( ) ( )( k B ( ) ( ) ( ) ( ) ( ) ( k T B ) 2 = ε F. ( ) π 2. ( ) 1+ π 2. ( k T B ) 2 = 2 3 Nε 1+ π 2. ( ) = ε /( ε 0 ).

Transcription

1 PHYS47-Statistical Mechanics and hermal Physics all 7 Assignment #5 Due on November, 7 Problem ) Problem 4 Chapter 9 points) his problem consider a system with density of state D / ) A) o find the ermi energy,, use the method of equation 98 and 99: N dd d N ) Use 96 µ ) π 6 Use 96, µ ) π 6 Alternative Solution: D )d N exp β )+ ξ Use f v ) v Γ v + µ Solving, k D / D / ) D ) k ) π / ) 6 / ) k ) π ) k ) π / ) 6 / ) k ) π D k ) + v v f Γ π + π x dx exp k x)+ 6 ξ + f k µ / + µ /k ) N k µ N + π k µ + + π k µ Now aylor expand to get µ π k C) Use 9), E gs D d rom 98, E E gs + π 6 D d k ) N + π +! f ) + π k µ + π k µ / k k + +, and part A, N E gs N 6, k ) N + π k

2 rom eqn 99, π k D )k π k π Using N, k Nπ /k π Note the question ask that the answers must be expressed in terms k, N,, and he above forms are ok if you show how the expressions are transformed into N Alternative Solution: lnξ PV k dd ) ln + exp β ) d ln + exp β ) Integrating by parts lnξ β d exp k β )+ lnξ k f ) β rom earlier N k f ) Use E lnξ f ) E Nk f β f and ) µ /k ) Γ k k x dx, exp x)+ f ) f ) µ /k + π k! µ +, f k +π! µ + E Nk f f ) Nk µ /k ) k +π µ + + π k µ + Now use the result of, µ π k E Nk k π k k +π including terms that will contribute up to k + π k Now aylor expand E N + π 6 +π π k N + π k E π N k k Nπ, where we are only

3 Problem ) Problem 6 Chapter 9 points) or a system where fermions have energy α α, with α a positive integer A) It is clear that N or N, J, 8 J 65eV ) At,k ev /K K 6eV asically, so this is a quantum gas It is clear that just like the D harmonic the density of state is constant and is simply D / Use equation 9, E gs D d E E gs + π 6 D k ) N + π 6 N N d N k ) N rom equation 99, π k D )k π N k Nπ Alternative Solution for D )d N k ) exp β )+ Γ ξ Use f v ) v Γ v + N k f ) k lnξ PV k + v v π rom 98, x dx exp k x)+ f 6 ξ + f µ k µ N dd ln + exp β µ /k )! + π k ) d ln + exp β ) Integrating by parts lnξ d β exp k β )+ Γ lnξ k f β ) f ) Use E lnξ rom earlier N k f E Nk f f + rom a) N x dx, exp x)+ β k k f Use f µ k f ) µ /k + π k! µ + E Nk µ k + π k µ +, and using µ E N + π k,c E V Nπ k /k Nπ

4 Problem ) Problem 7 Chapter 9 points) Heating by adiabatic expansion Initially, an ideal gas of N fermions is confined, by an insulated wall that does not allow heat exchanged, to a volume, V i, and has, effectively, temperature ero he volume is increased adiabatically no heat added or removed) suddenly to V When the gas comes to equilibrium it is a classical ideal gas A) What is the final temperature of the gas? rom 9, energy is E 5 N, and 9, / / h N m 8π V E N 5/ h / i 5 V i m 8π At final equilibrium, since no work is done and heat is not exchanged, the equipartition theorem gives, E Nk N 5/ / 5 V i h m 8π / / h 5 m 8π / N V i /, which is independent of the final volume, V ) Derive an equation for the ratio, V /V i and explain your reasoning? In class we discussed the Sackur-etrode equation that shows that the change in entropy / V of a classical ideal gas is ΔS S S i Nk ln V i, which for a classical i adiabatic process, the temperature does not change, i, ΔS S S i Nk ln V V i Note that the textbook makes similar arguments in section 4 to derive equation ut for this system the initial entropy should be ero, if it is a quantum system at ero temperature, as required by the third law of thermodynamics ut if the system is a classical ideal gas, the initial entropy is greater than ero, unless the temperature is really low his means that the actual change in entropy is ΔS > Nk ln V V i leading to inequality V V i < exp ΔS Nk, in this adiabatic quantum to classical transition, where ΔS is the actual entropy change C) he title of this Heating by adiabatic expansion is an oxymoron Explain the sense in which the title is a contradiction in terms, and the sense in which the title is a legitimate use of words Answer: In adiabatic process, no heat are exchanged, hence it should be possible to heat the system However it is true that there is no heat flow from the outside, or flow out of the system, so the process is a true adiabatic process here also is an increase in temperature due to the transition from quantum to classical system

