last orbital‘s the negative forces between electrons could be established in a noticeable magnitude. ∴ Repulsive potential energy WR= 𝑩 𝒓𝑵 ∴ The total potential energy of a crystal can be expressed as, W(r) = - 𝑨𝒛𝟏𝒆𝒁𝟐𝒆 𝟒𝝅€𝟎𝒓𝟐 + 𝑩 𝒓𝑵 → (1) For univalent crystal z1= z2=1 W(r) = 𝑨𝒆𝒆 𝟒𝝅€𝟎𝒓𝟐 + 𝑩 𝒓𝑵 At equilibrium, 𝒅𝑼 𝒅𝒓 𝒓=𝒓𝟎 =0 −𝑨𝒆𝟐 𝟒𝝅€𝒐 𝒅𝒓𝟎 −𝟏 𝒅𝒓 + 𝑩𝒅𝒓𝟎 −𝒏 𝒅𝒓 = 0
- 8. 8 𝑨𝒆𝟐 𝟒𝝅€𝟎𝒓𝟎 𝟐 - 𝑩 𝒓𝟎 𝑵+𝟏 = 0 𝑩𝑵 𝒓𝟎 𝑵+𝟏 = 𝑨𝒆𝟐 𝟒∏€𝟎𝒓𝟎 𝟐 B = 𝑨𝒆𝟐𝒓𝟎 𝑵+𝟏−𝟐 𝟒𝝅€𝟎𝑵 B = 𝑨𝒆𝟐𝒓𝟎 𝑵−𝟏 4𝝅€𝟎𝑵 ∴ Minimum potential energy, 𝐖(𝐫) 𝒓=𝒓𝟎 = 𝑼𝒎𝒊𝒏= − 𝑨𝒛𝟏𝒆𝒁𝟐𝒆 𝟒𝝅€𝟎𝒓𝟐 + 𝑩 𝒓𝑵 𝑼𝒎𝒊𝒏= −𝑨𝒆𝟐 𝟒𝝅€𝟎𝒓𝟎 + 𝑨𝒆𝟐𝒓𝟎 𝑵−𝟏 𝟒𝝅€𝟎𝑵𝒓𝟎 𝑵 = −𝑨𝒆𝟐 𝟒𝝅€𝟎𝒓𝟎 + 𝑨𝒆𝟐 𝑵𝟒𝝅€𝟎𝒓𝟎 = −𝑨𝒆𝟐 𝟒𝝅€𝟎𝒓𝟎 𝟏 − 𝟏 𝑵 𝑼𝒎𝒊𝒏= 𝑨𝒆𝟐 𝟒𝝅€𝟎𝒓𝟎 𝟏 𝑵 − 𝟏 In kilometers 𝑼𝒎𝒊𝒏= 𝑨𝒆𝟐𝑵𝑨 𝟒𝝅€𝟎𝒓𝟎 𝟏 𝑵 − 𝟏 Here 𝑵𝑨 → Avogadro Number 1.7. Introduction of Crystallography: Matter exists in three different states namely solids, liquids and gaseous states. In liquid and gaseous states the atoms, molecules or ions move from one place to another and there is no fixed position of atoms in the substance. In solids the position of atom or molecules are fixed and they are may or may not be present at regular intervals of distances in three dimensions. If the atoms or molecules in a solid are periodically at regular intervals of distances in 3 dimensions then such a solids are known as crystalline solids. If the atoms or molecules does not have such periodicity in the substances is known as Amorphous solid. Fig: (a)crystalline solid (b) Amorphous solid If the periodicityof atoms or molecules are extended throughout the solid such a solid known as single crystalline solid.
- 9. 9 Fig: Single crystalline solid If the periodicity of atoms or molecules are up to small regions called grains. And these grains are large in number of different sizes in the solid. Such a materials are known as polycrystalline solids. Fig: polycrystalline solid. Crystalline Solids Amorphous solids The crystalline solids posses‘ regular arrangement of ions or atoms or molecules. The solids have different physical properties (thermal, electrical conductivity, refractive index) in different directions. These are anisotropic in nature. These solids posses elasticity. These posses more density. Eg: calcite, quartz, gold, silver, Al etc... These solids posses complete random arrangement of ions or atoms or molecules. These solids have some physical properties in all the directions. These are isotropic in nature. These solids posses plasticity except rubber. These posses less density. Eg: glass, plastic etc. 1.8. Space lattice (or) Crystal lattice: The atomic arrangement in a crystal is called crystal structure. In crystalline materials, the atoms or molecules are arranged periodically in regular intervals of distances. To explain crystal symmetrieseasily, it is convenient to imagine a points in space about which these atoms are located such points in space are called lattice points and the totality of all such lattice points form space lattice orcrystal l lattice. (or) The geometric arrangement of lattice points, which describes the three dimensions arrangements of atoms or molecules or ions in a crystal, is called space latticeor crystal lattice. If the points are arranged in a single or single column, then the lattice is solid to be linear lattice. If it is arranged in a two dimension then the lattice is called planar lattice. 1.9.1. Basis: The crystal structure is formed by associating with every lattice point a set of assembly of atoms or molecules or ions identical in composition, arrangement, orientation is called as Basis. (Or) A group of atoms or molecules identical in composition is called Basis. Lattice + basis → Crystal structures
- 10. 10 Fig: Space lattice + basis = crystal structure In crystalline solids like aluminium, sodium, copper the basis is single atom. In case of NaCl, NaBr, KCl, KBr the basis is diatomic and in CaF2 the basis is triatomic. 1.9.2. Unit Cell: The smallest portion of space which can generate the complete crystal structure by repeating its own Dimensions in various directions is called a ―Unit Cell‖. (Or) In every crystal some fundamental grouping of particles is repeated. Such fundamental grouping of particles is called a ―Unit Cell‖. Primitive Unit Cell It is defined as Unit cell which contains only one lattice point per unit cell. Multiple Unit Cell (or) Non-Primitive Cell: The unit cell which contains more than one lattice point per unit then it is known as Multiple Unit Cell (or) Non-Primitive Unit Cell. Fig: Unit cell 1.10. Lattice parameters of a Unit Cell: The basic lattice parameters are interfacial angles and primitives. Primitives: The intercepts on x,y,z-axis in a unit cell are called primitives . They are denoted by a,b,c. The primitives will give the knowledge of the size of the unit cell. The interfacial angles give the shape of the unit cell. α β γ ------> interfacial angles a, b, c -------> primitives X,Y,Z -------> crystallographic axis Fig: Unit cell and its lattice parameters
- 11. 11 1.11. Crystal Systems: The crystal systems are divided into seven typed based on the lattice parameters i.e. primitives and Interfacial angles, The different crystal systems are 1). Cubic system 2). Tetragonal system 3). Orthorhombic system 4). Monoclinic system 5). Triclinic system 6). Trigonal system 7). Hexagonal system 1) Cubic system:- In this crystal system all the three edges of the unit cell are equal and right angles to each othera=b=c; α= β= γ=900 Eg :- NaCl , CaF2 etc. Fig: Cubic Crystal Systems 2) Tetragonal system: In this system all the two edges of the unit cell are equal while the third is different. The three axes are mutually perpendicular. a=b≠c; α= β= γ=900 Eg: - TiO2, NiSO4, Sno4 etc. 3). Orthorhombic system: In this crystal system the three edges are different but three axis areperpendicular each other a ≠ b≠ c ; α = β = γ =900 Eg: BaSO4, KNO3 4). Monoclinic system: In this system the edges of the unit cell are different , two axis at right angles and third axis is obliquely inclined. a ≠ b ≠ c; α = β = γ = 900 ≠ γ Eg: Na2B4O710H2O (Borax), CaSO42H2O (Gypsum )
- 12. 12 5). Triclinic system: In this system the edges of unit cell are different and the 3 axis are obliquely inclined to each other a ≠ b ≠ c; α ≠ β ≠ γ ≠ 900 Eg :K2Cr2O7 ,CuSO45H 2O 6). Trigonal system: In this crystal system all the 3 edges of the unit cell are equal and are equallyInclined to each other at an angle other than 900. a = b = c; α = β = γ ≠ 900 Eg: Bismuth, calcite. 7). Hexagonal system: In this system the two axes of the unit cell are equal in length lies in one plate at 1200 with each other and the third axis is perpendicular to this plane . a=b≠c; α=β= 900 ; γ=1200 Eg: SiO2, Mg, Zn 1.12. Bravais Lattices: According to crystal system Bravais showed that there are 14 ways of arranging points in space lattice such that all the lattice points have exactly the same environment. These 14 different lattice types collectively called as Bravais lattices. The different space lattice formed by Primitive (P), Body centered (I), Face centered (F), Base centered(C). S.No Crystal system Bravais lattice No. of lattice system Axial length Interfacial angles examples 1 2 3 4 5 6 7 Cubic Tetragonal Orthorhombic Monoclinic Triclinic Trigonal Hexagonal P,I,F P,I P,I,F,C P,C P P P 3 2 4 2 1 1 1 a=b=c a=b=c a=b=c a=b=c a=b=c a=b=c a=b=c α= β= γ=900 α= β= γ=900 α= β= γ=900 α= β= γ=900 α= β= γ=900 α= β= γ=900 α= β= γ=900 NaCl, TiO2 KNo3 CaSo42H2o K2Cr2O7 calcite SiO2
- 13. 13 1.13 𝐒𝐢𝐦𝐩𝐥𝐞𝐂𝐮𝐛𝐢𝐜 (𝐒𝐂) 𝐒𝐭𝐫𝐮𝐜𝐭𝐮𝐫𝐞: The unit cell contains 8 atoms located at 8 corners is called simple cubic structure. In this structure the all atoms touch each other at their edges. Effective number of atoms in unit cell is 1. i.e. 8 corners have 8 atoms; each corner atom is shared by 8 unit cell. So the share of the each atom towards unit cell is1/8. Therefore n = 8x1/8 N=1. If we consider an atom at one corner as centre it is surrounded by 6 equidistant neighboring atoms. Hence the coordination number of simple cubic is 6. Nearest neighbor distance: In this structure atoms touch each other at their edges. Hence a‘=2r, Where ‗a‘ is Lattice constant. Packing fraction: It is defined as the ratio of volume of all atoms in the unit cell to the volume of unit cell. Volume of the all atoms in the unit cell Therefore packing fraction= ----------------------------------------------- Volume of the unit cell. = nx 𝟒 𝟑 𝛑𝐫𝟑 𝒂𝟑 = 1x 𝟒 𝟑 𝛑𝐫𝟑 𝟖𝒓𝟑 = 𝝅 𝟔 =0.52 = 52% Eg: polonium. 1.14 Body Centered Cubic(BCC) Structure: The structure which 8 atoms are located at 8 corners and one atom is located at centre of body i.e. unit cell such a structure is called body centered cubic structure (BCC). In this structure the corner atoms may not touch each other. But they are in contract with the atom which is located at the centre of the unit cell. Effective number of atoms in BCC is 2 . i.e. 8 atoms are located at 8 corners; one atom is located at centre of the body. So, share of each corner atom towards 8 unit cell is 1/8. Therefore n = (8x 𝟏 𝟖 )+1 = 2. Coordination Number: -If we consider one atom at the centre it is surrounded by 8 equidistant neighboring atoms. Hence the coordination number is 8. Nearest neibhor Distance (a): From the ∆ ABC, (AC) 2 = (AB) 2 + (BC) 2 (AC) 2 = a2 + a2 (AC) 2 =2 a2
- 14. 14 From the ∆ ACD, (AD) 2 = (AC) 2 + (CD) 2 (4r) 2 = 2 a2 + a2 16r2 = 3 a2 4r = √3 a r = √3 a/4 a = 4r/√3. Packing fraction = nx 𝟒 𝟑 𝛑𝐫𝟑 𝒂𝟑 = 2x 𝟒 𝟑 𝛑𝐫𝟑 𝟔𝟒𝒓𝟑 𝟑 𝟑 = √ 𝟑𝝅 𝟖 = 0.68 = 68%. Eg: Na, Fe etc. 1.15 Face Centered Cubic (FCC) structure: The structure in which 8 atoms are located at 8 corners, 6 atoms are located at the centre of the solids.Such a structure is called FCC. In this structure the corner atoms may not touch each other but they are In contract with the atoms located on its faces. Effective Number in F.C.C is 4. i.e. 8 atoms are located at 8 corners, 6 atoms on sides shared by two sides. So they, n = (8× 𝟏 𝟖 )+ (6× ½) = 4. Coordination number is 12. If we consider one atom at a face there are 12 atoms of same distance. Therefore, Coordination Number in FCC is 12. Nearest Distance (a): From the ∆ ABC, (AC) 2 = (AB) 2 + (BC) 2 16r2 = a2 + a2 16r2 = 2 a2 8r2 = a2 2√2r = a r = a/2√2. ∴ Packing fraction =nx 𝟒 𝟑 𝛑𝐫𝟑 𝒂𝟑 = 4x 𝟒 𝟑 𝛑𝐫𝟑 𝟏𝟔 𝟐𝒓𝟑 = 𝝅 𝟑 𝟐 = 0.74 = 74%. Eg: Cu, Al, Ag etc. 1.16.1. Miller Indices: It is possible for defining a system of parallel and equidistant planes which can be imagined to pass through the atoms in a space lattice, such that they include all the atoms in the crystal. Such a system of
