Kinetic Molecular Theory of Matter • Heat capacity of gases and metals • Pressure of gas • Average speed of electrons in semiconductors • Electron noise in resistors Positive metal ion cores Free valence electrons forming an electron gas Fig. 1.7: In metallic bonding the valence electrons from the metal atoms form a "cloud of electrons" which fills the space between the metal ions and "glues" the ions together through the coulombic attraction between the electron gas and positive metal ions. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Ideal gas approximation. Pressure Square Container Microscopic model Area A a Face B Face A vy Definition: P (pressure) = force per unit area vx Gas atoms Pressure (P) is caused by the collisions of molecules with walls a a Fig.1.15: The gas molecules in the container are in random motion. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Our task is to show that Temperature (T) is related to average speed of molecules Ideal Gas Assumptions: 1. molecules are in constant and random motions, all directions are equivalent 2. no interactions between molecules but collisions 3. all collisions are elastic 4. molecule size is negligible 5. Newtonian mechanics is valid Connecting the pressure with the average speed of gas molecules Step 1. Change in Momentum of a Molecule Δp = 2mvx N Δp = change in momentum, m = mass of the molecule, vx = velocity in the x direction Step2. Force produced by unique molecule Δp 2mv x mv x F= = = Δt 2a / v x a 2 Step3. Calculating pressure 2 2 2 2 mN v x mN v x1 + v x2 + ... + v x N Total _ force mv x1 + mv x2 + ... + mv x N = 3 = P= = 2 3 a a a N V 2 Step4. Averaging speed v x + v y + v z = v 2 = 3v x 2 2 2 2 2 2 Nmv 2 1 2 P= = ρv 3V 3 Ν = total number of molecules ρ = density of gas v = velocity of molecules Pressure ÙTemperature Ú Ú Volume Ideal Gas Equation ⎛ N ⎞ ⎟⎟ RT PV = ⎜⎜ ⎝ NA ⎠ P is the pressure, i.e. force per unit area N is the total number of molecules in volume V R is the gas constant (8.3144 J mol-1 K-1) NA is the Avogadro’s number (6.022 ×1023 mol-1) Experimental law – macroscopic point of view Gas Pressure in the Kinetic Theory 2 ⎞ ⎛ m v 1 2 ⎟ P = ρ v2 = N ⎜ 3 3 ⎜⎝ 2 ⎟⎠ Ideal Gas Equation ⎛ N ⎞ ⎟⎟ RT PV = ⎜⎜ ⎝ NA ⎠ P = gas pressure, N = number of molecules, m = mass of the gas molecule, v = velocity, V = volume, ρ = density. Mean Kinetic Energy per Atom 1 2 3 KE = mv = kT 2 2 k =R/NA Boltzmann constant, T = absolute temperature 0 K ≈ -273 0C Gas constant Boltzmann’s constant R = 8.3144 J mol-1 K-1 k = 8.61 × 10-5 eV K-1 R = k × NA R – macroscopic (technical, thermodynamical etc.) calculations k – microscopic (atomic) calculations Mole (a gram molecule) = a quantity of a substance equal to the molecular weight of a substance expressed in grams Example: Carbon has atomic mass of 12 ⇒ a mole of carbon is 12 grams Mole contains 6.0220 × 1023 species (atoms, molecules, etc.) NA = 6.0220 × 1023 - an Avogadro's number KE = Internal Energy per Mole for a Monatomic Gas 1 2 3 mv = kT 2 2 ⎛ 1 2⎞ 3 U = N A ⎜ mv ⎟ = N A kT ⎝2 ⎠ 2 U = total internal energy per mole, NA = Avogadro’s number, m = mass of the gas molecule, k = Boltzmann constant, T = temperature Molar Heat Capacity at Constant Volume dU 3 3 Cm = = N Ak = R 2 2 dT Cm = specific heat per mole at