Unit 7 (B) Solid state Physics

Unit 7 (B) Solid state Physics hermal Properties of solids: Zeroth law of hermodynamics: If two bodies A and B are each separated in thermal equilibrium with the third body C, then A and B are also in thermal equilibrium. First law of thermodynamics: It is also called the law of conservation of energy. he law state that, the amount of heat supplied to a system is equal to the algebraic sum of the change in the internal energy of the system and the amount of external work done by the system. dq = du + dw Second law of thermodynamics: 1) By Kelvin statement: It is impossible to obtain a continuous supply of work from a body by cooling it to a temperature lower than that of the surroundings. 2) By clasius statement: It is impossible to make heat flow from a body at a lower temperature to a body at a higher temperature without doing external work on the substance. Amount of heat converted into work Efficiency () = otal heat absorbed = H 1 H 2 h 1 = Q 1 Q 2 Q 1 = 1 - H 2 h 1 = 1 - Q 2 Q 1 = 1 2 1 = 1-2 1 or or or SUDY MAERIAL AVAILABLE FOR PGRB PHYSICS 10 UNI NOES, PRACICE QUESIONS, PREVIOUS YEAR QUESION PAPERS. CONAC:9444249503 he second law of thermodynamics gives the condition under which heat can be converted into work. Carnot s Engine is based on the above concept. Carnot s heorem: All reversible heat engines working between the same two temperature limits have the same efficiency. No engine can be more efficient than a Carnot s reversible engine working between the same two temperatures. All reversible heat engines have more carnot engint has maximum efficiency. So carnot engine is a reversible one. Eg:- Find the efficiency of a carnots engine working between 127 0 C and 27 0 C 1 = 273 + 127 = 400 K 2 = 273 + 27 = 300 K 1

= 1-2 = 1-300 = 0.25 K 1 400 % of efficiency = 25% Find the efficiency of the carnots engine working between the steam point and the ice point. 1 = 273 + 100 = 373 K 2 = 273 + 0 = 273 K = 1-2 = 1-273 = 100 1 373 373 % of efficiency = 100 x 100 = 26.81% 373 Entropy : It is a thermodynamical property and is defined by the relation. ds = δq (or) 2 δq S 1 S 2 = 1 he entropy remains constant during an adiabatic process. When heat is absorbed during a process, there is increase in entropy and when heat is rejected during a process there is decrease in entropy. he unit of entropy is Joule /K or K cal / K From the relation, δs = δq δq Heat added or subtracted Absolute emperature otal change in entropy in a reversible cycle is zero (carnot s cycle) Entropy of a system increases in all irreversible processes. Vapourisation (Heat Absorbed) Melting (Heat Absorbed) Gas Liquid Solid Condensation (Heat Released) Freezing (Heat Released) Sublimation (Heat Absorbed) Deposition (Heat Released) From the above Entropy increases Entropy Decreases here is a change in entropy when a substance goes from one state to the another state. 2

hird law of hermodynamics: (Nernst Heat heorem) A absolute zero temperature the entropy tends to zero and the molecules of a substance or a system are in perfect order (well arranges) Entropy is a measure of disorder of molecules of the system. Zero point energy: he energy of the molecules at absolute zero temperature is called zero point energy. With the decrease in entropy, the disorder decreases. Extensive variables depends upon the mass or size of the substance. Eg: Mass, volume, entropy, length, area, electric charge, Intensive variables are independent of mass or size of the system. Eg: Pressure, temp, viscosity, density emf, surface tension. Maxwell s hermodynamical relations. From the 2 laws, by hermodyndamics, Maxwell was able to derive 6 fundamental thermodynamical relations. he state can be specified by any pair of quantities. Pressure (P) * Volume (V) PV Mechanical emperature () * Entropy (S) hermal (S) From the I law of thermodynamics, Q = du + w From II law of thermodynamics, S = dq Using I law SUDY MAERIAL AVAILABLE FOR Q = du + w (or) Q = du + P dv (or) 10 UNI NOES, PRACICE QUESIONS, du = q - p dv -------------- (1) PREVIOUS YEAR QUESION PAPERS. Using II law, CONAC:9444249503 S = dq q = ds -------------- (2) Equation (3) is combination of (1) & (2). S V y x x y - y x x = S x y y x - x y V y x his is the general expression for Maxwell s thermo dynamical relation, Maxwell s I thermodynamical relation:- Put x = S and Y = V S V =1 ; =1 => s = x ; = v= y x y y V S = x = - x y (or) Maxwell s I thermodynamical relation. S V P 3 V S

