solid stae.pdf

Solid State Physics

Solid State Physics Ananta Charan Pradhan NIT Rourkela

February 11, 2016

Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

OUTLINE

1

Solid State Physics Introduction to Crystals Band theory of Solids


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Solid State Physics

Amorphous and crystalline solids Solid consists of atoms, ions, or molecules packed closely together with strong forces of attraction. Solid maintains both a well-defined volume and shape. Classified as crystaline and amorphous depending upon the arrangement of atoms or molecules. Amorphous 1. Short range 2. No sharp melting point 3. Irregular Shape 4. Isotropic 5. Less rigid e.g., glass, rubber, plastics, cement

Crystaline 1. Long range 2. Sharp melting point 3. Regular shape 4. Anisotropic 5. More rigid e.g., Copper, quartz, mica, sugar, diamond


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Solid State Physics

Amorphous and Crystalline solids


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Solid State Physics

Crystal Defects

Actual crystals are never perfect. Defects: missing of atoms, atoms out of place, irregularities in the spacing of rows of atom and the presence of impurities in the crystal are defects. Nature and concentration of defects highly influence the properties of the crystal. Simplest types: 1 2

Point defects (Vacancies & Interstitials) Line defects (Dislocation)


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Solid State Physics

Types of Crystal Defects Point defects a. Vacancy b. Interstitial

a

b

c. Substitutional impurity

c

d.Interstitial Impurity

d

Line defect or Dislocation a. Dislocation line

b. Dislocation force

c. edge & d. screw dislocation


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Solid State Physics

Forces in Crystals Electrical forces are responsible in binding the atoms and molecules giving different solid structures. Forces are two types : attractive (to keep the atoms together) and repulsive (comes into play when the atom is compressed). Cohesive enegry: It is the energy that must be supplied to the solid to separate its constituents into neutral free atoms at rest. Classification of solids based on bonding or cohesion: 1. Ionic crystals (e.g. NaCl) 2. Covalent crystals (e.g. Diamond and other corresponding structures). 3. Metallic crystals (e.g. Na). 4. Van der Waals crystals (e.g. Argon and other noble elements).


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Solid State Physics

Ionic Crystals Need atleast two atoms; one with low ionization and other with high electron affinity. In an ionic crystal the ions (+ve and -ve) assemble themselves in an equilibrium configuration in which the attractive forces between positive and negative ions balance the repulsive forces between the ions. In an ionic crystal each ion is surrounded by as many ions of the opposite sign as can fit closely, which leads to maximum stability. Ionic crystals thus can be described as an ensemble of hard spheres which try to occupy a minimum volume while minimizing electrostatic energy at the same time. Ionic crystals are insulators and has no directionality. e.g., NaCl and CsCl.


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Solid State Physics

Ionic crystals The number of nearest neighbors for a constituent particle is defined as coordination number. In other words the number of spheres which are touching a particular sphere is known as its coordination number. Unit Cell:A unit cell is the smallest unit of volume that contains all of the structural and symmetry information to build-up the macroscopic structure of the lattice by translation. NaCl : fcc, lattice points=(1/8)×8+(1/2)×6=4, Coordination # 6 CsCl : bcc, lattice points = (1/8)×8+1=2, Coordination # 8


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Solid State Physics

Cohesive Energy of Ionic Crystals


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Solid State Physics


Consider a Na+ in NaCl. Each Na+ ion in NaCl has 6 nearest Cl+ neighbors and at a distance r. 6e2 The PE of the Na+ ion due to these 6 Cl− ions is U1 = − 4πǫ 0r √ The next nearest neighbors are 12 Na+ ions, each one the distance 2r away 2 and the PE of a Na+ ion due to 12 Na+ ions is U2 = + 4πǫ12e√2r 0 When the the summation is continued over all the + and - ions in a crystal of infinite size, the result is 12 e2 e2 e2 (6 − √ + ...) = −1.748 4πǫ = −α 4πǫ Ucoulomb = − 4πǫ 0r 0r 0r 2 This result holds good for Cl− as well. The quantity α is called the Madelung constant of the crystal. It has the same value for all crystals of the same structure. Simple crystal structures have α between 1.6 and 1.8.


