hermal Physics Homework #10 (optional) (19 points) 1 Your work on this homework set will not be accepted after 3 p.m. on December 4 (W) 1. (3 points) A common mistake people make when they try to calculate a partition function and a partition function of a simple system Consider a simple microscopic model for which the energy eigenstates are labeled by a quantum number n and the corresponding energy eigenvalues are given by the following table: ε n n ------------------------------------------------ 1 0 2 α 3 α 4 2α 5 6 3α 7 3α 8 3α he partition function is then given by Z( ) =1 + 2 exp α / k B [ ( )] + exp 2α / ( k B ) [ ]+ 4exp[ 3α / ( k B ) ], but not by Z( ) =1 + exp α / k B [ ( )] + exp 2α / ( k B ) [ ] + exp 3α / k B [ ( )]. Don t forget that the partition function is the sum of the Boltzmann factors for all the eigenstates (i.e., the sum over the quantum number n), not the sum of the Boltzmann factors for all the distinct energy eigenvalues. Even when two or more eigenstates have the same energy eigenvalue (i.e., these eigenstates are degenerate), the Boltzmann factor for each eigenstate must contribute to the partition function.
Consider another microscopic model with the following nine energy eigenstates: 2 s ε s 1 0 2 δ 3 2δ 4 δ 5 δ 6 2δ 7 δ 8 2δ 9 2δ where ε s is the energy eigenvalue for eigenstate s. (a) (1 point) Find the partition function for this model as a function of temperature. Hint: 1 + 4a + 4a 2 = 1+ 2a ( ) 2. (b) (1 point) Find the Helmholtz free energy F of this model as a function of temperature. (c) (1 point) Find the internal energy U of this model as a function of temperature. As hints, the Mathematica plots for ˆ F F δ and ˆ U U δ as functions of ˆ k B δ are shown below:
2. (7 points) he Helmholtz free energy for the blackbody radiation at temperature 3 When you are asked to sketch a function f(x) by hand below, make sure: (i) Label each axis with the symbol for the corresponding physical quantity. (ii) he label for each axis must also include the SI unit for the physical quantity. (iii) Clearly indicate the location of the origin for each graph. (iv) Clearly show whether the slope at x = 0 of the curve for the function is zero, positive, or negative. Electromagnetic radiation inside a container of volume V is in thermal equilibrium with the container at temperature. he container acts as a heat reservoir for the radiation. he Helmholtz free energy for the radiation is given by bv( k B ) 4 / 3, where b is a positive constant. (a) (1 point) Sketch the Helmholtz free energy as a function of for constant V. (b) (2 points) Find the internal energy of the radiation as a function of and V. Sketch the internal energy as a function of for constant V. (c) (2 points) Find the pressure of the radiation as a function of and V. Sketch the pressure as a function of for constant V. (d) (2 points) Find the entropy of the radiation as a function of and V. Sketch the entropy as a function of for constant V. 3. (4 points) he Einstein model of a crystalline solid By the end of nineteenth century, the experimental data for an elemental solid such as diamond had shown that the molar heat capacity at constant volume c v remained close to 3R at high temperatures while it approached zero as the temperature was decreased toward the absolute zero. his low-temperature behavior was especially puzzling because statistical mechanics combined with classical mechanics could not explain such behavior. In 1907, just two years after his so-called annus mirabilis (i.e., 1905, when he submitted his Ph. D. dissertation proposing a new way of determining the molecular size and published four papers: one on the Brownian motion, two on the special relativity, and one on the photoelectric effect. Well, wow!), Einstein proposed a model for a crystalline solid insulator (i.e., no free electrons) to explain the puzzling low-temperature behavior of c v.