5 Problem 4) D ermi Electron Gas in the High and Low emperature Limit: In class we showed that N V λ f /, λ h / πmk )) /, and exp βµ ) P k λ f 5/, E Nk f 5/ f /, with V A) 5 points) Eq 74 gives the classical chemical potential is µ k ln Nλ, show that in the classical ideal gas limit µ <, and that for very dilute gas the chemical potential is a large negative number Hence explain in no more than two sentences why < in the classical ideal gas limit V V In dilute limit, V/N is large, is large, andln Nλ Nλ > µ k ln V Nλ V Since exp βµ ) exp ln Nλ V Nλ ) 5 points) Show that in the classical ideal gas limit, the D ermi Ideal gas relations given above becomes PV Nk, E /)Nk, and µ k ln V N λ HIN: use properties of f v ) for low E Nk f 5/ ) + +, 5/ 5/ f / f / ) Using the appendix, f 5/ / + / +, where E Nk f 5/ f / ) Nk Nk C) 5 points) Use the free-energy relations in the appendix to show E S Nµ PV rom appendix, Helmholt free energy E S ; Gibbs G + PV Nµ, + PV Nµ E S + PV Nµ E S Nµ PV D) 5 points) In class we show that at low temperature, µ π k, and E 5 N + 5π k Use result of C to find heat capacityc E V and entropy S at low temperature up to linear order in temperature Comment on whether your results obey the third law of thermodynamics

6 E 5 N + 5π k, as required by the third law of hermodynamics or the entropy, start with, N V λ f / V λ N and f / P k λ f 5/ PV k V λ f 5/ N f 5/ Nk π k Note as,, and E f / Nk f 5/ ), which means PV E rom C) E S Nµ PV S E + PV Nµ S 5 E Nµ Using µ π k and E 5 N + 5π k, S 5 E 5 N + 5π k N π k, S π k Note as, S, as required by the third law of hermodynamics Nk NOE O MARKER: In original version, I forgot the factor N in the relation for the energy E 5 N + 5π k Please do not take off marks from students Problem 5) D Ultra-relativistic ermion gas with dispersion relation ap, a is a constant In this problem neglect the spin in all calculations A) 5 points) Show that the density of state is D As always start with quantum counting ap dp d /a, V 4π h a f / V h 4πp dp, with dispersion relation V h 4πp dp V h a 4π d, and D ) V 4π h a ) 5 points) Show N d exp β )+ D ) N V b m β f m ), where b, m and m are constants that you are expected to determine

7 N V 4π h a N 4πV k ha d Write x β, exp β )+! Γ x d 8πV exp k x)+ ha f ) Hencem m, and b 8πV h a C) 5 points) he heat capacity of the D Ultra-relativistic ermion gas are f 4 ) Nk f ) 9 f f ), C P f 4 ) Nk f ) +6 f f 4 ) ) original version was f ) wrong)using the properties of the ermi-dirac unction given in the appendix show that the heat capacities obey the third law of thermodynamics HIN: Use lowtemperature expansion in the appendix, and since is large, keep only the first term in the expansion ξ rom appendix, f v ) v + v v ) π Γ v + 6 ξ + At very low, keep only the first term, f ξ ; f ξ! ; f ξ 4,ξ βµ 4 4! f 4 ) 4 /4) /6 Nk ξ /6 ξ / f 9 f f ξ 9 ξ ), as required by third law of thermodynamics NOE O MARKER: Due to error do not grade the part below onc P or C P f 4 Nk f +6 f f 4 ) f ) ξ 4 /4) ξ /6) +6 ξ /) ξ 4 /4) ξ /6) Appendix from Statistical Mechanics, rd edition, Pathria and eale Grand-canonical Ensemble Grand partition function, Ξ N i exp βe i ), exp βµ i Probability of state i with energy E i and particle number N i, P i N i exp βe i Ξ Average Energy E lnξ β and average particle number N N lnξ i Grand Potential Ω g k lnξ PV ree Energy Helmolt E S ; GibbsG + PV Nµ ; Ω g Nµ

8 Shannon s heorem S k p i lnp i i ermi-dirac D) Degenerate gas Average number of particle in state of energy, n D exp β )+ N PV k, exp βµ ) ) exp β )+ d exp β )+ D, PV k lnξ ln + exp β ) dd ln + exp β ermi-dirac unctions f v ) x v dx Γ v) exp x)+ Γ v) v )! exp x)x v dx, Γ m+ m!, m,,,! ; Γ m+ ii5i m π,m,,, Γ / m π Useful for classical limit, low, f v ) + v 4 v 4 + v Low temperature limit, exp βµ f v ξ ) v Γ v + Usingξ v ln v 5/, f 5/ ) + v v π 6 ν, some values are, 8 5π / v /, f / ) 4 π / v /, f / ) π /,ξ βµ ln : ξ + v v ) v ) v ) 7π 4 6 ln ) 5/ + 5π 8 ln ln ) / + π 8 ln ln 4 ln 5/ π + + +,,, ξ + 4

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