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Solid State Physics

Cohesive Energy continue .... Repulsive energy Urepulsive = + rBn r−n implies a short range force that increases as the interionic distance decreases. αe2 Utotal = Ucoulomb + Urepulsive = − 4πǫ + rBn 0r At the equillibrium separation r0 of the ions, U is a minimum and dU/dr = 0 when r = r0 . Hence αe2 nB ( dU dr )r=r0 = 4πǫ0 r 2 − r n+1 = 0 0

0

2

αe rn−1 B = 4πǫ 0n 0 The total potential energy at equillibrium separation is given by αe2 U0 = − 4πǫ (1 − n1 ) 0 r0 We must add this amount of energy per ion pair to separate an ionic crystal into individual ions. The average value of the exponent is n ∼ 9 which can be found from the observed compressibility of the crystal. The repulsive force varies sharply with r.


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Solid State Physics



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Solid State Physics

Covalent crystals

Covalent crystals arise from the sharing of electrons by adjacent atoms. Each atom in the covalent bond contribute an electron to the bond. Because there are no delocalized electrons, covalent solids do not conduct electricity. The rearranging or breaking of covalent bonds requires large amounts of energy; therefore, covalent solids have high melting points. e.g., allotropes of C (Diamond, graphite). Cohesive energies are greater in covalent crystals than in ionic crystals. As a result covalent crystals are harder. (diamond, boric nitride (2nd hardest))


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Solid State Physics

Covalent crystals: Graphite Graphite and diamond are identical chemically but are different physically. 1

Opaque and metallic

2

Hexagonal structure, coordination number = 3

3

Layers of C atoms in a hexagonal Sheet like structure in which each atom is joined to three others by weak cavalent bonds. One electron per carbon atom is free to circulate through the network. Weak van der Waals forces bond the layers together.

4

Conductor of electricity.

5

Graphite is very soft.


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Solid State Physics

Covalent crystals: Diamond

1

diamonds are transparent and brilliant (refractive index = 2.41, internal reflection).

2

Tetrahedron structure and coordination number 4.

3

Each carbon atom is strongly bonded to four adjacent carbon atoms.

4

Bad conductor of electricity.

5

Diamonds are the hardest.


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Solid State Physics

van der Waals Bond All atoms and molecules (even inert gas atoms) exhibit weak, short range (r−7 ) attraction for one another due to van der Waals (molecular) forces. Arose to explain departures of real gases from ideal gas law. Van der Waals (molecular) bonding originates from the fluctuation of electron distribution between two distorted inert gas atoms. Van der Waals force responsible for the condensation of gases into liquids and the freezing of liquids into solids in the absence of ionic, covalent, metalic bond. friction, surface tension, viscosity, adhesion, cohesion and so on are arise from these forces. Polar molecule can attract non-polar molecule due to this force as shown in the figure below.


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Solid State Physics

van der Waals Bond continue .....

The van der Waals attraction between two molecules the distance r apart is proportional to r−7 , so that it is significant only for molecules very close together. Alignment of polar molecules with ends of opposite sign adjacent and attractive force between a polar and a non-polar molecule is due to the van der Waals. Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

van der Waals Bond continue ..... The electric field E a distance r from a dipole of moment p is given by E=

3(p.r) 1 p r] [ − 4πǫ0 r3 r5

(1)

The field E induces an electric dipole moment p’ in non-polar molecule which is proportional to E in magnitude. Induced dipole moment p’=αE, where α is called the polarizability of the molecule.The energy of the induced dipole in the electric field E is

U = −p′ .E = −αE.E = −

α p2 3p2 3p2 9p2 ( 6 − 6 cos2 θ − 6 cos2 θ + 6 cos2 θ) (2) 2 (4πǫ0 ) r r r r

So the interaction energy, U =−

α p2 (1 + 3cos2 θ) 6 2 (4πǫ0 ) r


(3)

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Solid State Physics

van der Waals Bond continue..... The potential energy of the two molecules that arise from their interaction is negative i.e., force between them is attractive. U ∝ r−6 , hence F = −dU/dr ∝ r−7

On the average, nonpolar molecules have symmetrical charge distribution, but at any instant the distributions are asymmetric. The fluctuation in the charge distributions of nearby molecules are coordinated which leads to an attractive force between them. van der Waal forces are much weaker than forces in ionic and covalent Molecular bonds. Molecular solids are soft, often volatile, have low melting temperatures, and are electrical insulators.