he Einstein model assumes: (1) he atoms are arranged on a crystalline lattice. Einstein didn t really have an evidence for atomic lattice structure inside a solid (in 1912, Laue published the theory and experimental results for x-ray diffraction by crystalline lattices, and in the following year, W.H. Bragg and W. L. Bragg (a father and a son) analyzed crystal structures of NaCl, KCl, and KBr, etc.), and I m not even sure whether he assumed a crystal structure for the atoms in his article. However, all we need to assume in this model is that the atoms stay around fixed locations inside the solid and that they are vibrating around these locations. (2) he atoms are distinguishable in this model because each atom stays around a particular lattice location and we can label each atom by its lattice location. herefore, there is no need for a 1/N! factor in the partition function. (3) In this model, we therefore have N thermally vibrating atoms or N non-interacting 3- dimensional harmonic oscillators. o make the model even simpler, we assume that these harmonic oscillators have a common frequency ω. Assigning one common frequency for all the atoms is obviously a drastic over-simplification, and to improve on this model, we must take into account a distribution of frequencies for the vibrations of the atoms. hat was partially accomplished by the Debye model, which we will discuss later. 4 he Einstein model = 3N non-interacting harmonic oscillators with frequency ω he energy eigenstates for the Einstein model Each energy eigenstate for the Einstein model is labeled by 3N integers: n (1) x,n (1) y,n (1) z ;n (2) x,n (2) y,n (2) (N z ;...;n ) ( x n N) ( N) { y n z }, (i) where n α is the quantum number for the oscillation of the i-th atom along the α-direction (α = x, y, or z) and n (i) α = 0, 1, 2, 3,...,. he energy eigenvalues for the Einstein model he energy eigenvalue is then given by { } = ε (i ) (i ) n α = hωn α E nα (i) N i =1 α =x, y, z N i =1 α = x, y, z = hωn x (1) + hωn y (1) + hωn z (1) +... + hωn x (N ) + hωn y (N ) + hωn z (N ) In a quantum mechanics course, you must have learned that the energy eigenvalue of a onedimensional harmonic oscillator is given by
# ε n = hω n + 1 & (. $ 2' 5 In the above energy eigenvalue, we set the ground state energy for each one-dimensional harmonic oscillator with n (i) α = 0 to be zero instead of ( 1 / 2)hω since we can set the zero energy wherever we want. In 1901, Planck first introduced discrete energy values for a harmonic oscillator in his study of blackbody radiation, and only six years later Einstein adopted this idea in his model for atomic vibrations in a solid insulator. he partition function for the Einstein model Since the total energy eigenvalue is simply a sum of the 3N independent energy eigenvalues as in the monatomic ideal gas model, applying the decomposition formula in Sec.12.9 of the text, we have Z, N harmonic oscillator: ( ) = Z 1 ( ) 3N, where Z 1 ( ) is the partition function for a single one-dimensional Z(, N) = & exp ε n / k B ' n =0 ( ) * [ ( )] = & exp E / ' n= 0 [ ( )] n 1 = [ 1 exp( E / ) ] 3N ( ) * 3 N 3N = & exp nhω / k B ' n= 0 [ ( )] ( ) * 3N where we have introduced the Einstein temperature defined by E hω k B that roughly divides the low-temperature regime from the high-temperature regime for this model, and we have also used the following formula: a n n=0 = 1 1 a ( if a <1). (a) (1 point) Using h ~ O( 10 34 J sec), ω ~ O( 10 13 10 14 rad / sec), k B ~ O 10 23 J / K ( ), ( ). estimate the order of magnitude for the Einstein temperature to be E ~ O 100 1000 K Of course, Einstein could not estimate this temperature from an experimental value for ω because such data did not exist yet. Instead, he regarded this temperature as a parameter to be determined by fitting the experimental data for c v ( ) with the theoretical curve for c v ( ) derived from his model. He used the data for diamond and obtained E 1320 K.
(b) (1 point) Find the Helmholtz free energy F(, n) for the Einstein model as a function of temperature and the mole number n of the atoms (use Nk B = nr to replace N with n). As a hint, a Mathematica plot for F / ( 3nR E ) as a function of ˆ / E is shown below. he Helmholtz free energy is a decreasing function of temperature. 6 F / ( 3nR E ) vs. / E U / ( 3nR E ) vs. / E (c) (1 point) Find the internal energy U(,n) for the Einstein model as a function of temperature and the mole number n of the atoms. As a hint, a Mathematica plot for U / ( 3nR E ) as a function of ˆ / E is shown above he internal energy is an increasing function of temperature and at high temperatures ( >> E ), U 3nR. (d) (1 point)find the molar heat capacity at constant volume c v ( ) for the Einstein model as a function of temperature. As a hint, a Mathematica plot for c v R as a function of ˆ / E is shown below. he molar heat capacity at constant volume is an increasing function of temperature: it converges to zero as approaches zero while at high temperatures ( >> E ), c v 3R. c V / R vs. / E As mentioned above, the small values for c v at low temperatures was a mystery until Einstein came up with this model. But the story didn t end there, because more accurate data for the lowtemperature molar heat capacity showed c v 3, clearly in disagreement with the Einstein model. he Debye model (1912) eventually explained this low-temperature behavior.
