Physics 301 11-Oct-2002 13-1Example: Magnetic Particles in a Magnetic FieldRecall the paramagnetic spin system we discussed in lecture 4. In this system, thereare magnets with orientations parallel or antiparallel to a magnetic field. In the parallelorientation, the energy is −mB = −E,wherem is the magnetic moment and B is themagnetic field. In the antiparallel orientation the energy is +mB =+E. In lecture 4,we worked out the relative numbers of parallel and antiparallel magnets and found that itdepended on the ratio of thermal to magnetic energies.Following the discussion in K&K, pages 127–129, suppose that we have the same kindof system, but in addition, the magnetic particles are free to move, so the aligned magnetswill be attracted to regions of high field strength while the antiparallel magnets will berepelled from regions of high field strength. Of course, in the regions of high field, onewould expect to find a greater fraction aligned even if the particles couldn’t move...Let n↑be the concentration of parallel and n↓be the concentration of antiparallel systems. Justas with an ideal gas, we expect that microscopic or internal contribution to the chemicalpotential should depend on the concentration,µ↑,int= τ logn↑nQand µ↓,int= τ logn↓nQ.We assume that we can treat the parallel and antiparallel magnets as distinct kinds of“particles.” To the internal chemical potential must be added the external potential duethe energy in the magnetic field,µ↑= τ logn↑nQ− mB ,µ↓= τ logn↓nQ+ mB .Now, the parallel and antiparallel magnets are in thermal equilibrium with each other andcan be changed into one another. That is, one can remove a particle from the parallelgroup and add it to the antiparallel group and vice-versa. When the system has come toequilibrium, at temperature τ , the free energy must be stationary with respect to changesin the particle numbers which means the chemical potentials of the two kinds of particlesmust be the same. Furthermore, we are allowing the particles to diffuse to regions of higheror lower field strength, and the chemical potential must be independent of field strength.So,µ↑= µ↓= Constant.This relation together with the previous equations are easily solved to yieldn↑(B)=12n(0)e+mB/τand n↓(B)=12n(0)e−mB/τ,Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 11-Oct-2002 13-2where we’re explicitly showing that the concentrations depend on B and n(0) is the com-bined concentration where B = 0. The combined concentration as a function of B isn(B)=n↑(B)+n↓(B)=n(0) cosh(mB/τ )=n(0)1+m2B22τ2+ ···.These relations show both effects we mentioned earlier. The higher the field strength,the greater the fraction of aligned magnets (as we already knew from lecture 4) and thegreater the concentration of magnets. The magnetic particles diffuse to regions of highfield strength.In figure 5.6, K&K show a plot of chemical potential versus concentration for sev-eral different field strengths. In problem 5 of chapter 5, we are asked for what value ofm/τ was this figure drawn. The key datum to extract from the plot is that at a givenchemical potential, the concentration increases by two orders of magnitude as B is in-creased from 0 to 20 kG. We can plug this directly into the previous expression to getm/τ =5. 30/(20000 G) = 0.000265 G−1. Note that we had to use the cosh form of theexpression, not the series, because mB/τ > 1. Problem 5.5 also asks how many Bohrmagnetons must be contained in each particle. A Bohr magneton (roughly the magneticmoment of an electron) is µB= e¯h/2mc where e and m are the charge and mass of anelectron. µB=0.927 × 10−20erg G−1. Doing the arithmetic, we obtain about 1200 mag-netons. The particles must contain 1200 paramagnetic molecules with a spin of ¯h/2anda magnetic moment of µB. They could also contain a more or less arbitrary number ofnon-magnetic molecules.Example: Impurity IonizationIn pages 143–144, K&K discuss an impurity atom in a semiconductor. The atom maylose a valence electron and become ionized. The energy required to remove an electronfrom the donor atom is I. The model for this impurity atom is a three state system:the ionized state has energy 0 and no electron is present. There are two bound states,both have energy −I and both have one electron present. One has the electron with spinup along some axis and the other has the electron with spin down. The grand partitionfunction isZ =1+e(µ + I)/τ+ e(µ + I)/τ,where the first term comes from the ionized state and the second and third terms accountfor the spin up and spin down bound states. The average number of (bound) electronsand the average energy areN =e(µ + I)/τ+ e(µ + I)/τ1+e(µ + I)/τ+ e(µ + I)/τ=2e(µ + I)/τ1+2e(µ + I)/τ,E =−Ie(µ + I)/τ− Ie(µ + I)/τ1+e(µ + I)/τ+ e(µ + I)/τ=−2Ie(µ + I)/τ1+2e(µ + I)/τ,Copyrightc 2002, Princeton University Physics Department, Edward J. GrothPhysics 301 11-Oct-2002 13-3The probability that the impurity atom is ionized isP (N =0)=11+2e(µ + I)/τ.If we don’t know the value of µ, we can’t actually calculate any of these averages or thisprobability. What sets the value of µ?Answer:µ is determined by the electron distributionin the rest of the semiconductor. (A subject we’ll get to in a few weeks!) Although we don’tknow µ at this point, we’re used to the idea that µ increases with increasing concentration.In the above expressions we see that increasing µ increases the mean number of particlesin the system, decreases the mean energy (energy goes down for a bound particle), anddecreases the probability of being ionized. All this is reasonable and might have beenexpected. The higher the concentration of electrons in the semiconductor, the harder it isfor the atom to give an extra electron to the semiconductor and become ionized!Example: K&K, Chapter 5, Problem 6In this problem we are asked to work with a 3 state system. The states are: (1) noparticle, energy is 0; (2) one particle, energy is still 0: (3) one particle, energy is ,soa particle can be absent, present with zero energy, or present with energy . The grandpartition function isZ =1+λ + λe−/τ,where λ =exp(µ/τ ). The three terms in this sum correspond to the three states enumeratedabove. The thermal average occupancy is just the average number of particles in the systemand isN =1Z