- 15. 15 planes is called crystal planes. Many different systems of planes could be identified for a given space lattice. The position of a crystal plane can be specified in terms of three integers called Miller indices.If these are enclosed in ―( )‖ as (h, k, l) then it represent a plane. If they are enclosed in ―[ ]‖ as [h, k, l] then it represent the direction of crystal.Steps to determine Miller Indices for a given plane as shown in the figure. Take a lattice point as origin ‗o‘ of crystallographic axis x, y, z in a space lattice. Let A, B, C be the crystal plane intercepts there axis of 4a, 4b, 3c. Step (1): Determine the coordinate of intercepts made by the plan along the crystallographic axis x y z 4a 4b 3c In general x y z pa qb rc (where p=4, q=4, r=3). Step (2):-- Divide the intercepts with lattice point translational distances along the axis. 4a/a 4b/b 3c/c 4 4 3 In general Pa/a qb/b rc/c P q r Step (3):-- Determine the reciprocals of these numbers i.e. 1/4 1/4 1/3 In general 1/p 1/q 1/r. Step (4):-- Reduce the reciprocals to smallest integers and enclosed them in ―( ) ―[by multiplying with L.C.M] 1/4× 12 1/4× 12 1/3× 12 3 3 4 (3 3 4) Miller Indices may be defined as the reciprocals of the intercepts may by the plane on crystallographic axis when reduced to smallest numbers Important features of Miller indices are, 1. When a plane is parallel to any axis, the intercept of the plane on that axis is infinity. Hence its Miller index for that axis is zero. 2. When the intercept of a plane on any axis is negative, a bar is put on the corresponding Miller index. 3. All equally spaced parallel planes have the same index number (h k l) Ex: The planes ( 1 1 2) and (2 2 4) are parallel to each other 1.16.2. Distance of Separation between Successive planes: Let us consider a rectangular coordinate system with origin ‗o‘ at one of the lattice point. Let (h k l ) be the miller indices of a plane A,B,C which makes intercepts at OA, OB, OC along x, y, z axis respectively. Let OM be the normal passing through the origin and meet the plane A, B, C at N. such that ON = d, the normal makes an angles α, β, γ along x, y, z axis respectively. Fig: Inter planar Spacing The intercepts OA, OB, OC are such that OA = a/h, OB = b/k, OC = c/l →(1) We have, Cos α = ON/OA = d1/(a/h) Cos β = ON/OB = d1/ (b/k) ` → (2) Cos γ = ON/OC = d1/(c/l) We know that, Cos2 α + Cos2 β + Cos 2 γ = 1 → (3) Submit eq (2) in eq (3),
- 16. 16 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 1 1 d a b c h k I h k l d a b c d1 2 = 𝟏 [ 𝒉𝟐 𝒂𝟐 + 𝒌𝟐 𝒃𝟐 + 𝒍𝟐 𝒄𝟐] d1 = 𝟏 𝒉𝟐 𝒂𝟐 + 𝒌𝟐 𝒃𝟐 + 𝒍𝟐 𝒄𝟐 Let ( h/2 k/2 l/2 ) be the miller indices of next plane A‘ B‘ C‘ making intercepts OA‘ OB‘ OC‘ along x, y, z axis respectively. The normal meets the 2nd plane A‘, B‘, C‘ at N‘ such that ON‘ = d2. The extension of d1 to d2the normal makes the same along α, β, γ with x, y, z axis respectively. The intercepts of OA‘, OB‘, OC‘ are such that OA‘ = 𝒂 𝒉 𝟐 OB‘ = 𝒃 𝒌 𝟐 , OC‘ = 𝒄 𝒍 𝟐 Cos α = ON‘/OA‘ = hd2/2 Cos β = ON‘/OB‗= kd2/2b → (4) Cos γ = ON‘/OC‘ = ld2/2c We know that, Cos2 α + Cos2 β + Cos 2 γ = 1 → (5) d2 2 𝒉𝟐 𝟒𝒂𝟐 + 𝒌𝟐 𝟒𝒃𝟐 + 𝒍𝟐 𝟒𝒄𝟐 d2 = 𝟐 𝒉𝟐 𝒂𝟐 + 𝒌𝟐 𝒃𝟐 + 𝒍𝟐 𝒄𝟐 Let the separation between the planes ABC & ABC be‗d ‗. d = d2− d1 =2d1−d1 = d1 d = 𝟏 𝒉𝟐 𝒂𝟐 + 𝒌𝟐 𝒃𝟐 + 𝒍𝟐 𝒄𝟐 For cubic system a = b = c. d = 𝒂 𝒉𝟐+𝒌𝟐+𝒍𝟐 For Tetragonal system a = b ≠ c. d = 𝟏 𝒉𝟐+𝒌𝟐 𝒂𝟐 + 𝒍𝟐 𝒄𝟐 1.17. Structure of Sodium Chloride (NaCl): Sodium chloride crystal is an ionic one. In the NaCl lattice, positively charged sodium ion and negatively charged chlorine ion are situated side by side. It consists of two FCC sub lattices. One of the chlorine ion has its origin at (0, 0, 0) point while sodium ion has its origin midway along the cube edge at (a/2, 0, 0) point as shown in figure. Due to the electro static force between sodium and chlorine ions, the two ions are attracted towards each other. When the two ions come closer their outer electron shells comes closer giving rise to strong forces of repulsion when attraction and repulsion balance, equilibrium is obtained.
- 17. 17 Fig : Structure of Sodium Chloride Each unit cell contains eight corner ions, six face centered ions, twelve ions at halfway of the edges and one at the center of the body. Number of Cl- ions per unit cell is (8X1/8) + (6X1/2) = 1+3 = 4 ions Number of Na+- ions per unit cell is (12X1/4) + 1 = 3+1 = 4 ions Each unit cell has four NaCl molecules. Each Na+ ion has six Cl- and each Cl- has six Na+ ions. The coordination number of NaCl for opposite kind is six and same kind is twelve. The nearest neighbor‘s distance is a/2 for opposite kind and a√2 for the same kind. KCl, KBr, MgO, AgBr etc., have the similar structure as NaCl. 1.18. Diamond Structure: Diamond possesses a structure which is a combination of two interpenetrating FCC sub lattices along the body diagonal at a length of quarter of cube edge. One sub lattice, say x, has its origin at the point (0, 0, 0) and other sub lattice has its origin quarter of the way along the body diagonal i.e., at the point (a/4, a/4, a/4). The fractions such as ¼, ½, ¾ denote the height above the base since the coordination number is 4, the structure is loosely packed. In the unit cell, in addition to the eight corner atoms, there are six face centered atoms and four atoms are located inside the unit cell. Each corner atom is shared by eight adjacent unit cells and each face centered atom is shared by two unit cells. Effective number of atoms per unit cell: In the unit cell in additional to the 8 corner atoms there are 6 face centered atoms and four more atoms located inside the unit cell. Each corner atom is shared by 8 adjacent unit cells and each face centered atom is shared by 2 cells hence the total number of atoms unit cell is (8 × 𝟏 𝟖 ) + (6 × 𝟏 𝟐 ) + 4 = 8 Fig: Diamond Structure Coordination number: In this lattice each atoms has 4 nearest neighbors with it forms covalent bonds. Thus the coordination number of diamond crystal is 4. Atomic Radius: From the diagram (XY) 2 = 𝒂𝟐 𝟏𝟔 + 𝒂𝟐 𝟏𝟔 (XY) 2 = 𝟐𝒂𝟐 𝟏𝟔 = 𝒂𝟐 𝟖 (XZ) 2 = (XY) 2 + (YZ) 2 (XZ) 2 = 𝒂𝟐 𝟖 + 𝒂𝟐 𝟏𝟔 4r2 = 𝟑𝒂𝟐 𝟏𝟔 r 2 = 𝟑𝒂𝟐 𝟔𝟒 r = 𝟑𝒂 𝟖 a = 𝟖𝒓 𝟑
- 18. 18 Nearest neighbor distance: r = 𝟑𝒂 𝟖 2r = 𝟑𝒂 𝟒 Atomic packing fraction: Packing fraction = 𝒏× 𝟒 𝟑 𝝅𝒓𝟑 𝐚𝟑 = 𝟖× 𝟒 𝟑 𝝅𝒓𝟑 𝟖×𝟖×𝟖×𝒓𝟑 𝟑 𝟑 = 𝟑𝝅 𝟏𝟔 = 0.34 = 34% Problems 1) In sodium crystal the equilibrium distance between ion is r0=2.81 A° and A=1.748 Taking n=9. Calculate the potential energy per ion pair (June 2008) Solution: We know that, W0 = −𝑨𝒆𝟐 𝟒∏€𝟎𝒓𝟎 𝟏 − 𝟏 𝑵 W0= 𝟗∗𝟏𝟎𝟗 𝟏.𝟕𝟒𝟖 𝟏.𝟔∗𝟏𝟎−𝟏𝟗 𝟐 𝟐.𝟖𝟏∗𝟏𝟎−𝟏𝟎 𝟏 − 𝟏 𝟗 W= -1.27× 𝟏𝟎−𝟏𝟖 Joule = −𝟏.𝟐𝟕×𝟏𝟎−𝟏𝟖 𝟏.𝟔×𝟏0−𝟏𝟗 eV = -7.97 eV 2) Calculate the binding energy of NaCl of which the nearest neighbor distance is 0.324nm. express the energy in eV and also in KJ/K mol. Modeling constant for NaCl=1.748 and n=9.5 (June 2010) Solution: We know that, W0 = −𝑨𝒆𝟐 𝟒∏€𝟎𝒓𝟎 𝟏 − 𝟏 𝑵 Joule = - (9×𝟏𝟎𝟗 ) 𝟏.𝟕𝟒𝟖∗ 𝟏.𝟔∗𝟏𝟎−𝟗 𝟐 𝟎.𝟑𝟐𝟒∗𝟏𝟎−𝟗 𝟏 − 𝟏 𝟗.𝟓 = -6.9eV The binding energy per mol Wk mol = NAW0= 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟔 ×6.9 eV = 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟔 × (6.9) × (𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 ) Joule = 667× 103 KJ/k mol 3) Calculate the interplanar spacing for (3 2 1) plane in a simple cubic lattice whose lattice contact a = 4.2× 𝟏𝟎−𝟏𝟎 m. Solution: (h k l) = (3 2 1) a = 4.2× 𝟏𝟎−𝟏𝟎 m. 𝒅𝒉𝒌𝒍 = 𝒂 𝒉𝟐+𝒌𝟐+𝒍𝟐 = 𝟒.𝟐×𝟏𝟎−𝟏𝟎𝐦. 𝟗+𝟒+𝟏 = 𝟒.𝟐×𝟏𝟎−𝟏𝟎𝐦. 𝟏𝟒 = 1.13× 𝟏𝟎−𝟏𝟎 m. 4) The atomic radius of copper is 1.278 A0 . It has atomic weight 63.54. Find the density of copper. Solution: we know that, a = 𝟒𝒓 𝟐 = 𝟒×𝟏.𝟐𝟕𝟖 𝟐 = 3.16 A0 ρ = 𝑵𝑴 𝑵𝒂𝟑 = 𝟒×𝟔𝟑.𝟓𝟒 𝟔𝟎.𝟐×𝟏𝟎𝟐𝟑 ×(𝟑.𝟔𝟏×𝟏𝟎−𝟖)𝟑 = 8.98 gm/cm3 = 8980 kg/ m3
- 19. 19 OBJECTIVE QUESTIONS 1) At equilibrium condition the atoms posses [ a ] a) Minimum potential energy b) Maximum potential energy c) Maximum kinetic energy d) Maximum total energy 2) Ionic bonds are mainly formed in [ b ] a) Organic compounds b) Inorganic compounds d) Metals d) None of these 3) The nature of binding for a crystal with alternate and evenly spaced positive and negative ion is [ a ] a) Ionic b) Covalent c) Metallic d) Dipole 4) Ionic solids have [ b ] a) Low melting point b) Moderate melting point c) High melting point d) none of these 5) What is the nature of binding in OH4? [ b ] a) Ionic b) Covalent c) Van der waals d) Metallic 6) Among the following the strongest bond is [ a ] a)Ionic bond b)Covalent bond c)Metallic bond d)Hydrogen bond 7) A crystal is [ c ] (a) A three dimensional representation of a solid (b) A three dimensional regular arrangement of U.C (c) A basis attached to the lattice point (d) All the above 8) The number of lattice points in a primitive cell is [ a ] (a) 1 (b) ½ (c) 2 (d) 3/2 9). The atomic radius of BCC lattice is [ a ] (a) 𝟑𝒂 𝟒 (b) 𝟑𝒂 𝟐 (c) a/2 (d) 𝒂 𝟐 𝟐 10). Ratio of number of atoms per unit cell for Sc, Bcc, Fcc crystal is [ a ] (a) 1:2:4 (b) 2:1:4 (c) 1:2:2 (d) 4:2:1 11). In a simple cubic lattice, 𝒅𝟏𝟎𝟎:𝒅𝟏𝟏𝟎: 𝒅𝟏𝟏𝟏 is [ d ] (a) 6:3:2 (b) 2:3:6 (c) 𝟐: 𝟑: 𝟔 (d) 𝟔: 𝟑: 𝟐 12). The Basis structure of Nacl is [ c ] (a) Simple cubic (b) BCC (c) FCC (d) HCP