c v approaches zero as the temperature approaches absolute zero, because when << E, ( ) = E / >>1 and therefore the Boltzmann factor for the first excited state for the ( ) hω / k B [ ] <<1 harmonic oscillation along a given direction for an atom is very small or exp hω / k B so that most atoms in the model would be in the ground state with zero energy and varying the temperature would not change the internal energy much (i.e., a small value for c v ). Clearly, it is the discreteness of the energy eigenvalue of a quantum harmonic oscillator or the energy gap hω between the ground state and the first excite state that is responsible for the low-temperature behavior of c v. It is remarkable that the macroscopic quantity c v had been a clue for the quantum nature at the microscopic level. In fact, the Einstein model promoted the idea of energy quantum originally proposed for the blackbody radiation by Planck. 7 Like the Einstein temperature in this model, for a given model we usually find a characteristic temperature 0 that separates the high-temperature regime from the lowtemperature regime. he characteristic temperature is also tied with a characteristic energy scale k B 0 and varies from system to system, which implies that a certain temperature deemed as a high temperature for one system may well be a low temperature for another. For example, 300 K is a low temperature for diamond with E 1320K, while the same 300 K is a high temperature for silver with E 160K. We should always ask what is the characteristic temperature or energy scale for a model or a particular phenomenon observed in a system at hand. 4. (5 points) he two-state model for the nuclei in solid gallium Consider a model for N nuclei in a solid piece of metallic gallium where each nucleus has its ground state at energy 0 and its first excited state at energy Δ. he first excited state is very close to the ground state and corresponds to a non-spherical charge distribution inside the nucleus with a non-zero nuclear quadrupole moment. Other excited states have energies much higher than Δ and will be neglected. We assume that these nuclei do not interact with each other (i.e., this is a system of N non-interacting nuclei.). Each energy eigenstate for the entire system of the N nuclei is then labeled by N integers: { n 1,n 2,n 3,..., n N }, where each n i is the quantum number for the i-th nucleus. n i = 0 corresponds to the ground state with energy ε i = 0, while n i = 1 corresponds to the first excited state with energy ε i = Δ. More succinctly, we can express ε i in terms of n i as ε i = Δn i. he energy eigenvalue for the N non-interacting nuclei is then given by
where E N ({ n i }) = ε i ε i = Δn i., i =1 8 Since each nucleus is fixed around a particular location inside the solid, the nuclei in this model are distinguishable (or we can label each nucleus) so that we do not need to worry about the 1/N! factor. (a) (2 points) Find the partition function for the nuclei as a function of the temperature and the number of nuclei N. (b) (1 point) Find the Helmholtz free energy F for the nuclei as a function of the temperature and the mole number n for the nuclei. Use Nk B = nr to replace N with n. Get rid of Δ in your final expression for the Helmholtz free energy by introducing the following characteristic temperature: Δ Δ / k B. As a hint, a Mathematica plot for F / ( nr Δ ) as a function of ˆ / Δ is shown below. he Helmholtz free energy is a decreasing function of temperature. F / ( nr Δ ) vs. / Δ U / ( nr Δ ) vs. / Δ (c) (1 point) Using your final expression for the Helmholtz free energy in terms of, n, R, and Δ, find the internal energy U of the nuclei as a function of and n. As a hint, a Mathematica plot for U / nr Δ ( ) as a function of ˆ internal energy is an increasing function of temperature. / Δ is shown above. he (d) (1 point) Using your final expression for the internal energy in terms of, n, R, and Δ, find the molar heat capacity at constant volume c v for the nuclei as a function of and n. As a hint, a Mathematica plot for c v / R as a function of ˆ / Δ is shown below.
9 c v / R vs. / Δ For >> Δ (i.e., in the high temperature regime), we get c v R # 4 $ Δ 2 & (. ' For <<120, 000 K, the molar heat capacity due to the conduction electrons in gallium is approximately γ, while for <<θ = 320 K the molar heat capacity due to the lattice waves in gallium is approximately D 3. If Δ << <<1 K, then the total molar heat capacity is given by c v R # 4 $ Δ 2 & ( + γ + D 3 R ' 4 # $ Δ 2 & ( + γ or c ' v 2 R 4 2 Δ + γ 3. where we have assumed ( R 4) ( Δ ) 2 + γ >> D 3. By using the experimental data for the molar heat capacity c v for gallium and plotting c v 2 γ = 0.596 mj / ( mol K 2 ). against 3, we find Δ = 4.55 10 4 K and