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Solid State Physics

Hydrogen Bond Strong type of van der Waals bond is called a hydrogen bond occurs between certain molecules containing hydrogen atoms. Hydrogen bonds occurs between H containing molecules due to permanent dipole moments (e.g. H2O, NH3, HCl). H2 O is non-symmetric, tetraheron structure and prone to H-bond. Each H2 O can form H-bond with four other H2 O, central molecule provides bridging proton for two bonds and the attached molcules provides the other two. The open bridge like structure of ice is due to H-bonding. Hence low density. H-bonding in biological matterials

Hydrogen bond of water and Ice structure.


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Solid State Physics

Metalic bond

Atoms in metals lose weakly bound electrons to form cations. Sea of delocalized electron gas surrounds the ions. Metallic bonds (electrostatic interactions between the ions and the electron cloud) hold the metallic solid together. Because outer electrons of metal atoms are delocalized and highly mobile, metals have electrical and thermal conductivity. The total energy of atoms is lower when they bound together than when they are separate atoms.(Explaination: Linear combination of atomic orbitals) PE of free electrons reduced and the KE is increased. Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

Summary of crystal bondings


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Solid State Physics

Ohm’s law Ohm’s law : I = VR Where V is the potential difference across the ends of a metal conductor and I is the resulting current, R is the resistance. Ohm’s law follows from the free electron model of a metal. Free electrons in metal move randomly and undergo frequent collisions and these collisions represent the scattering of electrons waves by the crystal irregularities. Collision time: τ = vλF where vF is the electron velocity corresponding to the Fermi energy ǫF , since only electrons at the top of their energy distribution can be accelerated and λ = mean free path. q vF = ( 2ǫmF ) and it is extremely high compared to the drift velocity (vd ) produced by electric field.(e.g., for Cu vF = 1.57 × 107 m/s and vd < 1 mm/s). Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

Drift Velocity

n = number of free electrons per unit volume Q = neAL = total charge in length L of the conductor t = vLd = time for this charge to pass through the cross section = neAvd Current, I = Qt = neAL L vd

The drift velocity of the electron is, vd =

I nAe


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Solid State Physics

Drift Velocity Example 10.2 Find the drift velocity vd of the free electrons in a copper wire whose cross-sectional area is A=1.0 mm2 when the wire carries a current of 1.0 A. Assume that each copper atom contributes one electron to the electron gas. Answer we know that, in Cu, n = N/V 8.94×103 kg/m3 mass/m3 = (63.5u)×(1.66×10 =atoms/m3 = mass/atom −27 kg/u) =8.48 × 1028 atoms/m3 = 8.5 × 1028 electrons/m3 , and here I = 1.0 A and A = 1.0 mm2 = 1.0 × 10−6 m2 . I Hence vd = nAe = 7.4 × 10−4 m/s But if the free electrons have so small a drift velocity, why does an electric appliance go on as soon as its switch is closed and not minutes or hours later? The answer is that applying a potential difference across a circuit very rapidly creates an electric field in the circuit, and as a result all the free electrons begin their drift almost simultaneously. Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

Resistivity Electric field E produced due to potential difference V across the ends of a conductor of length L is E = V /L and the force, F = eE on the free electrons F = eE whoes acceleration is a = m m Drift velocity eE λ eEλ vd = aτ = ( )( ) = (4) m vF mvF I But we also know = nAe ´ . Substituting this in the above equation, we get ³ 2vd´³ 2 λ A Eλ = ne I = nAe mvF mvF L V Resistance of metal conductor

³ mv ´ L F (5) nAe2 λ A The quantity in parentheses is known as the resistivity ρ of the metal and is mvF a constant for a given sample at a given temperature and ρ = nAe 2λ R=


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Solid State Physics

Resistivity The scattering of free electron waves in a metal that leads to its electric resistance. Again the scattering of free electron waves depends on crystal imperfections as well as temperature. Thus resistivity of a metal ρ = ρi + ρt , where ρt ) depends on temperature and ρi depends on the concentration of crystal impurity.