- 20. 20 CHAPETER-2 X-RAY DIFFRACTION AND DEFECTS IN CRYSTALS 2.1. Introduction to X – ray Diffraction: X-Rays are electromagnetic waves like ordinary light, therefore, they should exhibit interference and diffraction .Diffraction occurs when waves pass across an object whose dimensions are of the order of their own wavelengths. The wavelength of X-rays is of the order of 0.1nm or 10 -8 cm so that ordinary devices such as ruled diffraction gratings do not produce observable effects with X-rays. Laue suggested that a crystal which consisted of a 3-dimensional array of regularly spaced atoms could serve the purpose of a grating. The crystal differs from ordinary grating in the sense that the diffracting centers in the crystal are not in one plane. Hence the crystal acts as a space grating rather than a plane grating. There are 3 main X-ray diffraction methods by which the crystal structure can be analyzed. Laue Method – for single crystal Powder Method- for finely divided crystalline or polycrystalline powder Rotating crystal Method - for single crystal 2.2. Bragg′s Law: The monochromatic x-ray beam of wavelength λ is incident at an angle θ to Bragg‘s planes. Let the inter planer 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 scattering for the X-ray incident on it. Fig: X-Ray Scattering by Crystal Apart of the incident x-ray beam AB incident an atom at B in plane 1 is scattered along the direction BC. Similarly apart of incident x-ray DE fall on atom at E in plane 2 and is scattered in the direction EF and it is parallel to BC.Let the beam AB and DE make an angle θ with the Bragg‘s planes. This angle θ is called the angle of diffraction or glancing angle. If the path difference between the rays ABC & DEF is equal to nλ, than the rays reflected from constructive plane or phase. So constructive interference takes place among the reflected ray BC and EF, then the path difference between the rays is, PE+QE=nλ From the Δle PBE, 𝐬𝐢𝐧 𝜽 = 𝑷𝑬 𝑩𝑬 = 𝑷𝑬 𝒅 PE = d 𝐬𝐢𝐧 𝜽 From the Δle QBE, 𝐬𝐢𝐧 𝜽 = 𝑬𝑸 𝑬𝑩 = 𝑬𝑸 𝒅 EQ = d 𝐬𝐢𝐧 𝜽 For constructive interference, PE + EQ = nλ d 𝐬𝐢𝐧 𝜽 + d𝐬𝐢𝐧 𝜽 = nλ 2d 𝐬𝐢𝐧 𝜽 = nλ The above equation is called Bragg‘s Law. 2.3. Laue method: S1& S2 are two lead screens in which two pin holes act as slits. X-ray beam from an X –ray tube is allowed to pass through these two slits S1& S2. The beam transmitted through S2will be a narrow pencil of X – rays . The beam proceeds further to fall on a single crystal such that Zinc blended (ZnS) which is mounted suitably on a support . The single crystal acts as a 3 – dimensional diffraction grating
- 21. 21 to the incident beam. Thus, the beam undergoes diffraction in the crystal and then falls on the photographic film. The diffracted waves undergo constructive interference in certain directions, and fall on the photographic film with reinforced intensity. In all other directions, the interference will be destructive and the photographic film remains unaffected. The resultant interference pattern due to diffraction through the crystal as a whole will be recorded on the photographic film (which requires many hours of exposure to the incident beam). When the film is developed, it reveals a pattern of fine spots, known as Laue spots. The distribution spots follow a particular way of arrangement that is the characteristic of the specimen used in the form of crystal to diffract the beam. The Laue spot photograph obtained by diffracting the beam at several orientations of the crystal to the incident beam are used for determining the symmetry and orientations of the internal arrangement of atoms, molecules in the crystal lattice . it is also used to study the imperfections in the crystal . 2.4. Powder Method (or) Debye Scherrer method: X-ray powder method is usually carried out for polycrystalline materials. The given polycrystalline material is grounded to fine powder and this powder is taken in capillary tube. This tube is made up of non diffracting material, and fixed at the centre of cylindrical Debye Scherrer cylindrical camera as shown in figure. The principle under this technique is that millions of tiny crystals in powder have completely random orientation all the possible diffraction phases are available for Bragg reflection to takes place. All the orientations are equally reflected ray will form a cone. whole axis likes along the direction of incident beam and whose semi vertical angle is twice the glancing angle for that particular planes. Fig: Powder method - apparatus The different cones intercept the film in a series of concentric circular from the radial of these arcs. The angle can be calculated and hence the spacing between the atoms can be evaluated as shown in figure. The photographic film is in cylindrical shape, whose axis is perpendicular to the beam. Let s be the distance of particular arc from the centre. Let R be the radius of camera. Then 4θ = 𝑺 𝑹 θ = 𝑺 𝟒𝑹 If S1, S2, S3, are the distances between symmetrical lines on the stretched film, then θ1 = 𝑺𝟏 𝟒𝑹 θ2 = 𝑺𝟐 𝟒𝑹 ………………………. Using these values of θn in Bragg‘s equation nλ =2d Sinθn.
- 22. 22 Where, n=1, 2, 3, … is the Order of diffraction d= Interplanar spacing θn = Angle of diffraction for nth order. The inter-planar spacing d can be calculated. 2.5. Introduction to Defects in crystals: The structural imperfections of crystal can be classified on the basis of their geometric structures as point defect, line defect, surface defect and volume defect. 2.6. Point defects: As the name indicates, these defects are at same point in the crystal .so these are called as point defects point defects are also called as zero dimensional defects. The common types of point defect in a solid material are grouped into 3 categories. 1. Lattice site defect: a) Vacancies (Schottky defect ) b) Interstitial vacancies (Frenkel defect ) 2. Compositional defect: a) Substitutional impurity defect b) Interstitial impurity defect 3. Electronic defect” 1. Lattice site defects: In this type of defect, some atoms may not present in their regular atomic sites (or) positions. They are, a) Vacancies: As shown in the diagram at a point one or two or three atoms are missed. This is referred as single or double or treble vacancies. The vacancies are formed due to imperfect pattern during crystallization or due to thermal vibration at high temperature. Fig: Vacancy Schottky defect: In an ionic crystal if a action vacancy exist, then in the very nearby place. On anion vacancy will also exist that is moved to surface of the crystal. So that change neutrality is maintained in vacancy region as shown in diagram. This is known as Schottky defect. Eg: Crystals such as NaCl, KCl, KBr etc, Fig: Schottky Defect a) Interstitial vacancies: If an atom is moved to an interstitial space in the crystal then the defect is known as interstitial defect. Frenkel defect In an ionic crystal, if a cation (positive ion) moves to an interstitial space so that a vacancy is formed in its atomic position Here change neutrality is maintained in the defect region as shown in the diagram. This type of defect is known as Frenkel defect.
- 23. 23 Eg; crystal such as CaF2, AgCl, AgI, AgBr etc. 1. 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 are in the sites of regular parent atoms are in the interstitial spaces. These defects are described as follows. a) Substitutional Impurity defect:As shown in diagram during crystallization free foreign atoms Occupies the regular parental positions or atomic sites. Eg: In extrinsic semiconductors either trivalent or pentavalent atoms he sites of silicon or germanium atoms. Fig: Substitutional impurity b) Interstitial impurity defect: The spaces between the parents the parental atoms in the crystal are known as interstitial spaces small sized (lower atomic number) atoms, such as hydrogen etc. may fit in these interstitial spaces. These atoms are known as interstitial atoms and the defect formed due to presence of interstitial atoms is known as interstitial impurity defect. Fig: Interstitial impurity defect 1. Electronic defect: At absolute zero in a purely covalent crystal (silicon) the electrons are tightly bound to the core and all are set to be in the value band. Absolute zero sum of the electrons likely to occupy. So in the crystal of pure silicon some temperature .so in the crystal of pure of silicon some of the electrons from the covalent bonds get thermally released and become free to move. In this way the deficiency of an electron creates a hole. Then the electrons & holes give rise to electronic imperfection. 2.7. Line defects: If a crystal plane ends somewhere in the crystal, that along the edge of the incomplete. Plane produces defect in the crystal called line defect the line defect is of two types, a) Edge dislocation b) Screw dislocation
- 24. 24 a) Edge dislocation: In a perfect crystal , the atoms arranged in both vertical and horizontal plane parallel to the side faces as shown in figure if one of these vertical planes does not extend to full length but ends in between within the crystal. Then it is called edge dislocation therefore in this defect, a line of atoms is not in proper removed. Figure: Edge dislocation The following points are observed.: i) In the perfect crystal, the atoms are in equilibrium position. ii) In the imperfect crystal, just above the discontinuity the atoms are prell hard and are in a stat of completion. iii) Just below the discontinuity, the atoms are pulled apart and are in a state of tension. iv) This distorted configuration extends all along the edge inside the crystal, perpendicular to the cross section. v) Edge dislocations are represented by perpendicular or T depending on whether the incomplete plane stale from the top or from the bottom of the crystal. These two configurations are referred as positive or negative edge dislocation respectively. Burger’s Vector: To understand the concept of burgers vector let us consider two crystal one perfect and the other edge dislocation as shown figure Figure: Burger’s Vector The burgers vectors of a dislocation are determined by tracing a burgers circuit. Let us consider the burgers circuit in case of perfect crystal. The burgers circuit is realized by drawing a rectangular from point p. starting from point p, we go up by 6 steps to a point Q then move towards right by 6 steps to a point R, further move down by 6 steps to a point S. Finally move towards left by 6 steps to reach the starting point p. This gives a closed circuit PQRS known as Burger circuit. b) Screw dislocation: Figure shows when a part of the crystal is displaced relative to the rest of the crystal. It is important to mention here that the displacement terminates with the crystal. Screw dislocation is defined as displacement of the atoms in one part of a crystal relative to the rest of the crystal. The screw dislocation forms a ramp around the dislocation line. The Buggers vector is also in the figure. Buggers vector in parallel dislocation line. Normally the real dislocations in the crystal are mixture of edge quell screw dislocation.