Maxwell s II thermidynamical relation Put x = and Y = V 0.- x V =1 ; =1 => x = ; = y= v y = 0 ; V = 0 y x S S V - V 0 V = 1 V V - S V = V V 1 --------------- Maxwell s II Relation S P V Maxwell s III hermodynamical relation: Put x = S ; y = P V --------------- Maxwell s III Relation S = S P V P S Maxwell s IV themodynamical relation: Put x = ; y = P S V S V. = V P Maxwell s V thermodynadmical relation: Put x = P ; y = V V = Maxwell s VI thermodynamical relation : x = ; y = S S = S V P - V S - S V P V = 1 V S S = 1 S P O R I V E => S P V 2 4 Maxwell s V relation 3 1 Iv 1 III II I Applications of Maxwell s hermodynamic relations: 1) Specific heat of Maxwell s hermodynamic relations: C P = Q s = Q 4

C P = S P q =.s 2) Specific heat at constant volume C V = Q V C V = S V C P C V = S P - S V Using Maxwell s II relation the above equation becomes, C P C V = V - V P For a perfect gas Equation of state is PV = R V = R ; V V P = R P C P C V = P R R R 2 = x V x P = PV C P C V = R2 = R R = Rydberg s const. R V V C P - C V = R -- Mayer s Relation Vanderwaals gas:. P + a V 2 = R (v b) C P - C V = R 1 + 2a VR Joule homson cooling : When a gas is allowed adiabatically through a porous plug from the region of constant high pressure to the region at constant low pressure, the temperature of escaping gas changes. his is called as Joule homson effect. he change of emperature, When +ve Heating -ve Cooling Zero No heating or cooling d = V C P (-1) dp is the Equation for the change in emperature due to Joule homson Effect. Where, = 1 V V P 5

Joule homson co-efficient () = H = 1 CP d = V CP.. V ( - 1) dp. P V Here dp is the fall in pressure and it is always a negative quantity. Results of Joule homson Effect. 1) If ( - 1) is +ve ; >1 and d is Negative Cooling Effect. 2) If ( - 1) is zero ; =1 and d is zero No Cooling or Heating. 3) If ( - 1) is -ve ; <1 and d is Positive Heating Effect. For perfect gas, d = 0 is zero. For Vander Waals gas, P + a V2 (v b) = R 1) Vc = 3b a = 27R2 c 2 64P c 2) c = 8a 27Rb a 3) Pc = b = R c 27b 2 8P c Critical co-efficient of a gas, R c = 8 = 2.667 P c V c 3 emperature of inversion i = 2a Rb Boyle s emp i = 2 B. i c = 27 4 = 6.75 B = a Rb emperature of inversion ( i ) emperature at which Joule homson Effect is zero and changes sign is known as emperature of inversion. Cases: i c = 6.75 i = 2a Rb 1) When > i or 2a < b --- Heating Effect. R 6

2) When < i Cooling Effect. 3) i is different for different gases. 4) Adiabatic Expansion must result in fall of emperature. 5) Adiabatic compression must result in increase in emperature. hermodynamic potentials, 1) Integral Energy (u) 2) Helmhortz free energy or Helmholtz work function F = u - S 3) Enthalpy H = u+pv 4) Gibb s function G = u+pv S Integral Energy (u) dq = du + dw Isobaric pressure constant du = dq dw du = dq Pdv Isochoric : Volume unchanged du =.ds - Pdv For an Adiabatic process, dq = 0 As, dq =.ds,.ds = 0 du - Pdv Enthalpy : Also known as otal heat H = u + Pv Change in Enthalpy, dh = du + pdv + vdp Now, du =.ds pdv dh =.ds pdv + pdv + vdp =.ds + vdp For reversible isobaric process, dp = 0 dh =.ds = dq he change in Enthalpy is equal to the heat absorbed Gibb s function (G) or Gibb s Free Energy G = u S + PV F = u S G = F + Pv his is the relation connecting Gibbs function and Helmholtz function Enthalpy ; H = G + S = Gibb s free Energy + Latent heat H = u + Pv ; u = H Pv Put u value in Gibb s Equation, G = u S + Pv 7

= H Pv S + PV Gibb s function G remains constant in an Isothermal Isobaric process. Relation between C P, C V and :- C P = H P = H C V = V V doubt First ds Equation is ds = C V d + Second ds Equation: ds = C P d - u H dv P dp Phase ransition: I order phase transition: he change of phase which takes place at constant temperature and pressure in which heat is either absorbed or evolved during the change of phase are called first order phase transitions. Here the entropy and density or volume changes. I order phase transition can be defined as the one in which the Gibb s function with respect to pressure and temperature change discontinuously at the transition time. Clausius Clapyron Latent heat equation: dp S = 2 S 1 dt V 2 V 1 S 2 S 1 = s = Q = L Where L is the Latent heat of Vapourisation dp dt = L (V 2 V 1 ) his equation is called as the Clausius, Clapeyron latent heat equation or first order phase transition equation. Eg: When heat is given to water at 100 0 C it changes from liquid to vapour state. he density of water is 1000kg/m 3. When there is a change of state from water to ice then volume, entropy density changes. Second order phase ransition:- It can be defined as the phenomenon that takes place with no change in entropy and volume at constant temperature and pressure. Eg: ransition of liquid Helium I to Helium II at point (2.19k) 2) ransition of a ferro magnetic material to a paramagnetic material at curie point. 3) ransition of a super conducting metal into an ordinary conductor in the absence of Magnetic field. 4) Order disorder ransition in chemical compounds and alloys 8