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Solid State Physics

Band theory of Solids Why band theory? The energy band structure of a solid determines whether it is a conductor, an insulator, or a semiconductor. Resistivity of Cu (conductor), ρ = 1.7 × 10−8 Ω.m

Resistivity of qartz (insulator), ρ = 7.5 × 1017 Ω.m A huge span in resistivity (order of 25).

Origin of band structure When the individual atoms brought together to form the solid, their velency electron wave function overlaps giving a band structure (1cc Cu = 1023 atoms). Closely spaced energy levels form an energy band that consists of a virtually continuous spread of permitted energies. The energy bands, the gaps between them, and the extent they are filled by electrons determine the electrical behaviour of solids. Ananta Charan Pradhan (NIT Rourkela)Solid State Physics February 11, 2016 29 / 55

Solid State Physics

Formation of band structure


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Solid State Physics

Formation of band structure: Na Na: 1s2 2s2 2p6 3s1 and r0 = 0.367 nm.


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Solid State Physics

Distribution of electron energies


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Solid State Physics

Conductor Sodium solid: Na: 1s2 2s2 2p6 3s1 . 3s can hold 2(2l+1) = 2 electrons. 3s band formed by N atoms can hold 2N e’s. 3s band is half filled and fermi energy (ǫF ) lies in the middle of the band. Supply of energy drifts the electron to the partially filled energy band making it a good conductor. Magnesium solid: Mg: 1s2 2s2 2p6 3s2 , filled 3s shell. 3s and 3p bands overlap as Mg atoms come close together. N Mg atoms can hold 8N electrons (3s+3p). With only 2N electrons in the band, it is only one-quarter filled and so Mg is a conductor. Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

Insulator Carbon: C: 1s2 2s2 2p2 and Si: 1s2 2s2 2p6 3s2 3p2 The 2s and 2p bands overlap at first and then at smaller distance the combined bands split into two bands each containing 4N electron. The lower valency band is completely filled 4N electrons and the conduction band is empty and the two bands are 6 eV wide (band gap). Fermi energy (ǫF ) is at the top of the valency band. At room temperature diamond has not enoogh energy (0.025 eV) to jump the 6 eV gap. 8 (NIT Rourkela)Solid State Physics Ananta Charan Pradhan

A field of 10 V/m needed to gain 8 eV

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Solid State Physics

Semiconductor Si: 1s2 2s2 2p6 3s2 3p2 . Eg = 1.1 eV in Si. Better than diamond at room temperature. At low temperatures a small number of its valence electrons have enough thermal energy to jump the forbidden band and enter the conduction band. Allow a small amount of current to flow when an electric field is applied. Thus silicon has a resistivity intermediate between those of conductors and those of insulators (0.1-60 ohm m).


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Solid State Physics

Impurity Semiconductors N-type Semiconductor Small amount of impurity can drastically change the conductivity of a semiconductor (e.g., As atoms in a Si crystal). Si: 1s2 2s2 2p6 3s2 3p2 & As: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 As has 5 valency e’s and Si has 4. The fifth electron needs very little energy - only about 0.05 eV in Si and about 0.01 eV in Ge to be detached and move about freely in the crystal. Fermi level lies in the middle of the gap for intrinsic semiconductor but addition impurity (N-type) shifts it towards conduction band due to the donor levels below the conduction band.


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Solid State Physics

Impurity Semiconductors P-type Semiconductor Si: 1s2 2s2 2p6 3s2 3p2 & Ga: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 As has 5 valency e’s and Si has 3. Presence of Ga leaves holes in the crystal structure and an electron needs little energy to enter into the hole and as it does leaves another hole in the crystal. Electron moves towards anode filling succesive holes. Fermi level lies in the middle of the gap for intrinsic semiconductor but addition impurity (P-type) shifts it towards valency band due to the acceptor levels below the conduction band.