- 25. 25 Figure: Screw dislocation Problems 1) Calculate the ratio d100:d110:d111 for a single cubic structure. Sol: d = 𝒂 𝒉𝟐+ 𝒌𝟐+ 𝑙𝟐 d100 = 𝒂 𝟏 d110 = 𝒂 𝟐 d111= 𝒂 𝟑 𝒂 𝟏 : 𝒂 𝟐 : 𝒂 𝟑 = 1: 𝟏 𝟐 : 𝟏 𝟑 = √6:√3:√2 2) The Bragg‘s angle in the 1st order for [2, 2, 0] reflection from Ni (BCC) is 38.2o . When x-rays of wavelength λ=1.54 Ao are employed in a diffraction experiment. Determine the lattice parameter ofNi. Sol: (h, k, l) = (2, 2, 0) Θ=38.2o λ=1.54 Ao a=? 2d sinθ = nλ 20×d×𝐬𝐢𝐧( 𝟑𝟖. 𝟐)=1×1.54×10-10 d=1.245Ao d= 𝒂 𝒉𝟐+ 𝒌𝟐+ 𝒍𝟐 1.245×10-12 = 𝒂 𝟖 a= 𝟖 × 1.245 × 10-12 a=3.52Ao 3) Monochromatic x-rays of λ=1.5Ao is incident on a crystal phase having an inter planar spacing of 1.6Ao . Find the highest order for which Bragg‘s reflection maximum can be seen. Sol: d=1.6 Ao , θ=90o , λ=1.5 Ao , n=? We know that,nλ = 2d sinθ For the highest order of diffraction n = nmax and θ = 90o nmax ×1.5 = 2(1.6) sin90o nmax = 2.13 ∴ Highest order of reflection is nmax=2 4) Calculate the glancing angle at (110) plane of a cubic crystal having axial lengtha=0.2nm.corresponding to the 2nd order diffraction maximum for the x-rays of wavelength 0.065nm
- 26. 26 Solution: (h, k, l) = (1, 1, 0) a = 0.26m λ=0.065nm We know that 2d sinθ = nλ d = 𝒂 𝒉𝟐+ 𝒌𝟐+ 𝒍𝟐 d= 𝟎.𝟐𝟔 𝟐 d=0.18 2(0.18)sinθ=2(0.065) 0.368sinθ=0.13 Sinθ= 𝟎.𝟏𝟑 𝟎.𝟑𝟔𝟖 Θ=200 41‘ 13‘‘ 5) The Bragg‘s angle for reflection from the (111) plane in a BCC crystal is 19.20 . For an X-ray of wavelength 1.54 A0 . Compute the cube edge of the unit cell. [June 2010] Solution: θ=19.2, order = 1 2d 𝐬𝐢𝐧 𝜽 = nλ 2d×sin (19.2) = 1(1.54) 2d × 0.32 = 1.54 d = 2.341 A0 We know that, d = 𝒂 𝟑 a = 4.05 A0 6) What is the angle at which the 3rd order reflection of x-rays of 0.79A0 wavelength can occur in a crystal of 3.04X10-8 cm? Sol: 2d 𝐬𝐢𝐧 𝜽= nλ λ=3.04×10-8 cm 𝐬𝐢𝐧 𝜽 = 𝟎.𝟕𝟗𝐗𝟏𝟎−𝟖 𝟐𝑿𝟑.𝟎𝟒𝑿𝟏𝟎−𝟖 θ = 22. 940 7) A beam of x-ray is incident on an ionic crystal with lattice spacing 0.313nm. Calculate the wavelength of x- rays, if the first order Bragg‘s reflection takes place at a glancing angle of 70 48‘ .[June 2011] Sol: 2d Sinθ = nλ Given data, d=0.313nm n=1 θ= 70 48‘ λ = 𝟐𝒅𝒔𝒊𝒏𝜽 𝒏 λ = 𝟐𝑿𝟎.𝟑𝟏𝟑𝑿𝟏𝟎−𝟗𝑿𝟎.𝟏𝟑𝟓𝟕 𝟏 λ = 0.0836nm. Objective Type Questions 1) X-ray are used for crystal diffraction studies because [ c ] a) They have higher penetrating power b) Crystals are transparent to x-rays c) The inter atomic spacing is of the order of x-ray wavelength d) They have high resolving power 2) Inter atomic spacing is of the order of [ a ] a) 2 to 3 A0 b) 2 to 3µm c) 2 to 3nm d) 2 to 3mm
- 27. 27 3) For simple cubic the ratio 𝟏 𝒅𝟏𝟎𝟎 : 𝟏 𝒅𝟏𝟏𝟎 : 𝟏 𝒅𝟏𝟏𝟏 is given by [ d ] a) 𝟏: 𝟏 𝟐 : 𝟑 b) 1: 𝟐: 𝟑 𝟐 c) 1:√2: 𝟏 𝟑 d)√1:√2:√3 4) For body centered cubic the ratio 𝟏 𝒅𝟏𝟎𝟎 : 𝟏 𝒅𝟏𝟏𝟎 : 𝟏 𝒅𝟏𝟏𝟏 is given by [ a ] a) 𝟏: 𝟏 𝟐 : 𝟑 b) 1: 𝟐: 𝟑 𝟐 c) 1:√2: 𝟏 𝟑 d) √1:√2:√3 5) For face centered cubic ratio 𝟏 𝒅𝟏𝟎𝟎 : 𝟏 𝒅𝟏𝟏𝟎 : 𝟏 𝒅𝟏𝟏𝟏 is given by [ c ] a) 𝟏: 𝟏 𝟐 : 𝟑 b) 1: 𝟐: 𝟑 𝟐 c) 1:√2: 𝟏 𝟑 d)√1:√2:√3 6) Substitutional impurity is [ a ] (a) Point defect (b) line defect (c) Surface defect (d) volume defect 7) Interstitial impurity is [ d ] (a) Surface defect (b) volume defect(c) Line defect (d) point defect 8) These are point defects [ d ] (a) Vacancies (b) interstitial atoms (c) substitutional (d) all of these 9) Point defects influence [ d ] (a) Mechanical properties (b) optical properties (c) Electrical properties (d) all of these 10) Frenkel and Schottky defect are exhibited [ c ] (a) by metals (b) ceramics (c) ionic crystals (d) b and c 11) Change in density is caused by [ b ] (a) Frenkel defects (b) Schottky defect (c) Edge dislocation (d) screw dislocation 12) One dimensional imperfections are called [ b ] (a) Point defects (b) line defect (c) Surface defect (d) volume defect 13) Screw dislocation belongs to [ b ] (a) Point defects (b) line defect (c) Surface defect (d) volume defect 14) Burger‘s vector define [ c ] (a) The magnitude of point defect (b) The orientation of different crystal planes vertically (c) The managitude and direction of dislocation (d) None of these 15) A grain boundary is ----------------------- [ c ] (a) A point defect (b) The combination of edge dislocation and screw dislocation (c) The region where the crystal orientation changes sharply (d) The region where the crystal orientation changes gradually 16) The number of vacancies in a crystal [ d ] (a) First increases and then decreases (b) Does not change with temperature (c) Decrease as the temperature increases (d) Increase as the temperature increases
- 28. 28 UNIT-II CHAPTER-3 PRINCIPLES OF QUANTUM MECHANICS 3.1.1Wave and particle duality: A particle has mass, it is located at some definite point, it can move from one place to another, it gives energy when slowed down or stopped. Thus, the particle is specified by 1. Mass m 2. Velocity v 3. Momentum p and 4. Energy E. A wave is spread out over a relatively large region of space , it cannot be said to be located just here and there, it is hard to think of mass being associated with a wave. Actually a wave is nothing but a rather spread out disturbance. A wave is specified by its (1) Frequency (2) Wavelength, (3) Phase of wave velocity, (4) Amplitude and (5) Intensity Considering the above facts, it appears difficult to accept the conflicting ideas that radiation has a dual nature, i.e., radiation is a wave which is spread out over space and also a particle which is localized at a point in space. However, this acceptance is essential because radiation sometimes behaves as a wave and at other times as a particle as explained below: (1) Radiations including visible light, infra-red, ultraviolet, X-rays, etc. behave as waves in experiments based on interference, diffraction, etc. This is due to the fact that these phenomena require the presence of two waves at the same position at the same time. Obviously, it is difficult for the two particles to occupy the same position at the same time. Thus, we conclude that radiations behave like wave.. (2) Planck‘s quantum theory was successful in explaining black body radiation, the photo electric effect, the Compton Effect, etc. and had clearly established that the radiant energy, in its interaction with matter, behaves as though it consists of corpuscles. Here radiation interacts with matter in the form of photon or quanta. Thus, we conclude that radiations behave like particle. 3.1.2de–Broglie Hypothesis: In the Newton time, matter and radiation both were assumed to consist of particles. With the discovery of phenomena like interference, diffraction and polarization it was established that light is a kind of a wave motion. In the beginning of 20th century some new phenomena (photoelectric effect, Compton effect, etc.) were discovered which could not be explained on the basis of wave theory. These phenomena were explained on the basis of quantum theory in which light quanta or photons are endowed corpuscular properties – mass (hν/c2 ), velocity v and momentum hν/c. But, when the photon theory was applied to phenomenon such as interference, diffraction, etc. (which have been fully explained on wave theory) it proved helpless to explain them. Thus, light has a dual nature, i.e., it possesses both particle and wave properties. In some phenomena corpuscular nature predominates while in others wave nature predominates. The manifest of properties depends upon the conditions under which the particular phenomena occur. But wave and particle never expected to appear together. Louis de- Broglie in 1924 extended the wave particle parallelism of light radiations to all the fundamental entities of Physics such as electrons, protons, neutrons, atoms and molecules etc. He put a bold suggestion that the correspondence between wave and particle should not confine only to electromagnetic radiation, but it should also be valid for material practices, i.e. like radiation, matter also has a dual (i.e., particle like and wave like ) character. In his doctoral thesis de-Broglie wrote that there is an intimate connection between waves and corpuscles not only in the case of radiation but also in the case of matter. A moving particle is always associated with the wave and the particle is controlled by waves. This suggestion was based on the fact that nature loves symmetry, if radiation like light can act like wave some times and like a particle at other times, then the material particles (e.g., electron, neutron, etc.) should act as waves at some other times. These waves associated with particles are named de- Broglie waves or matter waves. 3.1.3Expression for de- Broglie wavelength: The expression of the wavelength associated with a material particle can be derived on the analogy of radiation as follows: Considering the plank‘s theory of radiation, the energy of photon (quantum) is
- 29. 29 E = h𝝊 = 𝒉𝒄 𝝀 → (1) Where c is the velocity of light in vacuum and 𝝀 is its wave length. According to Einstein energy – mass relation E = mc2 → (2) 𝝀 = 𝒉 𝒎𝒄 = 𝒉 𝒑 → (3) Where mc = p is momentum associated with photon. If we consider the case of material particle of mass m and moving with a velocity v , i.e momentum mv, then the wave length associated with this particle ( in analogy to wave length associated with photon ) is given by 𝝀 = 𝒉 𝒎𝒗 = 𝒉 𝒑 → (4) Different expressions for de-Broglie wavelength (a) If E is the kinetic energy of the material particle then E = 𝟏 𝟐 mv2 = 𝟏 𝟐 𝒎𝟐𝒗𝟐 𝒎 = 𝒑𝟐 𝟐𝒎 ⟹ p2 = 2mE or p = 𝟐𝒎𝑬 Therefore de- Broglie wave length 𝝀 = 𝒉 𝟐𝒎𝑬 → (5) (b) When a charged particle carrying a charge ‗q‘ is accelerated by potential difference v, then its kinetic energy K .E is given by E = qV Hence the de-Broglie wavelength associated with this particle is 𝝀 = 𝒉 𝟐𝒎𝒒𝑽 → (6) For an electron q = 1.602×10-19 Mass m = 9.1 X 10-31 kg ∴ 𝝀 = 𝟔. 𝟔𝟐𝟔 × 𝟏𝟎−𝟑𝟒 𝟐 × 𝟗. 𝟏 × 𝟏𝟎−𝟑𝟏 × 𝟏. 𝟔𝟎𝟐 × 𝟏𝟎−𝟏𝟗𝑽 = 𝟏𝟓𝟎 𝑽 = 𝟏𝟐.𝟐𝟔 𝑽 A0 → (7) 3.1.4Properties of Matter Waves: Following are the properties of matter waves: (a) Lighter is the particle, greater is the wavelength associated with it. (b) Smaller is the velocity of the particle, greater is the wavelength associated with it. (c) When v = 0, then 𝝀 = ∞ , i.e. wave becomes indeterminate and if v = ∞ then 𝝀 = 0. This shows that matter waves are generated only when material particles are in motion. (d) Matter waves are produced whether the particles are charged particles or not (𝝀 = 𝒉 𝒎𝒗 is independent of charge). i.e., matter waves are not electromagnetic waves but they are a new kind of waves. (e) It can be shown that the matter waves can travel faster than light i.e. the velocity of matter waves can be greater than the velocity of light. (f) No single phenomenon exhibits both particle nature and wave nature simultaneously.