K = - 1 V = + 1 V v p v Isothermal compressibility P Volume coefficient of expansion S 1 S 2 = Q =L S 1 S 2 = L Latent heat S 1 S 2 => S 1 S 2 = 0 For second order phase transitions, there is no change in entropy and volume. S 1 S 2 ds 1 ds 2 dp d = CP 2 CP 1 V ( 2 1 ) dp = P 2 CP 1 d V ( 2 1 ) Also, --------- (2) ------------------ (1) his is the equation for second order phase transition Equation (1) and (2) are called as Ehrenfests equations. hese Ehrenfests equation represents the condition of equilibrium between the two phases. Dulong and Petit s law he atomic heat of all the Elements in the solid state is a constant E = 3R he law can also be stated as, he product of the specific heat and the atomic weight of all the Elements in the solid state is a constant. Whose value is 6. Atomic heat = Specific heat x Atomic weight Specific heat at constant volume, C v = de = d (3R) = 3R d d C v = 3R Where C V is the atomic heat of substance whose value is 6. Dulong and petits law is 1) Atomic heat = 6 2) CV = 6 3) de d = 6 Specific heat decreases with decrease in temperature Specific heat at absolute zero Zero Specific heat Increases with increase in emperature Atomic heat is Maximum at the value 6. 9

Production of low emperature:- OK = -273.16 0 C O 0 C = -273.16K he lowest emperature corresponds to Ok (-273.16 0 C) is called the absolute zero temperature. emperature upto - 65 0 C can be obtained with KoH and ice. With liquid helium boiling under normal pressure temperature of -268.9 0 C can be reached. By boiling liquid Helium under reduced pressure temperature of the order of 1K could be obtained (ie 1K = -272 0 C) 1 + OK = - 273.16 0 C +1 = - 272.16 0 C With liquid helium isotope (He 3 ) boiling under reduced pressure a temperature of below 0.4k can be reached. emperature below 0.4K can be reached by Adiabatic demagnetization method. Debye, Macdougall produced temperature below 1K with the help of galonium sulphate. Adiabatic demagnetization (1 esla = 10 4 Gauss) he paramagnetic specimen (salt is suspended in a vessel surrounded by liquid helium in Dewar flask. Liquid Helium is boiled under reduced pressure and it is placed in the dewar flask. N S Paramagnetic substance Liquid Helium Dewar flask Liquid Helium Hydrogen he purpose of liquid Helium is to boil the substance. Liquid Hydrogen is placed below the Dewar flask so as to maintain the high temperature. he magnetic field provided when applied the temperature rises. When the magnetic field is removed the temp. reduces in paramagnetic substances. he emp is calculated by curie la Curie law - = C 10

Haas was able to produce emperature upto 0.002k using a double sulphate of potassium and aluminium. According to Curie s law Intensity of magnetization, I H I = C H C Curie constant I of I mole of paramagnetic substance, M = IV = C H V = CV H Change in emp, = CV 2C H (H 2 2-H 1 2) C H = Q H C H Specific heat of a substance at constant magnetic field H. If the magnetic field is reduced from H 1 = H to H 2 = 0. hen, change in emperature = CV CV (0 H2) = 2C H 2C H H2. = CV 2C H H2. emperature of paramagnetic substance decreases on decreasing the magnetizing field H. In the above, Greater in H (Initial field), lower is (Initial temperature) greater is the temperature fall. Cooling produced by Adiabatic demagnetization is explained using S (emp- Entropy) During Adiabatic demagnetization the lower temp of the order of 10-4 can be reached. Measurement of very low temperature Human body = 36.98 0 C or 98.4 0 F. Helium vapour pressure thermometer is used to measure temp upto 0.7K For lower emperature the graph between saturated vapour pressure and temp is extrapolated. Normal graph Intrapolated y Extended graph Extrapolated Phonons Quantum of Energy. x he energy of a lattice vibration or an Elastic wave is quantized and the quantum of this energy is termed as phonon. Sound waves Acoustic phonons. hermal vibrations hermally Excited phonons. Energy of phonon = h. - Angular frequency of the mode of vibration. otal energy for n number of phonon s is, h E = (e h/k 1) 1 E = nh 11

