Some Important Results of Statistical Mechanics

Some important results of Statistical Mechanics Ionization Equilibrium - Saha’s Equation Consider the ionization/recombination reaction for an atom A: A ↔ i+ + e− (1) The Law of Mass Action for this reaction is,  n e ni qe qi = na qa Q q = V  (2) When calculating the various partition functions, it is very important to measure all energies from the same reference. For this particular case, suppose we arbitrarily assign zero energy to an atom at rest and to an ion at rest. Since the energy needed to produce an electron-ion pair is eVi , we then have to assign the energy eVi to an electron at rest, or eVi + R i . The translational partition function for an electron,  Qtr. e = 2πme kT h2 3/2 V was derived based on the set R i for the free electron in a box. Since Q = have, Qtr e =  − gi e eVi +R i kT  = i 2πme kT h2 3/2  i i gi e− kT , we now eVi V e− kT (3) Including now the spin degeneracy g = 2 of a free electron,  2πme kT qe = 2 h2 3/2 eVi e kT (4) Similarly, for ions and atoms, including their electronic excitation degeneracy gi , ga ,  q i = gi  qa = ga 2πmi kT h2 2πma kT h2 3/2 (5) 3/2 (6) and since mi  ma , Equation (1) yields,  3/2 eV n e ni gi 2πme kT − kTi = 2 e na ga h2 (7) which is called the Saha equation for ionization equilibrium. As developed, the equation applies to ionization of an atom. Molecules and molecular ions have other degrees of freedom (vibration, rotation), and the equation needs to be modified 1 accordingly. But the modifications are formally very simple: replace gi , ga by the respective Internal Partition Functions: vib rot Qint = Qexc i i Qi Qi · · · and then, In MKS units, exc vib rot Qint m ; = Qm Qm Qm · · · (m for molecule) (8)  3/2 eVi ne ni 2πme kT Qint i = 2 int e− kT 2 Qm nm h (9) eVi ne n i Qint = 2 iint × 4.84 × 1021 T 3/2 e− kT nm Qm (10) Vi eVi The dominant T dependence is in e− kT = e− T (eV ) . For moderately low temperatures (up to a few eV ), this factor is extremely sensitive: d ln(e−Vi /T ) Vi 1 = T d ln T In quasi-neutral plasma ni = ne , and Saha relates ne to na . The full equilibrium composition requires an extra condition, typically the pressure, P = (ne + ni + na )kT = (2ne + na )kT Useful properties for selected atoms: Element M (g/md) Vi (V) H 1.008 13.6 He 4.003 24.6 Li 6.94 5.39 N 14.01 14.6 O 16 13.6 Ne 20.18 21.6 Na 23.00 5.14 A 39.94 15.8 K 39.10 4.34 Xe 131.3 12.1 Cs 132.9 3.89 Hg 200.6 10.4 E12 (V)(1st exc.) g0atom 10.2 2 19.8 1 1.85 2 2.38 4 1.97 ∼9 16.6 1 2.10 2 11.5 1 1.61 2 8.32 1 1.39 2 4.67 1 2 g0ion ξ1 (Outer shell electrons) 1 1 2 2 1 1 ∼9 3 4 4 ∼6 6 1 1 ∼6 6 1 1 4 6 1 1 2 2 Free electrons in metals As we saw already, electrons obey Fermi-Dirac statistics: Ni = gi α+βε i e +1 α=− where μ kT and β= 1 kT We assume, to first order that electrons do not interact among themselves, so only translational degrees of freedom will be considered. Let us analyze the case of very low temperature, in the limit T → 0, and define the “Fermi Energy” as μ = εF , so we have, Ni = gi e εi −εF kT +1 The exponential part of this expression becomes extremely large or very small whenever εi = εF , so we identify two regions of interest depending on the energy with respect to the Fermi level: 1. When εi < εF then Ni = gi Since there can only be one particle in each state (electrons obey the exclusion principle), all states are occupied. We say the electron gas is “degenerate”. 2. When εi > εF then Ni = 0 All states are empty. This situation can be illustrated with the following diagram: Occupation index Ni 1 gi Fermi Sea εF εi Even at T = 0 there are particles with energies larger than zero (εi < εF ) contrary to the classical description of an ideal gas that has zero energy for zero temperature. What is the Fermi Energy? We know that all particles are contained in the states with energies lower than εF , and that there is only one particle per state. Assuming that levels are closely spaced between each other, so the summation can be replaced with an integral, the total number of particles becomes, ˆ εF N= dg 0 3 After solving Schrödinger’s equation for a free particle in a box with volume V = Lx Ly Lz = L3 we find that the quantum number n and the energy are related by,  L n= π 2mε 2 n2 = n2x + n2y + nz2 where Earlier we saw that the degeneracy is the