- 30. 30 (g) 3.1.5 Distinction between matter waves and electromagnetic waves: S. No Matter Waves Electromagnetic Waves 1 Matter waves are associated with moving particles (charged or uncharged) Electromagnetic waves are produced only by accelerated charged particles. 2 Wavelength depends on the mass of the particle and its velocity,𝝀 = 𝒉 𝒎𝒗 Wavelength depends on the energy of photon 3 Matter waves can travel with a velocity greater than the velocity of light. Travel with velocity of light c= 3×108 m/s 4 Matter wave is not electromagnetic wave. Electric field and magnetic field oscillate perpendicular to each other. 5 Matter wave require medium for propagation, i.e, they cannot travel through vacuum. Electromagnetic waves do not require any medium for propagation, i.e., they can pass through vacuum 3.2.1Davisson and Germer’s Experiment: The first experimental evidence of matter waves was given by two American physicists, Davisson and Germer in 1927. The experimental arrangement is shown in figure 3.1(a). The apparatus consists of an electron gun G where the electrons are produced. When the filament of electron gun is heated to dull red electrons are emitted due to thermionic emissions. Now, the electrons are accelerated in the electric field of known potential difference. These electrons are collimated by suitable slits to obtain a fine beam which is then directed to fall on a large single crystal of nickel, known as target T which is rotated about an angle along the direction of the beam is detected by an electron detector (Faraday cylinder) which is connected to a galvanometer. The Faraday cylinder ‗c‘ can move on a circular graduated scale s between 290 c to 900 to receive the scattered electrons. Fig 3.1(a) Davisson and Germer‘s experimental arrangement for verification of matter waves First of all, the accelerating potential V is given a low value and the crystal is set at any orbital azimuth (θ). Now the Faraday cylinder is moved to various positions on the scale‘s‘ and galvanometer current is measured for each position. A graph is plotted between galvanometer current against angle θ between incident beam and beam entering the cylinder [Figure3.1(b)]. The observations are repeated for different acceleration potentials.
- 31. 31 Fig. 3.1(b) Variation of Galvanometer current with variation of angle θ between incident beam and beam entering the cylinder It is observed that a ‗bump‘ begins to appear in the curve for 44 volts. Following points are observed. (a) With increasing potential, the bump moves upwards. (b) The bump becomes most prominent in the curve for 54 volts at θ = 500 . (c) At higher potentials, the bumps gradually disappear. The bump in its most prominent state verifies the existence of electron waves. According to de- Broglie, the wavelength associated with electron accelerated through a potential V is given by 𝝀 = 𝟏𝟐.𝟐𝟔 𝑽 A0 . Hence, the wavelength associated with an electron accelerated through 54 volt is𝝀 = 𝟏𝟐.𝟐𝟔 𝟓𝟒 = 1.67 A0 From X-ray analysis, it is known that a nickel crystal acts as a plane diffraction grating with space d = 0.91 A0 [see Figure 3.1(c)]. According to experiment, we have diffracted electron beam at θ = 500 . The corresponding angle of incidence relative to the family of Bragg plane θ1 = 𝟏𝟖𝟎−𝟓𝟎 𝟓𝟒 = 650 Using Bragg‘s equation (taking n=1), we have 𝝀 = 2dsinθ= 2(0.91A0 ) 𝐬𝐢𝐧 𝟔𝟓0 This is in good agreement with the wavelength computed from de-Broglie hypothesis. Fig. 3.1(c) Bragg planes in Nickel crystal As the two values are in good agreement, hence, confirms the de-Broglie concept of matter waves. 3.2.2G .P. Thomson′s Experiment: The experimental arrangement is as shown in Figure 3.2. High energy electron beam produced by the cathode ‗C‘ is accelerated with a potential up to 50kV. A fine pencil of accelerated beam is obtained by allowing it to pass through a narrow slit S and is made to fall on a very thin metallic film F of gold (or silver or aluminum). The electron beam is scattered in different directions by the metallic film and incident on photographic plate P. The entire apparatus is exhausted to a high vacuum so that the electrons may not lose their energy in collision with the molecules of air. Fig3.2: (a) G.P.Thomson‘s Apparatus, (b) pattern recorded on the photographic plate Since ordinary metals like gold are micro crystalline in structure, the diffraction pattern produced by them are similar in appearance to the X- ray diffraction powered pattern and consist of a series of well- defined concentric rings about a central spot as shown in figure 3.2(b). To make sure that this pattern is due to electrons and not due to any possible X - rays generated, the cathode rays in the discharge tube are deflected by magnetic field. It was observed that the diffraction pattern observed on the fluorescent screen placed instead of photographic plate also shifted. This confirmed that the pattern is due to the electrons. Thomson calculated the wavelength of the de-Broglie waves associated with the cathode rays using the equation 𝝀 = 𝟏𝟐.𝟐𝟔 𝑽 A0 and determined the spacing between the planes in the foils using Bragg‘s equation. The results obtained are in good agreement with those from X-ray studies.
- 32. 32 3.3 Heisenberg′s Uncertainty Principle: As a direct consequence of the dual nature of matter, in 1927, Heisenberg proposed a very interesting principle known as uncertainty principle. If a particle is moving, based on classical mechanics, at any instant we can find its momentum at any position. In wave mechanics, a moving particle can be regarded as a wave group. The particle that corresponds to this wave group may be located anywhere within the group at any given time. In the middle of the group, the probability of finding the particle is more but the probability of finding the particle at any other point inside the wave group is not zero. Narrower the wave group higher will be the accuracy of locating the particle. At the same time, one cannot define the wavelength 𝝀 of the wave accurately when the wave group is narrower. Since ( 𝝀 = 𝒉 𝒎𝒗 ), measurement of particles momentum (mv = 𝒉 𝝀 ) also becomes less accurate. Fig3.3 (a) A narrow de – Broglie wave group. Since the position can be precisely determined, Δ x is small. As the measurement of is less accurate, Δ p is large. (b) A wide de –Broglie wave group. Measurement of λ is accurate and hence Δ p is small whereas, since wave group is wide, Δ x is large On the other hand, when we consider a wide wave group, wavelength 𝝀can be well defined hence measurement of momentum becomes more accurate. At the same time, since the width of the wave group is large, locating the position of the particle becomes less accurate. Thus the uncertainty principle can be stated as ―it is impossible to know both exact position and exact momentum of an object at the same time‖. If ∆x and ∆p are the uncertainties in the position and momentum respectively when they are simultaneously measured, then ∆x ∆p≥ 𝒉 𝟒𝝅 Another form of the uncertainty concerns energy and time. In an atomic process let energy ‗E‘ be emitted during the time interval ∆t, If the energy is emitted in the form of electromagnetic waves, we cannot measure frequency 𝝊 of the waves accurately in the limited time available . Let the minimum uncertainty in the number of waves that we count in a wave group be one wave. Since frequency = 𝒏𝒖𝒎𝒃𝒆𝒓𝒐𝒇𝒘𝒂𝒗𝒆𝒔 𝒕𝒊𝒎𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 ∆ 𝝊 = 𝟏 ∆𝐭 Hence the corresponding uncertainty in energy ∆E = h ∆𝝊 ∆E ≥ 𝒉 ∆𝒕 or ∆E ∆t ≥ 𝒉 A more precise calculation based on the nature of wave of wave groups modifies this result to ∆E ∆t ≥ 𝒉 𝟒𝝅 This gives uncertainty in the measurement of energy and time of a process. Thus the more generalized statement of Heisenberg‘s uncertainty principle is ―It is impossible to specify precisely and simultaneously the values of both members of particular pair of physical variables that describe the behavior of an atomic system‖. 3.4.1 Schrodinger's time independent wave equation: Schrodinger developed a differential equation whose solutions yield the possible wave functions that can be associated with a particle in a given situation. This equation is popularly known as Schrodinger equation. The equation tells us how the wave function changes as a result of forces acting on the particle. One of its forms can be derived by simply incorporating the de-Broglie wavelength expression into the classical wave equation. If a particle of mass ‗m‘ moving with velocity v is associated with a group of waves, let 𝝍 be the wave function of the particle. Also let us consider a simple form of progressing wave represented by the equation 𝝍 = 𝝍0𝐬𝐢𝐧(𝝎𝐭 − 𝐤𝐱) → (1) Where 𝝍 = 𝝍 (x, t)𝝍0 is amplitude Differentiating eq (1) partially with respect to ‗x‘, we get 𝝏𝝍 𝝏𝒙 = -K𝝍0𝐜𝐨𝐬(𝝎𝐭 − 𝐤𝐱)