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Solid State Physics

ENERGY BANDS: ALTERNATIVE ANALYSIS

Discussed: Bringing together isolated atoms to form a solid has the effect of broadening their energy levels into bands of allowed electron energies. Electrons in the Solid: 1

Free Electrons

2

Nearly free Electrons → ENERGY BANDS

3

Tightly bound Electrons

Reciprocal lattice : Fourier transform of direct lattice (k-space). 2π (λ → K = 2π λ and a → a )


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Solid State Physics

ENERGY BANDS: ALTERNATIVE ANALYSIS Alternative Approach: An electron in a crystal moves in a region of periodically varying potential that cause diffraction. Difraction limits the electron to specific allowed energy bands. de Broglie Wavelength of a free electron λ =

h p

Low energy electrons : λ > a (lattice spacing) High energy electrons (ǫF ): λ ≈ a (lattice spacing) =⇒ diffraction of electrons.


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Solid State Physics

Bragg reflection Bragg reflection of electron from the atomic planes of crystal: nλ = 2a sinθ n=1,2,3,.... General Approach : Electron waves in a crystal is explained in terms p nπ of k = 2π λ = ~ and the Bragg’s formula in terms of k is k = a sinθ

Bragg reflection in 2D square lattice Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

Bragg reflection in crystal 1

Reflection occurs when component of k in x-direction, kx = component of k in y-direction ky = nπ a .

2

k<

3

4

5

π a

nπ a

and

=⇒ electron moves freely through the lattice in any direction.

When k = πa =⇒ electron prevented from moving in the x or y direction by reflection. The more k exceeds πa , the more limited the possible directions of √ motion, until when k = πa sin 45◦ = 2 πa the electrons are reflected, even when they move diagonally through the lattice. Bragg plane: Perpendicular bisector to the line joining to the origin of the reciprocal space to any reciprocal lattice point. Loci of k-values that are bragg reflected and hence may be considered as the reflecting planes. 2

p The energy of a free electron is related to its momentum p by E= 2m and to 2 2 ~ k its wave number k by E = 2m Ananta Charan Pradhan (NIT Rourkela)Solid State Physics

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Solid State Physics

Brillouin Zones in 2D First Brilouin zone : The region in k-space that low k-electrons can occupy without being diffracted or Region which does not cross any Bragg plane (− πa < kx (&ky ) < πa ). Second Brilouin zone : The region which crosses the 1st Bragg plane but not the 2nd Bragg plane. Third Brilouin zone : The region which crosses the 2nd Bragg plane but not the 3rd Bragg plane.


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Solid State Physics

Energy Contour of Brillouin Zones Energy and wave number in the 1st BZ: E =

~2 k2 2m

Reflection ocurs from the boundary of a BZ in k-space by virtue of the interaction of the electron with the periodic array of positive ions that occupy the lattice point.


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Solid State Physics

Origin of band gaps At k = πa , E has two values, the lower belonging to the 1st brillouin zone and the higher to the 2nd BZ. The gap between the possible energies in the first and 2nd BZ corresponds to the forbidden band. The same pattern continues succesively higher BZs.


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Solid State Physics

Origin of band gaps


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Solid State Physics

Standing waves at BZ boundary only electrons which have k = /a feel the periodic potential, and they form standing wave patterns. Possible solution of Schrodinger’s eqn which gives standing waves are − 2 ψ − = A sin( πx a ) and charge density = −e|ψ | + 2 ψ + = A cos( πx a ) and charge density = −e|ψ |

Potential energy of an electron is greatest at midway between the ions and least at the ions themselves.


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Solid State Physics

Energy discontinuity by standing waves Two solutions at k= πa gives rise to band gap i.e., the difference between the expectation values of these 2 energy levels at BZ boundary. R Eg = U(x)[|ψ + |2 − |ψ − |2 ]dx


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Solid State Physics

Electron energy distributions in BZ An Example: The distribution of electron energies at the Brillouin zones.


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Solid State Physics

Forbidden Band for different k There is possibility of crystals having no forbidden band depending on values of k.


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Solid State Physics

Electron energy contours and fermi levels Electrical behaviour of solid depends on the degree of occupancy of its energy bands as well as on its band structure.


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solid stae.pdf

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