number of quantum states that share the same energy level. In this case we see that several combinations of the three numbers nx , ny , and nz will yield a particular n. In fact (see figure), every point over the spherical surface in the n-space (in the positive octant where the domain of the n’s coincide, since ni > 0 for i = x, y, z) will have the same energy. The elementary volume in this space will then be the elementary part of the degeneracy, nz dn dg = 2 n 4πn2 dn 8 where the “2” comes from the spin of electrons (“up” or “down”), while the “8” is there since we are only interested in the octant where the n’s exist. nx Combining the last two expressions we see that, ny ∂g V dg = dε = gε dε = 2 ∂ε 2π  2m 2 3/2 √ εdε which can be integrated directly, ˆ 0 εF V dg = 2 3π  2m 2 3/2 3/2 εF = N Finally we solve for the Fermi energy,  εF = 3π 2N V 2/3 2 2m Example: For Cu, N # electrons # atoms NA 6.02 × 1026 kmol−1 = 8940 kg/m3 = 8.5 × 1028 m−3 = = ρ= Vol V Vol w 63.5 kg/kmol For this example, the Fermi energy results in εF = 7.04 eV ≈ 80, 000◦ K. 4 Fermi energy ~2kT This means that T does not need to be very small (for instance, room temperature, at 300K) for the electron gas to be very degenerate. In general, at temperatures larger than zero, the particle distribution is modified just slightly, and will look similar to the figure above. Of course one could argue that not all particles in the system have the Fermi energy, perhaps many of them will have energies significantly smaller. To resolve this issue we can calculate the mean energy per particle ε̄ = E0 /N , where E0 is the total energy at zero temperature, which can be found by direct integration, ˆ E0 = ˆ εF εF εdN = 0 εdg 0 Using previous expressions, written in terms of the Fermi energy, 3 −3/2 E0 = N ε F 2 ˆ εF 0 3 ε3/2 dε = N εF 5 Then, the mean energy is, ε̄ = E0 3 = εF N 5 This value is not too far from the Fermi energy (just 3/5 of it). This means that most particles have large energies even at zero temperature. This situation can be better understood with a diagram. First we note that (for closely spaced energy levels) the F.D. distribution can be written as (when T = 0), dN = ∂N dε = Nε dε = gε dε ∂ε therefore Nε = gε Where Nε is the number of particles per unit energy. It is clear that the electron population grows with the square root of the energy up to the Fermi level, where it suddenly goes to zero. Now, we consider also the existence of free electrons outside the metal, these particles may come from the material surface after being extracted from it. We define eφ as the work required to extract an electron from the metal surface at the Fermi level (the work function). 5 ε Free electrons (outside the metal) eϕ εF Those electrons outside of the metal can be described by the classical approximation of a diluted gas, and therefore their chemical potential and partition function (for translational and spin degrees of freedom) are given by, No electrons here! μext = −kT ln Q N and, Fermi electrons (inside the metal)  s tr Q=Q Q =2 2πmkT h2 3/2 V We note that these expressions are written at the zero energy level of the free electrons outside of the metal. In other words, free electrons (energetically speaking) are located at an energy μext + εF + eφ with respect to the zero energy level in the Fermi system. To make the analysis consistent, we shift the energy levels of the electrons outside of the metal to match the Fermi level, this is simply done by making μext = −eφ. In this way, the equations above can be used to write, Nε   3/2  2πmkT 1 eφ = kT ln 2 h2 n∗e with n∗e = N V where n∗e is the free electron density. The flux of electrons (number of particles that cross an arbitrary surface, per unit time, per unit area) is given by, n∗ c̄e Γe = e 4  with c̄e = 8kT πme It is important to realize this is the flux of electrons coming from outside of the metal. In equilibrium, an equal flux must be leaving the metal. The (bi-directional) electron current density is then given by, je = eΓe = eφ eφ 4πeme (kT )2 e− kT = 120 × 104 T 2 e− kT Cm−2 3 h and if electrons are withdrawn from the outside, only the emission part remains. This is known as the Richardson-Dushman thermionic emission law. Thermionic emission enhanced with a normally applied electric field To increase the extraction rate from the metal surface, we could apply an electric field E normal to it. To quantify this situation we compute the work W (potential energy) of the extracted electron assuming that the only force that binds it to the surface is that of its image charge inside the material. 