- 33. 33 Again differentiating equation (1) with respect to ‗x‘ 𝝏𝟐𝝍 𝝏𝒙𝟐 = −K2 𝝍0 sin (𝝎t-kx) 𝝏𝟐𝝍 𝝏𝒙𝟐 = − k2 𝝍 𝝏𝟐𝝍 𝝏𝒙𝟐 + k2 𝝍 = 0 → (2) Since k = 𝟐𝝅 𝝀 , 𝝏𝟐𝝍 𝝏𝒙𝟐 + 𝟒𝝅𝟐 𝝀𝟐 𝝍 = 0 → (3) Eq (2) or Eq (3) is the differential form of the classical wave equation. Now, incorporating de- Broglie wavelength expression 𝝀 = 𝒉 𝒎𝒗 in to eq (3), we get 𝝏𝟐𝝍 𝝏𝒙𝟐 + 𝟒𝝅𝟐𝒎𝟐𝒗𝟐 𝒉𝟐 𝝍 = 0 → (4) The total energy E of the particle is the sum of its kinetic energy k and potential energy V i.e., E = K + V But K = 𝟏 𝟐 mv2 ∴ E = 𝟏 𝟐 mv2 + V 𝟏 𝟐 mv2 = E – V m2 v2 = 2m (E - V) → (5) Substituting eq (5) in eq (4), we get 𝝏𝟐𝝍 𝝏𝒙𝟐 + 𝟖𝝅𝟐𝐦(𝐄−𝐕) 𝒉𝟐 𝝍 = 0 → (6) In quantum mechanics, the value 𝒉 𝟐𝝅 occurs most frequently. Hence we denote ђ = 𝒉 𝟐𝝅 using this notation, we have 𝝏𝟐𝝍 𝝏𝒙𝟐 + 𝟐𝐦(𝐄−𝐕) ђ𝟐 𝝍 = 0 → (7) For simplicity, we have considered only one dimensional wave extending eq(7) for a3 – dimensional wave 𝝏𝟐𝝍 𝝏𝒙𝟐 + 𝝏𝟐𝝍 𝝏𝒚𝟐 + 𝝏𝟐𝝍 𝝏𝒛𝟐 + 𝟐𝐦(𝐄−𝐕) ђ𝟐 𝝍 = 0 → (8) Where 𝝍 (x, y, z); here, we have considered only stationary states of 𝝍after separating the time dependence of 𝝍 The Laplacian operator is defined as 𝛁2 = 𝝏𝟐 𝝏𝒙𝟐 + 𝝏𝟐 𝝏𝒚𝟐 + 𝝏𝟐 𝝏𝒛𝟐 → (9) Hence eq (10) can be written as 𝛁2 𝝍 + 𝟐𝐦(𝐄−𝐕) ђ𝟐 𝝍 = 0 → (10) This is Schrodinger wave equation. Since time factor doesn‘t appear, eq(8) or eq(10) is called ‗time independent Schrodinger wave equation‘ in three dimensions. 3.4.2 Physical significance of wave function𝝍: (1) The wave function 𝝍has no direct physical meaning. It is a complex quantity representing the variation of matter wave. (2) It connects the practical nature and its associated wave nature statically. (3) |𝝍 |2 (or 𝝍𝝍 * if function is complex) at a point is proportional to the probability of finding the particle at that point at any given time. The probability density at any point is represented by |𝝍|2 . (4) If the particle is present in a volume dxdydz, then |𝝍 |2 dxdydz =1 If a particle is present somewhere in space
- 34. 34 𝝍𝟐 𝐝𝐱𝐝𝐲𝐝𝐳 = 𝟏 ∞ −∞ 𝐎𝐫 𝝍𝝍 ∗ 𝐝𝐱𝐝𝐲𝐝𝐳 = 𝟏 ∞ −∞ The wave function satisfying the above condition is said to be normalized. 3.5.1 Particle in Infinite square potential well: A free electron trapped in a metal or charge carriers trapped by barriers trapped by the potential barriers of a double hetero junction can be approximated by an electron in an infinitely deep one- dimensional potential well. Consider one – dimensional potential well of width L as shown in fig. Let the potential V = 0 inside well and V = ∞ outside the well. Fig.3.4 Square potential well infinite height The time independent Schrödinger wave equation in one dimensional case 𝒅𝟐𝝍 𝒅𝒙𝟐 + 𝟐𝐦(𝐄−𝐕) ђ 𝝍 = 0 → (1) For a particle present inside the well where V=0 and 𝝍 = 𝝍(𝒙) 𝒅𝟐𝝍 𝒅𝒙𝟐 + 𝟐𝐦𝐄 ђ𝟐 𝝍 = 0 → (2) Let the general solution of eq (2) be 𝝍(𝒙) = A 𝐬𝐢𝐧 𝒌𝒙 + B 𝐬𝐢𝐧 𝒌𝒙 → (3) Where A and B are constants which can be determined from boundary conditions 𝝍(𝒙) = 0 at x = 0 And 𝝍(𝒙) = 0 at x = L → (4) Since 𝝍(𝒙) = 0 at x = 0 0= A 𝐬𝐢𝐧 𝒌(𝟎) + B 𝐜𝐨𝐬 𝒌(𝟎) ⟹ B = 0 → (5) Since 𝝍(𝒙) = 0 at x = L 0 = A 𝐬𝐢𝐧 𝒌𝑳 Which means A =0 or 𝐬𝐢𝐧 𝒌𝑳 = 0 since both A and B cannot be zero, A≠ 0. If A = 0, then 𝝍 = 𝟎 everywhere. This means that the particle is not in the well. The only meaningful way to satisfy the condition is 𝐬i𝐧 𝒌𝑳 = 0, or kL = nπ ; n = 1,2,3,… ∴ k = 𝐧𝛑 𝑳 → (6) Thus, eq (3) simplifies to 𝝍(𝒙) = A 𝐬𝐢𝐧 𝐧𝛑 𝑳 𝐱 → (7) Differentiating 𝝍 in eq (7) 𝒅𝝍 𝒅𝒙 = A 𝐧𝛑 𝑳 𝐜𝐨𝐬 𝐧𝛑 𝑳 𝐱 Again Differentiating, we get 𝒅𝟐𝝍 𝒅𝒙𝟐 = − A 𝐧𝟐𝛑𝟐 𝑳𝟐 𝐬𝐢𝐧 𝐧𝛑 𝑳 𝐱 𝒅𝟐𝝍 𝒅𝒙𝟐 = − 𝐧𝟐𝛑𝟐 𝑳𝟐 𝝍 = 0
- 35. 35 𝒅𝟐𝝍 𝒅𝒙𝟐 + 𝐧𝟐𝛑𝟐 𝑳𝟐 𝝍 = 0 → (8) Comparing eq (2) and eq (8), we get 𝟐𝐦𝐄 ђ𝟐 = 𝐧𝟐𝛑𝟐 𝑳𝟐 = k2 E = 𝐧𝟐𝛑𝟐ђ𝟐 𝟐𝒎𝑳𝟐 n is called the quantum number. Thus we obtain an important result. The particle cannot possess any value of energy as assumed in classical case, but it possesses only discrete set of energy values. The energy of the nth quantum level, En = 𝐧𝟐𝛑𝟐ђ𝟐 𝟐𝒎𝑳𝟐 = 𝐧𝟐𝐡𝟐 𝟖𝒎𝑳𝟐 (since ђ = 𝒉 𝟐𝝅 ) → (9) The wave functions and the corresponding energy levels of the particles are as suggested in Figure 3.5 Fig.3.5 Ground state and first two excited states of an electron in a potential well: a) the electron wave functions and b) the corresponding probability density functions. The energies of these three states are shown on the right. We are still left with an arbitrary constant ‗A‘ in eq (7). It can be obtained by applying normalization condition i.e.; the probability of finding the particle inside the box is unity. 𝝍𝒙 𝟐 𝒅𝒙 = 𝟏 𝑳 𝟎 𝑨𝟐 𝒔𝒊𝒏𝟐 𝒏𝝅𝒙 𝑳 𝒅𝒙 𝑳 𝟎 = 1 A2 𝟏 𝟐 𝑳 𝟎 𝟏 − 𝒄𝒐𝒔 𝟐𝒏𝝅 𝒍 𝒙 𝒅𝒙 = 𝟏 𝑨𝟐 𝟐 [x − 𝑳 𝟐𝝅𝒏 𝒔𝒊𝒏 𝟐𝝅𝒏𝒙 𝑳 ]L 0 = 1 ⟹ 𝑨𝟐 𝟐 𝑳 − 𝟎 − 𝟎 − 𝟎 = 𝟏 𝑨𝟐𝑳 𝟐 = 1 or A = 𝟐 𝑳 → 10) ∴The normalized wave function is 𝝍n = 𝟐 𝑳 sin 𝒏𝝅𝒙 𝑳 → (11) 3.5.2Particle in Three dimensional potential box: Let us consider the case of a single particle, i.e., a gas molecule of mass m, confined within a rectangular box with edges parallel to X, Y and Z axes as shown in figure. Fig 3.6 Three dimensional potential box
- 36. 36 Let the sides of the rectangular box be a, b and c respectively. The particle can move freely within the region 0 < x< a, 0 < y < b and 0 < z < c, i.e., inside the box where potential V is zero, i.e., V (x, y, z) = 0, for 0 < x< a V (x, y, z) = 0, for 0 < y < b and V (x, y, z) = 0, for 0 < z < c The potential rises suddenly to have a large value at the boundaries, i.e., the potential outside the box is infinite. The Schrodinger wave equation inside the box is given by 𝝏𝟐𝝍 𝝏𝒙𝟐 + 𝝏𝟐𝝍 𝝏𝒚𝟐 + 𝝏𝟐𝝍 𝝏𝒛𝟐 + 𝟐𝒎 ℏ𝟐 E𝝍 = 0 → (1) This is partial differential equation in three independent variables and may be solved by the method of separation of variables. The solution of equation (1) is of the form 𝝍 (x, y, z) = X (x) Y(y) Z (z) → (2) Where X (x) is a function of x alone, Y(y) is a function of y alone and Z (z) is a function of z alone. Differentiating 𝝍 in eq (2) partially with respect to ‗x‘, we get 𝝏𝝍 𝝏𝒙 = 𝝏𝑿(𝒙) 𝝏𝒙 Y(y) Z (z) → (3) Again differentiating eq (3) partially with respect to ‗x‘, we get 𝝏𝟐𝝍 𝝏𝒙𝟐 = 𝝏𝟐𝑿(𝒙) 𝝏𝒙𝟐 Y(y) Z (z) 𝝏𝟐𝝍 𝝏𝒙𝟐 = 𝝏𝟐𝑿(𝒙) 𝝏𝒙𝟐 𝐗 (𝐱) 𝐘(𝐲) 𝐙 (𝐳) 𝑿(𝒙) 𝝏𝟐𝝍 𝝏𝒙𝟐 = 𝟏 𝑿 𝝏𝟐𝑿 𝝏𝒙𝟐 𝝍 → (4) Where X = X(x) and 𝝍 = 𝝍 (x, y, z) Similarly, 𝝏𝟐𝝍 𝝏𝒚𝟐 = 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 𝝍 → (5) And 𝝏𝟐𝝍 𝝏𝒛𝟐 = 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 𝝍 → (6) Substituting eqs.(4),(5) and (6) in eq (1), we get 𝟏 𝑿 𝝏𝟐𝑿 𝝏𝒙𝟐 𝝍 + 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 𝝍 + 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 𝝍 + 𝟐𝒎 ℏ𝟐 E𝝍 = 0 𝟏 𝑿 𝝏𝟐𝑿 𝝏𝒙𝟐 + 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 + 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 + 𝟐𝒎 ℏ𝟐 E = 0 → (7) This equation can be written as 𝟏 𝑿 𝝏𝟐𝑿 𝝏𝒙𝟐 = − 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 − 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 − 𝟐𝒎𝑬 ℏ𝟐 → (8) The left hand side of eq (8) is a function of x alone, while the right hand side is a function of y and z and is independent of x. both sides are equal to each other. Both sides are equal to each other. This is possible only when they are separately equal to a constant quantity, i.e. 𝟏 𝑿 𝝏𝟐𝑿 𝝏𝒙𝟐 = 𝒌𝒙 → (9) And − 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 − 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 − 𝟐𝒎𝑬 ℏ𝟐 = 𝒌𝒙 Or 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 = − 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 − 𝟐𝒎𝑬 ℏ𝟐 − 𝒌𝒙 → (10) In eq (10), the left hand side is a function of y alone while right hand side is a function of z and is independent of y. If the above equation is to be satisfied, both sides must be equal to a constant say𝒌𝒚, i.e. 𝟏 𝒀 𝝏𝟐𝒀 𝝏𝒚𝟐 = 𝒌𝒚 → (11) and − 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 − 𝟐𝒎𝑬 ℏ𝟐 − 𝒌𝒙 = 𝒌𝒚 Or 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 = − 𝟐𝒎𝑬 ℏ𝟐 − 𝒌𝒙 − 𝒌𝒚 → (12)
- 37. 37 Again, we have 𝟏 𝒁 𝝏𝟐𝒁 𝝏𝒛𝟐 = 𝒌𝒛 → (13) And − 𝟐𝒎𝑬 ℏ𝟐 − 𝒌𝒙 − 𝒌𝒚 = 𝒌𝒛 Or 𝟐𝒎𝑬 ℏ𝟐 = −𝒌𝒙 − 𝒌𝒚 − 𝒌𝒛 → (14) For convenience, introduce 𝒌𝒙 = − 𝟐𝒎𝑬𝒙 ℏ𝟐 , 𝒌𝒚 =− 𝟐𝒎𝑬𝒚 ℏ𝟐 and 𝒌𝒛 = − 𝟐𝒎𝒛 ℏ𝟐 Now, the differential equations in x, y and z coordinates may be written as 𝝏𝟐𝑿 𝝏𝒙𝟐 + 𝟐𝒎𝑬𝒙 ℏ𝟐 X = 0 → (15a) 𝝏𝟐𝒀 𝝏𝒚𝟐 + 𝟐𝒎𝑬𝒚 ℏ𝟐 Y = 0 → (15b) And 𝝏𝟐𝒁 𝝏𝒛𝟐 + 𝟐𝒎𝑬𝒛 ℏ𝟐 Z = 0 → (15c) The general solution of eq [(15a)] will be a sine function of the arbitrary amplitude, frequency and phase, i.e., X(x) = A sin (Bx + C) → (16) Where A, B and C are constants whose values are determined by boundary conditions. 𝝍 2 represent the probability of finding the particle at any point within the box. Therefore, 𝑿(𝒙) 2 which is a function of x coordinates only represents the probability of Finding the particle at any point along the X-axis. As the potential is very high at the walls of the box, the probability of finding the particle at the walls will be zero, i.e., 𝑿(𝒙) 2 = 0 when x = 0 and x= a 𝑿(𝒙) = 0 when x = 0 and x= a Using the above boundary conditions in eq (16), we have 0 = A sin (0 + C); A≠ 0 ∴ Sin C = 0 and 0 = A 𝒔𝒊𝒏 (Ba +C) Or 0 = 𝐬𝐢𝐧 𝑩𝒂. 𝐜𝐨𝐬 𝑪+ 𝐜𝐨𝐬 𝑩𝒂. 𝐬𝐢𝐧 𝑪 0 = 𝐬𝐢𝐧 𝑩𝒂 𝐜𝐨𝐬 𝑪 ∴Sin Ba = 0 [since 𝐬𝐢𝐧 𝑪 = 0, 𝐜𝐨𝐬 𝑪 is not zero] ↔ Ba = 𝒏𝒙π Or B = 𝒏𝒙𝛑 𝒂 → (17) ∴ X(x) = A Sin 𝒏𝒙𝛑 𝒂 x → (18) Applying the normalization condition between x=0 to x=a, we have 𝑿(𝒙) 𝒂 𝟎 2 dx = 1 𝐀𝐒𝐢𝐧 𝒏𝒙𝛑 𝒂 𝐱 𝒂 𝟎 2 dx = 1 𝑨𝟐 𝐒𝐢𝐧𝟐 𝒏𝒙𝛑 𝒂 𝐱 𝒂 𝟎 dx = 1 𝑨𝟐 𝟐 𝟏 − 𝑪𝒐𝒔 𝟐𝒏𝒙𝛑 𝒂 𝐱 𝒂 𝟎 𝒅x = 1 𝑨𝟐 𝟐 [ 𝒙 − 𝐒𝐢𝐧 𝟐𝒏𝒙𝛑 𝒂 𝟐𝒏𝒙𝛑 𝒂 ]𝟎 𝒂 = 1 𝑨𝟐 𝟐 [a −0] = 1 𝑨𝟐 = 𝟐 𝒂 or A = 𝟐 𝒂 Therefore, X(x) = 𝟐 𝒂 𝐒𝐢𝐧 𝒏𝒙𝛑𝐱 𝒂 → (19) Similarly, we can solve equations [15(b)] and [15(c)] to obtain
- 38. 38 Y(y) = 𝟐 𝒃 𝐒𝐢𝐧 𝒏𝒚𝛑𝐲 𝒃 → (20) Z (z) = 𝟐 𝒄 𝐒𝐢𝐧 𝒏𝒛𝛑𝐳 𝒄 → (21) The complete wave function 𝝍𝒏𝒙 , 𝒏𝒚 , 𝒏𝒛 has the form 𝝍𝒏𝒙 , 𝒏𝒚 , 𝒏𝒛 = 𝟐 𝒂 𝐒𝐢𝐧 𝒏𝒙𝛑 𝒂 𝐱 𝟐 𝒃 𝐒𝐢𝐧 𝒏𝒚𝛑 𝒃 𝐘 𝟐 𝒄 𝐒𝐢𝐧 𝒏𝒛𝛑 𝒄 𝐙 = 𝟐 𝟐 𝒂𝒃𝒄 𝐒𝐢𝐧 𝒏𝒙𝛑 𝒂 𝐱 𝟐 𝒃 𝐒𝐢𝐧 𝒏𝒚𝛑 𝒃 𝐘 𝟐 𝒄 𝐒𝐢𝐧 𝒏𝒛𝛑 𝒄 𝐙 → (22) The wave function of a particle in a finite box and its probability density are shown in Figure 3.7. Fig 3.7 a) wave functions and b) probability densities of a particle in three dimensional box Differentiating eq (19) twice with respect to ‗x‘, we get 𝝏𝟐𝑿 𝝏𝒙𝟐 = −[ 𝝅𝒏𝒙 𝒂 ]𝟐 𝟐 𝒂 𝐒𝐢𝐧 𝒏𝒙𝛑𝐱 𝒂 = −[ 𝝅𝒏𝒙 𝒂 ]𝟐 X(x) → (23) Substituting eq (23) in eq [15(a)], we get −[ 𝝅𝒏𝒙 𝒂 ]𝟐 X(x) + 𝟐𝒎𝑬𝒙 ℏ𝟐 X (x) = 0 Or 𝟐𝒎𝑬𝒙 ℏ𝟐 = [ 𝝅𝒏𝒙 𝒂 ]𝟐 Or 𝑬𝒙 = 𝝅𝟐𝒏𝒙 𝟐 𝒂𝟐 𝒉𝟐 𝟒𝝅𝟐 𝟏 𝟐𝒎 𝑬𝒙 = 𝒉𝟐𝒏𝒙 𝟐 𝟖𝒎𝒂𝟐 → (24) Similarly, 𝑬𝒚= 𝒉𝟐𝒏𝒚 𝟐 𝟖𝒎𝒃𝟐 → (25) And 𝑬𝒛 = 𝒉𝟐𝒏𝒛 𝟐 𝟖𝒎𝒄𝟐 → (26) The allowed values of total energy are given by E = 𝑬𝒙 + 𝑬𝒚 + 𝑬𝒛 = 𝒉𝟐 𝟖𝒎𝟐 [ 𝒏𝒙 𝟐 𝒂𝟐 + 𝒏𝒚 𝟐 𝒃𝟐 + 𝒏𝒛 𝟐 𝒄𝟐 ] → (27) Where𝒏𝒙, 𝒏𝒚 and 𝒏𝒛 denote any set of three positive numbers. When the box is a cube, i.e., a = b= c, the energy expression is given by E = 𝒉𝟐 𝟖𝒎𝒂𝟐 [ 𝒏𝒙 𝟐+ 𝒏𝒚 𝟐+ 𝒏𝒛 𝟐 𝒂𝟐 ] → (28) With𝒏𝒙, 𝒏𝒚, 𝒏𝒛 = 1,2,3……..