constant volume (J K-1 mole-1), U = total internal energy per mole, R = gas constant Definition: Heat capacity is the rise of internal energy per unit temperature Degree of freedom (DF) = the method to absorb energy Examples: Monoatomic gas has 3 DF Diatomic gas has 5 DF Maxwell’s principle of equipartition of energy : Each DF has an average energy ½ kT Monoatomic gas has 3 DF. E=3/2 kT Cm=3/2 R Diatomic gas has 5 DF. E=5/2 kT Cm=5/2 R Solids have 6 DF. E=3kT Cm=3 R (Dulong - Petit rule) TRANSLATIONAL MOTION ROTATIONAL MOTION Ix= 0 vx vz x z Iz y vy Iy y axis out of paper Fig. 1.16: Possible translational and rotational motions of a diatomic molecule. Vibrational motions are neglected. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca 1 2 1 2 1 2 mv x + mv y + mvz 2 2 2 Monoatomic gas: E= Diatomic gas: 1 2 1 2 1 2 1 1 2 E = mv x + mv y + mvz + I yω y + I zω z2 2 2 2 2 2 KE = 3 kT 2 U = 5 kT 2 Iy,z = momenta of inertia, ωy,z = angular velocities E= (a) 1 2 1 2 1 2 mv x + mv y + mv z + 2 2 2 + 1 1 1 K x x2 + K y y2 + K z z 2 2 2 2 y x Kx,y,z = spring constant, x, y, z = extensions of springs z (b) Fig. 1.17 (a) The ball-and-spring model of solids in which the springs represent the interatomic bonds. Each ball (atom) is linked to its neighbors by springs. Atomic vibrations in a solid involve 3 dimensions. (b) An atom vibrating about its equilibrium position stretches and compresses its springs to the neighbors and has both kinetic and potential energy. 6 U = kT = 3kT 2 From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Dulong - Petit Rule U = 3kT dU -1 -1 Cm = = 3R = 25 J K mol dT R = Na × k Cm = specific heat per mole at constant volume (J K-1 mole-1), U = total internal energy per mole, T = temperature, R = gas constant Relative number of molecules per unit velocity (s/km) What is the distribution of molecule speeds ? 2.5 v* va v vrms 2 1.5 1 3 KE = kT 2 298 K (25 °C) v* va v vrms 1000 K (727 °C) 0.5 0 0 500 1000 1500 2000 Speed (m/s) Fig. 1.21: Maxwell-Boltzmann distribution of molecular speeds in nitrogen gas at two temperatures. The ordinate is dN/(Ndv),the fractional number of molecules per unit speed interval in (km/s)-1 From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca How to determine the speed of molecule ? Relative number of molecules per unit velocity (s/km) What is the influence of temperature ? 2.5 v* va v vrms 2 1.5 1 298 K (25 °C) v* va v vrms 1000 K (727 °C) 0.5 0 0 500 1000 1500 2000 Speed (m/s) Fig. 1.21: Maxwell-Boltzmann distribution of molecular speeds in nitrogen gas at two temperatures. The ordinate is dN/(Ndv),the fractional number of molecules per unit speed interval in (km/s)-1 From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Maxwell-Boltzmann Distribution for Molecular Speeds ⎛ mv 2 ⎞ ⎛ m ⎞3/ 2 2 ⎟⎟ ⎟ v exp⎜⎜ − nv = 4π N⎜ ⎝ 2π kT ⎠ ⎝ 2kT ⎠ nv = the velocity density function, N = total number of molecules, m = molecular mass, k = Boltzmann constant, T = temperature, v = velocity Maxwell-Boltzmann Distribution for Molecular Speeds ⎛ mv 2 ⎞ ⎛ m ⎞3/ 2 2 ⎟⎟ ⎟ v exp⎜⎜ − nv = 4π N⎜ ⎝ 2π kT ⎠ ⎝ 2kT ⎠ E=½mv2 nv = the velocity density function, N = total number of molecules, m = molecular mass, k = Boltzmann constant, T = temperature, v = velocity Maxwell-Boltzmann Distribution