6 This energy is formed by two parts: that required to bring the particle from infinity to a position x from the metal surface under the image force plus the one that arises from applying the electric field, ˆ x −e2 W = F dx + eE x with F = 4π0 (2x)2 ∞ metal E +e -e -x +x x therefore, W = e2 + eEx 16π0 x In terms of the potential, we have, φE = − W e =− − Ex e 16π0 x We take the derivative of this potential and set it to zero to find the position xm where φE is maximum,  xm = e 16π0 E The net effect of E is to decrease the potential barrier by,  φE,max = − eE 4π0 thus modifying Richardson’s law which we rewrite here as, je = eΓe = e 4πeme 2 − kT (kT ) e h3 7 “ q φ− eE 4π0 ” Black Body radiation We now turn to Bose-Einstein statistics since photons do not obey Pauli’s exclusion principle. Furthermore, the Lagrange multiplier α identified with the conservation of the total number of particles in a system will vanish here, since photons are not conserved in number. Energy on the other hand, is conserved. For example, a photon with energy 2hν can be absorbed by the material surface while two photons with energy hν can be emitted. If the energy levels are closely spaced, then we can write the B.E. distribution as, dN = dg e hv kT −1 ∂N ∂g 1 dν = hv dν ∂ν e kT − 1 ∂ν or or Nν dν = gν dν e hv kT −1 where Nν and gν are the number of particles and degeneracies per unit frequency, respectively. The wave equation solution for photons in a “box” of sides Lx , Ly and Lz is proportional to (sin kx x)(sin ky y)(sin kz z). The solution vanishes at x, y, z = 0, but also must vanish at every boundary, so we require (consider all sides of the box of length L), kx Lx =nx π ky Ly =ny π kz Lz =nz π and since Lx = Ly = Lz = L k 2 = kx2 + ky2 + kz2 2 n = nx2 + n2y + we have that kL = nπ n2z As with the analysis of free electrons in metals, the degeneracy of this system is related to the number of ways in which the quantum numbers that form n can be arranged to give the same energy. So we have once more dg = 2(4πn2 dn)/8, but this time the “2” comes from the two possible polarizations of electromagnetic waves. Now we relate n with the frequency, k= 2π 2π = ν λ c and since k= nπ L then n= 2L ν c In this way, the degeneracy can be written as dg = (8πV /c3 )ν 2 dν and the number of particles per unit frequency is therefore, Nν = 8πV ν2 hν c3 e kT − 1 and the number density (per unit frequency) is nν = Nν V With this, we can calculate Planck’s Formula, which is the energy per unit volume and per unit frequency, uν = hνnν = 8πh ν 3 hν c3 e kT −1 which depends only on temperature, a result not predicted by classical mechanics. The spectral emission (energy flux) is obtained from, qν = nν c 2πh ν 3 hν = 2 hν 4 c e kT − 1 which can be integrated to obtain the overall emission, 8 ˆ q= 0 ∞ 2πh qν dν = 2 c ˆ ∞ 0 ν 3 dν e hν kT −1 and by making x = hν/kT , we can write the emission as, 2πh q= 2 c  kT h 4 ˆ ∞ 0 x3 dx ex − 1 The integral in this expression can be evaluated numerically or by expanding the exponential in power series, integrating each element and then finding to what value the new integrated series converges to. This value is π 4 /15. We finally find the radiative power law, q = σT 4 where the Stefan-Boltzmann constant σ is given by, σ= 2π 5 k 4 −8 W = 5.67 × 10 15c2 h3 m2 K2 9 The Maxwellian distribution function for velocities From the classical approximation of a diluted gas, we already found that, Ni = N − εi gi e kT Q dN = N −ε e kT dg Q For closely spaced energy levels, We assume only translational degrees of freedom are relevant,  Q= 2πmkT h2 3/2 V and ε = π 2 2 n2 2m V 2/3 Once more, the degeneracy is given by the points on the sphere in n-space which have the same energy, so dg = (4πn2 dn)/8, and as the energy is ε = 12 mu2 , then, m h dg = 4πV 3 u2 du Defining f (u) as the number of particles per unit u interval and per unit volume, f (u) = dN/V 4πu2 du We then find that, f (u) = n m 2πkT 3/2 mu2 e− 2kT where n = N V A result that we had previously “discovered” through Kinetic Theory. 10