- 39. 39 CHAPTER – 4 ELEMENTS OF STATISTICAL MECHANICS & ELECTRON THEORY OF SOLIDS 4.0Introduction to Elements of Statistical mechanics: Up to the end of seventeenth century, the physical phenomena were explained with the help of ordinary laws of classical mechanics. Consider the following situations in which the system consists of a large number of particles. i) Consider the case of a gas consisting of 1023 molecules. In order to obtain the complete information about the motion of this system, we have to solve 1023 equations or more. Moreover, it is not possible to have a complete knowledge regarding the positions and velocities of all the molecules. Therefore, it is impossible to apply ordinary laws of mechanics to such a physical system. ii) Consider the radioactive decay of an atom. In this case, one cannot say which atom of the radioactive material will decay first and when. iii) Consider the case of an atom with two electrons present in it. To consider its behavior, no one has solved it completely by the use of ordinary laws of mechanics. Therefore, it is impossible to solve the problem of a system consisting of a large number of particles with the help of ordinary laws of mechanics. Such problems have been successfully solved by the method of statistical mechanics. Statistical mechanics is a branch of science which deals with the relationship between the overall behavior of a system of many particles and the properties of the particles. It is not concerned with the dynamics or interactions of individual particles, but it is concerned with the happening to the entire system. Since many phenomena in the physical world involve systems of very large number of particles, statistical approach is being made. Statistical mechanics can be applied to classical systems such as molecules in a gas as well as photons in a cavity and free electrons in a metal. 4.1. Phase Space: Let us consider a system consisting of N particles distributed in a given volume V. when we consider a static system, all the particles are fixed at various points in space. With three coordinates X, Y and Z mutually perpendicular to each other, we can specify the location of any particles in three dimensional space. Since there are N particles, 3N coordinates give complete information about such a system. The three dimensional space in which the location of a particle is completely specified by the three coordinates is known as position space. A small volume dV in position space is given by dV = dxdydz If the system is dynamic, the particles move with various velocities and possess moments. Hence to specify such a system, in addition to three position coordinates x, y and z, we need three components of momentum px, py and pz which are given by px = m.vx, py = m.vy and pz = m.vy Where vx, vy and vz are the three velocity components. The three dimensional space in which the momentum of a particle is completely specified by the three momentum coordinates px, py and pz is known as momentum space. A small volume element d𝜞 is given by d𝜞 = dpx.dpy.dpz A combination of the position space and momentum space is known as phase space. A point phase space is completely specified by six coordinates x, y, z, px, py and pz. Since there are N particles, 6N coordinates provide complete information regarding the position and momentum of all the N particles in the phase space in a dynamic system. A small volume dτ is given by dτ = dxdydz.dpxdpydpz i.e., dτ = dV d𝜞 Total phase space volume of the system is obtained by integrating the above equation i.e., τ = 𝒅𝝉 = 𝒅𝑽 𝒅𝜞 τ = V𝜞 Where V is the physical volume occupied by all the particles of the system and Π is the momentum space volume occupied by all the particles. 4.2.0 Ensembles: A system is defined as a collection of a number of particles. An ensemble is defined as a collection of a large number of macroscopically identical, but essentially independent systems. By the term
- 40. 40 microscopically identical we mean that each of the systems constituting an ensemble satisfies the same macroscopic conditions, e.g., volume, energy, pressure, total number of particles etc. By the term independent systems we mean that the systems constituting an ensemble are mutually non-interacting. In an ensemble the systems play the same role as the non-interacting molecules in a gas. 4.2.1The Canonical ensemble: In the canonical ensemble, the assemblies are having the same volume V, number of particles N and are in thermal contact with each other so that they are in thermal equilibrium and have the same temperature T. Fig4.1.The canonical ensemble 4.2.2. The Grand Canonical Ensemble: The grand canonical ensemble is a collection of independent assemblies having the same temperature T, volume V and the chemical potential μ. Fig4.2. the Grand Canonical ensemble 4.2.3. The Micro Canonical Ensemble: The ensemble in which a system has the same fixed energy, fixed volume and also the same number of particles is called micro canonical ensemble. Fig4.3.The Micro Canonical ensemble
- 41. 41 4.3.0 Statistical Distribution: Statistical mechanics determines the most probable way of distribution of total energy E among the N particles of a system in thermal equilibrium at absolute temperature. It establishes how many particles are likely to have the energy Є1, how many particles are likely to have the energy Є2, and so on. In statistical mechanics, one finds the number of ways W in which the particles can be arranged among the available states. If g(Є) is the number of states of energy Є and f(Є) is the distribution function i.e., the probability of occupancy of each state of energy Є, then the number of particles of energy Є is given by n(Є) = g(Є) f(Є) When the energy distribution is continuous g (Є) is replaced by g (Є) dЄ. Three kinds of distributions are possible corresponding to three different kinds of particles. Maxwell- Boltzmann Statistics Bose – Einstein Statistics Fermi – Dirac Statistics 4.3.1 Maxwell – Boltzmann Statistics: Maxwell – Boltzmann Statistics is also known as classical statistics. The main assumptions of Maxwell – Boltzmann statistics are: (i) The particles are identical and distinguishable (ii) The volume of each phase space cell chosen is extremely small and hence chosen volume has very large number of cells. (iii) Since cells are extremely small, each cell can have either one particle or no particle. (iv) The system is isolated i.e., the total number of particles of the system and their energy remain constant. (v) The particles are distributed among the energy levels that are spread from zero to infinity on the energy scale i.e., energy levels are continuous. (vi) Molecules of gases follow Maxwell – Boltzmann statistics. (vii) The number of particles with energies between Є and Є + dЄ in a sample of a gas that contains N molecules is given by n(Є) dЄ = 𝟐𝜫𝑵 (𝜫𝒌𝑻) 𝟑 𝟐 Є𝒆 −Є 𝒌𝑻 dЄ 4.3.2 Bose – Einstein Statistics: Bose – Einstein Statistics is also known as Quantum statistics. In Bose – Einstein Statistics, the following assumptions are made. (i) The particles are identical, indistinguishable and have integral spin. These particles are called Bosons. (ii) Bosons obey uncertainty principle. (iii) Any number of Bosons can occupy a single cell in phase space. (iv) Bosons do not obey Pauli‘s exclusion principle. (v) Wave functions representing Bosons are symmetric. i.e., Ψ(2,1) = Ψ(1,2) (vi) The wave functions of Bosons do overlap slightly. i.e., Weak interaction exists. (vii) Energy states are discrete. (viii) The probability that a Boson occupies a state of energy Є is given by f B-E(Є) = 𝟏 𝒆 (𝜶+ Є 𝒌𝑻 ) −𝟏 (ix) This is called Bose – Einstein distribution function. (x) The examples of particles obeying Bose- Einstein statistics are Photons, phonons, pions, gluons, He4 etc. 4.3.3Fermi – Dirac Statistics: Fermi – Dirac Statistics is also known as Quantum statistics. The main assumptions are: (i) The particles obeying Fermi –Dirac statistics are identical, indistinguishable and have half integral spin. These particles are known as Fermions. (ii) Fermions obey Pauli‘s exclusion principle. (i.e. There cannot be more than one particle in a single cell in phase space). (iii) Fermions obey uncertainty principle. (iv) Energy states are discrete. (v) Wave functions representing Fermions are antisymmetric. i.e., Ψ(2,1) = -Ψ(1,2) (vi) Weak interaction exists between the particles. (vii) Electrons, protons, neutrons and He3 are the examples of Fermions. (viii) The probability that a Fermion occupies a state of energy Є is given byfF-D (Є) = 𝟏 𝒆 𝜶+ Є 𝒌𝑻 + 𝟏
- 42. 42 4.4.0 Black body and its radiation: A perfectly black body is one which completely absorbs all the radiations of all wavelengths incident on it. Since it neither reflects nor transmits any radiation, it appears black whatever the color of incident radiation is. According to Kirchhoff‘s law, a body which is capable of absorbing radiation must also be capable of emitting radiation of all possible wavelengths. So, a perfectly black body is a good absorber as well as a good radiator. When it is heated to a suitable temperature, it emits radiations of all wavelengths (continuous spectrum). The wavelength of emitted radiation by a black body depends only on its temperature and is independent of the material of the body. The distribution of energy in black body radiation for different wavelengths and at various temperatures was determined experimentally by Lummer and Pringsheim 1899. The intensity of radiation corresponding to different wavelengths is measured at different temperatures and is plotted as shown in figure. Fig4.4 Energy distribution in black body radiation From this study, it is observed that: 1) The emission from a black body at any temperature is composed of radiation from all wavelengths. 2) At a given temperature, the energy is not uniformly distributed. As the temperature of the black body increases, the intensity of radiation for each wavelength increases. This shows that the total amount of energy radiated per unit area per unit time increases with increase of temperature. 3) The total energy of radiation at any temperature is given by the area between the curve corresponding to that temperature and the horizontal axis. 4) The amount of radiant energy emitted is small at very short and very long wavelengths. At a particular temperature, the spectral radiancy Eλ is maximum at a particular temperature. 5) The wavelength corresponding to the maximum energy represented by the peak of the curve shifts towards shorter wavelengths as the temperature increases. This is called Wien‘s displacement law. According to this law, λmT = constant. 4.4.1 Laws of black body radiation: Wien’s law Wien showed that the maximum energy point in the black body energy distribution shifts towards shorter wavelengths when the temperature of the body is raised. He showed that λmT = constant Where λm is the wavelengthcorresponding to maximum energy emission from a black body at absolute temperature T. He also showed that the maximum energy emitted by a black body is proportional to the fifth power of its absolute temperature. (Eλ) max ∞ T5 or (Eλ) max/ T5 = constant. Wien by applying Maxwell‘s law for distribution of velocities and the principle of equipartition of kinetic energy gave the expression for Eλ as Eλ = C1λ-5 𝒆 −𝑪𝟐 𝝀𝑻 Where C1 and C2 are constants. This law agrees with experimental values at shorter wavelengths.