for Translational Kinetic Energies nE = ⎛ 1 ⎞ 3 / 2 1/ 2 ⎛ E⎞ N ⎜ ⎟ E exp ⎜ − ⎟ ⎝ kT ⎠ π ⎝ kT ⎠ 2 nE = number of atoms per unit volume per unit energy at an energy E, N = total number of molecules per unit volume, k = Boltzmann constant, T = temperature. Boltzmann Energy Distribution ⎛ E⎞ nE = C exp⎜ − ⎟ ⎝ kT ⎠ N nE = number of atoms per unit volume per unit energy at an energy E, N = total number of atoms per unit volume in the system, C = a constant that depends on the specific system, k = Boltzmann constant, T = temperature Number of atoms per unit energy, nE Where lies the average kinetic energy ? Average KE at T1. T1 Average KE at T2 T2 > T1 EA Energy, E Fig. 1.22: Energy distribution of gas molecules at two different temperatures. The number of molecules that have energies greater than EA is the shaded area. This area depends strongly on the temperature as exp(-EA/kT). From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca What is “thermal equilibrium”? E= 1 2 1 2 1 2 mv x + mv y + mv z + 2 2 2 + 1 1 1 K x x2 + K y y2 + K z z 2 2 2 2 SOLID GAS M KE=3/2 kT 1 2 1 2 1 2 mv x + mv y + mvz 2 2 2 1 1 + I yω y2 + I zω z2 2 2 E= V v m Gas Atom KE=3/2 kT Fig. 1.23: Solid in equilibrium in air. During collisions between the gas and solid atoms, kinetic energy is exchanged. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca What is “thermal equilibrium”? SOLID GAS KE=3/2 kT M V KE=3/2 kT v m Gas Atom Heat = amount of energy transferred from kinetic energy of the atoms in solid to the kinetic energy of gas molecules In equilibrium heat transfer = 0 Mechanical “noise” Compression Equilibrium Extension Δx m Δx = 0 Compression Mean displacement Δx = 0 Instantaneous potential energy Δx < 0 PE(t)=½K(Δx)2 m Mean potential energy ½K(Δx)2=½kT Δx> 0 Extension m t Fig.1.24: Fluctuations of a mass attached to a spring due to random bombardment by air molecules (Δx )rms kT = K From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca K = spring constant, T = temperature, (Δx)rms = rms value of the fluctuations of the mass about its equilibrium position. Electrical noise A B B A v=0V v = -3 μV Voltage, v(t) B A Time v = +5 μV Fig.1.25: Random motion of conduction electrons in a conductor results in electrical noise. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Average value = 0 Average power = ? Root Mean Square Noise Voltage Across a Resistance 1. Instantaneous energy Current v(t) Electron Flow R E(t)=½ C V(t)2 2. Mean energy C Time 1 1 E = CV (t ) 2 = kT 2 2 3. RMS voltage kT V (t ) 2 = C Electron Flow Current 4. Bandwidth Fig.1.26: Charging and discharging of a capacitor by a conductor due to the random thermal motions of the conduction electrons. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Vrms = [4kTRB]1/2 B= 1 2πRC Johnson resistor noise equation R = resistance, B = bandwidth, Vrms = root mean square noise voltage, k = Boltzmann constant, T = temperature Root Mean Square Noise Voltage Across a Resistance E(t)=½ C V(t)2 Instantaneous energy 1 1 E = CV (t ) 2 = kT 2 2 Mean energy and equipartition principle V (t ) 2 = B= kT C 1 2πRC RMS voltage Bandwidth Vrms = [4kTRB]1/2 Johnson resistor noise equation R = resistance, B = bandwidth, Vrms = root mean square noise voltage, k = Boltzmann constant, T = temperature