Here n = 1 (e h/k 1) 1 he number of phonons can be increased or decreased by raising or lowering the temp. respectively. Frequency of phonon waves are of the order 10 4 to 10 13 hz. As phonons are also termed as indistinguishable particle. Einstein s theory of specific heat of solids:- Heat sis radiated in the form of discrete particles called the photons. Each particle has the energy = h where is the frequency of heat radiation h Planck s constant. When temp is raised the atoms execute SHM. Each Atom of solid has three degrees of freedom like monoatomic gas molecule. Mean energy per degree of freedom is,. h e h/k 1 Energy of each atom = 3h e h/k 1 Energy content of 1 gram of atom of solid consisting of N atoms, 3h E = e h/k 1 Atomic heat at constant volume, C v = de = 3NK eh /K h 2 d (e h /K 1) 2 K If = h = Einstein emp K NK = R - gas constant for a gram atom C V = 3R In = h K e / (e / 1) 2 ( )2 ----------------------------------------------------------- (1) ; Let = E = 3 x 10 9 K K = 1.38 x 10-23. hen E = he ; E Einstein s emp K E Einstein frequency for that solid Equation (1) is the Einstein equation for the Atomic heat of the solid at constant volume. Atomic heat is a function of emperature. e / At high emp (e / 1) 2 ( )2 becomes 1. C V = 3R which resembles the Dulong and petit s law. Atomic Heat decreases with the decrease in emperature and at Absolute zero temp. tends to zero i.e. C V = 0 Experimental curves for Atomic heat and emperature shows that the curves are same for all the substances. 12

Special case : At high emperature h K C V = 3R becomes small value e ( h /K (e h /K 1).)2 = ( h By e = 1 + x 1! +x 2 K )2 2! +..... e h/k (e h/k 1). ( h K )2 = 1 +( h K ) + 1 2! (h K )2 +.. (e h/k 1) 2 = (1 + ( h K ))2. C V = 3R eh/k ( h K )2 As h K (h K )2 0 C V = 3R [e h/k ] = 3R [e 0 ] C V = 3R(1) which resembles the Dulong and Petit s law with experimental value. 2) At low temperature: As 0 ; h. K C V = 0 hus the Atomic heat tends to zero as emperature approaches to absolute zero. When = 0 ; C V = 0 = Max ; C V = 3R C V ranges from 0 to 3R Atomic heat of different elements differs because of E the characteristic frequency or Einstein s frequency. For elements like Copper, Aluminium, Iron etc. the atomic heat at low temp decreases more rapidly. Debye s heory of specific heat of solids SUDY MAERIAL AVAILABLE FOR PGRB PHYSICS 10 UNI NOES, PRACICE QUESIONS, PREVIOUS YEAR QUESION PAPERS. CONAC:9444249503 By Einstein Assumption:- Vibrations of all atoms are simple Harmonic and they all have one and the same frequency. By Debye s assumption:- Any solid is capable of vibrating elastically in many different modes the frequency varying from one mode 13

he number of modes of vibrating of solids is limited. When soild is subjected to continuous elastic vibrations, 2 kinds of vibrations are set up. a) Longitudinal vibrations b) ransverse vibrations Number of modes of longitudinal vibrations per unit volume with the frequency, and +d is No. of modes = 42 d C e 3 C e Velocity of longitudinal vibrations. In case of transverse vibrations, No. of modes = 82 d or 42 d x 2 C t 2 C t 3 Here C t Velocity of ransverse vibrations. 1 + 1 C e 3 C t 3 m otal No. of vibrations = 4v 2 d SUDY MAERIAL AVAILABLE FOR 0 V Volume, m mode of frequency. PGRB PHYSICS otal thermal energy for 1 gram atom of solid is 10 UNI NOES, PRACICE QUESIONS, m PREVIOUS YEAR QUESION PAPERS. E = 9N h 3 d m 3 e h/k 1 CONAC:9444249503 0 Let characteristic emp 0. = h m ; k m = k Specific heat of 1 gram atom at constant volume or Atomic heat is / C V = de = 3R d 12( )3 x 3 dx 1 e x 1 e / 1 0 he above equation is the Debye s equation for atomic heat of a mono atomic solid. x = h. K C V = 3Rf (/) Where f ( ) is the Deby s function At high emp, C V = 3R he atomic heat of all substances tends to a maximum value equal to 3R. At low emperature near absolute zero. C V = 77.94 x 3R ( )3. C V 3 --------- Deby s 3 law Deby s 3 law states that, At low temperature the atomic heat of solid is directly proportional to cube of the absolute temperature. Atomic heat is zero at absolute zero. he curves showing the variation of specific heat with temperature, is same for all the elements. Also the behavior of elements regarding the variation is similar. 14