- 43. 43 Rayleigh-Jean’s Law: According to Rayleigh-Jean‘s law, the energy distribution in the thermal spectrum is given by Eλ = 𝟖𝝅𝒌𝑻 𝝀𝟒 This law agrees with experimental values at longer wavelengths only. Planck’s radiation law: In 1901, Max Plank derived a theoretical expression for the energy distribution in the black body on the basis of quantum theory of radiation. He made the following assumptions: 1) A black body radiator contains simple harmonic oscillators of possible frequencies. 2) The oscillators cannot emit or absorb energy continuously. Emission or Absorption of energy takes place in discrete amounts, i.e., energy of oscillator is quantized. 3) The energy of an atomic oscillator of frequency ν can have only certain values like 0, hν, 2hν, 3hν… This is an integral multiple of a small unit of energy hν called the quantum or photon. Let the total energy E be distributed among N number of particles. Let there be N0, N1, N2, N3, Nr…etc oscillators having energy Є, 2 Є, 3Є, + …. rЄ, ….. Etc. respectively. Now we have N = N0 + N1+ N2+ ….. + Nr + …. . → (1) And E =0xN0 + Є N1+ 2 Є N2+ ….. + rЄ Nr + …. . → (2) Where Є = hν According to Boltzmann‘s distribution formula, the number of oscillators having energy rЄ is given by Nr = N0𝒆( −𝒓Є 𝒌𝑻 ) → (3) Where k is Boltzmann‘s constant. Substituting the values of N1, N2, N3 … from eq (3) in eq (1), we get N = N0 𝒆(− Є 𝒌𝑻 ) + N0𝒆(− 𝟐Є 𝒌𝑻 ) + N0𝒆(− 𝟑Є 𝒌𝑻 ) + ….. + N0 𝒆(− 𝒓Є 𝒌𝑻 ) + ….. → (4) = N0 [1 + 𝒆(− Є 𝒌𝑻 ) + 𝒆(− 𝟐Є 𝒌𝑻 ) + 𝒆(− 𝟑Є 𝒌𝑻 ) + ….. +𝒆(− 𝒓Є 𝒌𝑻 ) + …..…. Let x = 𝒆(− Є 𝒌𝑻 ) N = N0 [1 + x + x2 + x3 + …… xr + ……… = 𝑵𝟎 𝟏−𝒙 → (5) [since 1 + x + x2 + x3 + …… xr + ……… = 𝟏 𝟏−𝒙 ] Substituting the values of N1, N2, N3 … from eq (3) in eq (2), we get E = N0 (0) + Є N0x + 2Є N0x2 + 3Є N0x3 + …….. rЄxr +…… = Є N0x [1+ 2x + 3x2 + …….. + rxr-1 + …….. = Є𝒙𝑵𝟎 (𝟏−𝒙)𝟐 [since 1+ 2x + 3x2 + …….. + rxr-1 + …….. = 𝟏 (𝟏−𝒙)𝟐 ] = Є𝒙 (𝟏+𝒙) 𝑵𝟎 (𝟏−𝒙) = 𝑵Є𝒙 𝟏−𝒙 [From eq (5)] = 𝑵Є 𝟏−𝒙 𝒙 = 𝑵Є 𝒙−𝟏−𝟏 → (6) Now the average energy of an oscillator is given by ⋶ = 𝑬 𝑵 = Є 𝒙−𝟏−𝟏 = 𝒉𝝂 𝒆 ( 𝒉𝝂 𝒌𝑻 ) −𝟏 → (7) The number of photons present in the wavelength range λ and λ+ dλ per unit volume (i.e. photon density) = 8π λ-4 kT → (8) Hence, total energy of photons within the wavelength range λ and λ+ dλ is given by
- 44. 44 E λ d λ = 8π λ-4 kT 𝒉𝝂 𝒆 ( 𝒉𝝂 𝒌𝑻 ) −𝟏 = 𝟖𝝅𝒉𝒄/𝝀 𝝀𝟒[𝒆 𝒉𝝂 𝒌𝑻−𝟏] E λ d λ = 𝟖𝝅𝒉𝒄 𝝀𝟓[𝒆 𝒉𝝂 𝒌𝑻−𝟏] = 𝟖𝝅𝒉𝒄 𝝀𝟓[𝒆𝝀 𝒉𝒄 𝒌𝑻−𝟏] → (9) This is called Planck‘s radiation law. Deduction of Wien‘s law and Rayleigh Jean‘s law from Planck‘s radiation law Case (i):For shorter wavelengths, i.e., when λ is very small, 𝒆 𝒉𝒄 𝝀𝒌𝑻>> 1, i.e., ‗1‘ in the denominator on the right hand side of eq(9) can be neglected in comparison to 𝒆 𝒉𝒄 𝝀𝒌𝑻. ∴ E λ d λ = 𝟖𝝅𝒉𝒄 𝝀𝟓 𝟏 𝒆 (𝒉𝒄 𝝀𝒌𝑻) E λ d λ = C1λ-5 𝒆 −𝑪𝟐 𝝀𝑻 d λ → (10) Where C1 = 8πhc and C2 = 𝒉𝒄 𝒌 are constants. This is Wien‘s law which agrees with experiment at short wavelengths. Case (ii): For longer wavelengths, i.e., when λ is very large, 𝒉𝒄 𝝀𝒌𝑻 is small and 𝒆 𝒉𝒄 𝝀𝒌𝑻 can be expanded as 𝒆 𝒉𝒄 𝝀𝒌𝑻 = 1 + 1 1! 𝒉𝒄 𝝀𝒌𝑻 + 𝟏 𝟐! ( 𝒉𝒄 𝝀𝒌𝑻 )𝟐 + …………. = 1 + 𝒉𝒄 𝝀𝒌𝑻 [neglecting higher order terms] Eλd λ = 𝟖𝝅𝒉𝒄 𝝀𝟓 𝒅𝝀 [𝟏+ 𝒉𝒄 𝝀𝒌𝑻 −𝟏] = 𝟖𝝅𝒌𝑻 𝝀𝟒 d λ → (11) This is Raleigh-Jean‘s formula which agrees with experiment at longer wavelengths. Therefore, Planck‘s radiation law is valid for all wavelengths. 4.5. Density of energy states: The density of states (DOS) of a system describes the number of energy states at each energy level that are available to be occupied. A high density of states at a specific energy level s means that there are many states available for occupation. If the density of states is zero, no states can be occupied at that energy level. The number of energy states with a particular energy value depends on how many combinations of the quantum number s result in the same value n. Let us consider a sphere of radius n (where n2 =nx 2 +ny 2 +nz 2 ) in three dimensional space as shown in figure4.5. Fig.4.5Density of energy states Every point (nx, ny, nz) within this space represents an energy state. Also as every integer represents one energy state, unit volume of this space contains exactly one state. Hence the number of states in any volume is equal to the volume expressed in units of cubes of lattice parameters. The number of energy states within a sphere of radius n, = 𝟒 𝟑 π𝒏𝟑 → (1) Since nx, ny, nz can have only positive integer values, we have to consider only one octant of the sphere. Hence available energy states = 𝟏 𝟖 [ 𝟒 𝟑 π𝒏𝟑 ] → (2) If we consider another sphere of radius (n+dn) within a small energy interval E+dE, the number of energy states having energy values between E and E+dE (i.e., in the energy interval dE) is given by x n E x n n x n dn x n E dE x n x n x n y n x n z n
- 45. 45 Z′ (E) dE = 𝟏 8 [ 𝟒 𝟑 π(𝒏 + 𝒅𝒏)𝟑 ] - 𝟏 𝟖 [ 𝟒 𝟑 π𝒏𝟑 ] = 𝟏 𝟖 x 𝟒𝝅 𝟑 [(𝒏 + 𝒅𝒏)𝟑 -𝒏𝟑 ] = 𝝅 𝟔 [𝒏𝟑 + 𝒅𝒏𝟑 + 𝟑𝒏𝟐 𝒅𝒏 + 𝟑𝒏𝒅𝒏𝟐 − 𝒏𝟑 ] = 𝝅 𝟔 [𝒅𝒏𝟑 + 𝟑𝒏𝟐 𝒅𝒏 + 𝟑𝒏𝒅𝒏𝟐 ] ~ 𝝅 𝟔 [𝟑𝒏𝟐 𝒅𝒏𝟐 ] [∵ 𝒅𝒏 is very small, higher order terms are neglected] = 𝝅 𝟐 [𝒏𝟐 𝒅𝒏] ∴ Z' (E) dE = 𝝅 𝟐 [𝒏(𝒏𝒅𝒏)] → (3) The expression for the energy of electron is E = 𝒏𝟐𝒉𝟐 𝟖𝒎𝑳𝟐 → (4) ⇰ 𝒏𝟐 = 𝟖𝒎𝑳𝟐 𝒉𝟐 E → (5) Or n = [ 𝟖𝒎𝑳𝟐 𝒉𝟐 ] 𝟏 𝟐 → (6) Differentiating eq (5) taking n and E as variables 2n dn = 𝟖𝒎𝑳𝟐 𝒉𝟐 𝒅E n dn= 𝟏 𝟐 𝟖𝑚𝑳𝟐 𝒉𝟐 𝒅E → (7) Substituting eq (6) and eq (7) in eq (3) Z' (E)dE = 𝝅 𝟐 [ 𝟖𝒎𝑳𝟐 𝒉𝟐 ] 𝟏 𝟐𝑬 𝟏 𝟐 𝟏 𝟐 [ 𝟖𝒎𝑳𝟐 𝒉𝟐 ] dE = 𝝅 𝟒 [ 𝟖𝒎𝑳𝟐 𝒉𝟐 ] 𝟑 𝟐𝑬 𝟏 𝟐 dE → (8) According to Pauli‘s exclusion principle, each energy level contains two electrons. This means each energy level will have two sub energy levels. Therefore, the above equation should be multiplied by 2. ∴ 𝒁′(𝑬) 𝒅𝑬 = 2 x 𝝅 𝟒 [ 𝟖𝒎𝑳𝟐 𝒉𝟐 ] 𝟑 𝟐𝑬 𝟏 𝟐 dE = 𝝅 𝟐 [ 𝟖𝒎 𝒉𝟐 ] 𝟑 𝟐𝑬 𝟏 𝟐𝑳𝟑 dE → (9) The number of energy states per unit volume, i.e., density of states is given by Z (E) dE= 𝐙′ (𝐄) 𝐝𝐄 𝑽 = 𝝅 𝟐 [ 𝟖𝒎 𝒉𝟐 ] 𝟑 𝟐𝑬 𝟏 𝟐 dE = 𝝅 𝟐 [ 𝟐𝟐(𝟐𝒎) 𝒉𝟐 ] 𝟑 𝟐𝑬 𝟏 𝟐 dE = 𝝅 𝟐 𝟖 𝟑 (2𝒎) 𝟑 𝟐𝑬 𝟏 𝟐 dE ∴ Z (E) dE = 𝟒𝝅 𝒉𝟑 (2𝒎) 𝟑 𝟐𝑬 𝟏 𝟐 dE → (10) 4.6.1 Fermi energy: 1. We have seen that quantization lead to the discrete energy levels. Pauli's exclusion principle explains the distribution of electrons amongst the allowed energy levels. It allows a maximum of two electrons (with spins in opposite directions) in any energy level. 2. Thus, a pair of electrons, one with spin up and the other with spin down occupy the lowest energy level. The next pair occupies the next level. This process goes on until all electrons in the metal occupy their positions. 3. But, there will be many more allowed energy levels available for occupation.