PHYSICS 112Homework 4Due in class, Tuesday, Feb. 7.MIDTERM: The midterm will be in class on Thursday, Feb. 9, and will cover the material on the firstthree homework assignments (plus the Debye theory of specific heats). You should review the solutionsfor these as part of your p re par at i on for the mi dte r m. The midterm will be c l ose d book but you canbring one sheet of your own notes.NOTE: To do some of these questions you will need the result for the chemical potential of a classicalideal gas. We will derive this a little later in the course but for now I just quote the result. It isµ(T ) = kBT ln(nVQ) , (1)where n = N/V is the density of atoms andVQ=2π¯h2mkBT3/2is called the “quantum volume” (m is the mass of the atom). The physical significance of VQ(whichis written as 1/nQin the book) is as follows. Particles of momentum p have a de Broglie wavelengthequal to 2π/k where k is given by p = ¯hk. Consider a particle whose kinetic energy is kBT (a typicalvalue). The correspon di ng “thermal de Broglie ” wavelength, λTis given by12m¯h2πλT2= kBT ,orλT=√πs2π¯h2mkBT. (2)Hence apart from the numerical factor of√π, the quantum volu me is the cube of the thermal de Brogliewavelength. It will figure extensively in our discussion of the ideal gas (bot h classical and quantum) alittle later in the course.The combination nVQis therefore the cube of the ratio of the thermal de Broglie limit to the typicalinterparticle spacing (remember that n is the inverse of the volume per particle). It i s there for e theaverage number of particles in a volume equal to VQ.A li t tl e late r in the course we shall see that the classical limit is whenλ ≡ exp(βµ) ≪ 1 ,or equivalently µ is l ar ge and negati ve. From Eq. (1), we see that for the ideal gas the classicallimit can be expressed asnVQ≪ 1 . (3)In other words the classical limit is when the average number of particles per quantum volume isvery much l es s than one, or in other words, the typical interparticle spacing is much greaterthan the thermal de Broglie wavelength. Since the thermal de Broglie wavelength is proportionalto T−1/2, see Eq. (2), the ideal gas is always classical at hi gh enough temperature, but quantum effectsbecome important at sufficiently low tem peratures. Quantum effects start to be important when λ ≃ 1,or e q ui val e ntly when nVQ≃ 1. Note that Eq. (1) is no longer valid at temperatures lower than this.I em phasi z e that µ(T ) and VQare temperature dependent.11. Heat capacity of photons and phonons. (Left over from HW 3.)Consider a solid with a Debye temperature equal to 100 K and with 1022atoms cm−3. Estimatethe temperature at which the photon contribution to the he at capacity would equal to the phononcontribution at 1 K.Note: Use the Debye approximation for the specific heat of the phonons. Note that C ∝ T3forboth photons and phonons because they have the same dispersion re l ati on ǫ ≡ ¯hω = ¯hck, where cis the speed of the wave. However, the speed of light is much greater than the spe ed of sound ina solid, which accounts for your answer being different from 1K. In fact Eq. (46), ( i n K i t te l andKroemer, 2nd Ed.), which gives the phonon energy in the Debye theory at low T , is equivalent tothe Stefan-Boltz mann law, Eq. (20). This can be see n by using the relation between the Debyetemperature θD, and the sound speed and atomic density, given in Eq. (44). The expressionfor phonons is 1.5 times as big because, for each wavevector, there are three polarizations forsound (one longitudinal and two transverse) whereas light only has two polarizations (there is nolongitudinally polarized light wave).2. Potential Energy in a Gravitational FieldIn class we will sh ow that concentration of molecul es in a gas in a gravitational field var i e s withheight h according ton(h) = n0exp(−mgh/kBT ) , (4)where g is the acceler ati on due to gravity, m is the mass of a molecul e , and n0is the density ath = 0. Find the total heat capacity per atom.Note: The main part of the question is to determine the average potential energy per atom. Theaverage kinetic energy per atom you may take to be (3/2)kBT (which will be discussed in class).You will need to average from h = 0 (the bottom of the gas) to h = ∞).3. Grand Partition Function for a Two-Level System(a) Consider a system which may be unoccupied with zero energy, or occupied by one particle ineither of two states, on e of energy 0 and the othe r of energy ǫ. Show that the grand partitionfunction (Gibbs sum) for this system isZ = 1 + λ + λ exp( − βǫ) ,whereλ = exp(βµ)is the “fugaci ty” (activi ty) and β = 1/kBT .Note: We assuming here that we cannot have two particles in the system at the same ti me .(b) Show that the average number of particles is given byhni = λ1 + exp(−βǫ)Z.(c) Show that the probability of the system being in the state with energy ǫ isλ exp(−βǫ)Z .(d) Determine the average energy of the system.2(e) Allow for the possibility that the orbital at 0 and that at ǫ can be each occupied by a particleat th e same time. Show that nowZ = 1 + λ + λ exp( − βǫ) + λ2exp(−βǫ) = (1 + λ) [1 + λ exp(−βǫ)] .Note: Because Z can be factored, we actually have two independent systems.4. States of Positive and Negative IonizationConsider a large number of hydrogen atoms fixed in space (so they are distingui shabl e ) . Supposethat they can e ach exist in one of four statesState Number of electrons EnergyGround 1 −12∆Positive ion 0 −12δNegative ion 212δExcited 112∆Show that the condition that the average number of electrons is unity can be expressed asλ2= exp(βδ) .5. Carbon Monoxide PoisoningIn carbon monoxide poisoning a CO molecule replaces the O2absorbed on a hemoglobin (Hb)molecule in the blood. To illustrate this effect, consider a simplified model in which each absorptionsite on the Hb may be vacant (with e ne r gy zero), or occupie d either by one molecule of O2, inwhich case it has energy ǫA, or by one molecule of CO, in which case it has energy ǫB.Let there be N Hb molecules in equilibrium with gaseous O2and CO. Let the activities (fugacities)be λ(O2) = 1 ×10−5and λ(CO) = 1 ×10−7(i.e. the fugacity of CO is less th an that of O2so theconcentration of CO will be smaller than that of O2). Also assume body temperature, T = 37◦C.(a) First assume that there is no CO (i.e. temporarily assume that λ(CO) = 0). Find the valueof ǫAsuch that