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All Solutions of Type: Statistical Mechanics
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GR8677 #15

Statistical Mechanics}Probability

Probability is mostly common sense and adhering to definitions.

The probability that a gas molecule or atom is in the small cube is P(in)=1E-6. The probability that it's not in that small cube is P(not)=1-1E-6. Assuming independent gas molecules or atoms, i.e., the usual assumption of randomness in Stat Mech, one gets, P(N gas atoms)=\left(1-1E-6\right)^N.

The answer is thus C.

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GR8677 #46

Statistical Mechanics}Blackbody Radiation

Recall the T^4 law

where \sigma is the energy density.

Apply that law. 10=T_1^4 and x=2^4 T_1^4, where mW is the units. Compare the two to get x=2^4 \times 10 mW =160 mW. This is choice (E).

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GR8677 #51

Statistical Mechanics}Specific Heat

Both Debye and Einstein assumed that there are 3N oscillators. (In fact, one can argue that the core of condensed matter begins with the assumption that a continuum piece of matter is basically a tiny mattress---a bunch of springs laden together.) Answer is thus (B).

However, Einstein was too lazy, and he decided that all 3N oscillators have the same frequency. Debye assigned a spectrum of frequencies (phonons).


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GR8677 #55

Statistical Mechanics}Fermi Temperature

When one deals with metals, one thinks of Fermi branding---i.e., stuff like Fermi energy, Fermi velocity, Fermi temperature, etc. So, \frac{1}{2}v_F^2\approx kT_F. Fermi stuff is based on the Fermi-Dirac distribution, which assumes that the particles are fermions. Fermions obey the Pauli-exclusion principle. (c.f. Bose-Einstein distribution, where the particles are bosons, who are less discriminating and inclusive than fermions. Both Bose-Einstein and Fermi-Dirac assume indistinguishable particles, but the Boltzmann distribution, which assumes the particles are distinguishable)

All the other choices are too general, since bosons can also satisfy them. (Moreover, the Born approximation is pretty much the fundamental assumption of all of QM---every single calculation you do involving the interpretation of mod square of wave functions as probability depend on the Born approx!)

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GR8677 #67

Statistical Mechanics}Partition Function

The problem gives three non-degenerate energies, so one can just directly plug this into the canonical(?) partition function to get,


where k is the Boltzmann constant, T is the (absolute) temperature.

Since,


For \epsilon >> kT, one can expand e^x \approx 1+x, and thus,

where one throws out the higher order \epsilon terms.

( in denominator) The average energy of each particle is U/3=\frac{4}{3}\epsilon, as in choice (C).

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GR8677 #87

Statistical Mechanics}Diatomic Molecules

Vibrational energy of a diatomic molecule goes to 0 at low temperatures. Thus, the springy dumbbell would be approximately a rigid dumbbell at a low-enough temperature. The other choices are too specific, and thus (E), the most general, must be it.

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GR8677 #88

Statistical Mechanics}Pressure


Recall the following results for low temperatures,

where one realizes that the Fermi temperature, T_F tends to be on the order of thousands for most material, and that the low temperature regime temperatures are definitely far less than the Fermi temperature; one thus has P_F>P_B,P_C. Moreover, classical effects occur at around T=300K, while the problem specifies the temperature domain for bosons to be far lower than that, thus one deduces that P_B < P_C. Therefore, P_F>P_C>P_B.

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GR8677 #79

Statistical Mechanics}Specific Heat

The specific heat at constant volume for high temperatures is c_v = 7/2 R. The specific heat at low temperatures is 3/2 R. Why?

There are three contributions to the specific heat of a diatomic gas. There is the translational, vibrational, and rotational. At low temperatures, only the translational heat capacity contributes U = 3/2 N k T \approx c_v T \Rightarrow c_V=3/2 Nk . At high temperatures, all three components contribute, and one has c_V=(3/2+1+1)Nk = 7/2 Nk.

The general formula is c_v = c_v(translational)+c_v(rotational)+c_v(vibrational)=Nk\left(3/2 + 1 + (h\nu/(k\theta))^2 \exp(h\nu/(k\theta))/(\exp(h\nu/(k\theta))-1)^2 \right)

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GR8677 #94

Statistical Mechanics}Internal Energy

The partition function is Z=\sum e^{-\epsilon_i/kT} = 1+e^{-\epsilon/kT}. Internal energy is given by U=\frac{NkT^2}{Z} \frac{\partial Z}{\partial T} \propto \frac{\epsilon e^{-\epsilon/kT}}{e^{-\epsilon/kT}+1}\propto \frac{\epsilon}{e^{\epsilon/kT}+1}, as in choice (D).

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GR8677 #35

Statistical Mechanics}Blackbody Radiation Formula

The blackbody radiation formula has u=\sigma T^4. The ratio of energy radiated is thus T_2^4/T_1^4. Since T_2=3T_1, the ratio is just 3^4=81, as in choice (E). (Obviously an increase since the final temperature is much higher than the initial.)

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GR8677 #49

Statistical Mechanics}Partition Functions

The partition function is given by the formula Z=\sum_i g_i e^{-\epsilon_i/kT}, where g_i enotes the degeneracy of the ith state.

Plug in the given information into the formula to get choice (E).

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GR8677 #63

Statistical Mechanics}Wien Law

Recall Wien's Law, \lambda T = 2.9 E-3. It relates the temperature with the wavelength at maximal intensity. The wavelength at the maximal intensity is approximately 2\mu m. Plugging this in, one finds that T = 2.9E-3/2E-6 \approx 1500K, as in choice (D).

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GR8677 #65

Statistical Mechanics}Approximations

At high temperatures, one has kT >> h\nu .

One expands the argument of the denominator according to e^x \approx 1+ x (for small x). The denominator becomes (1+h\nu/kT -1)=h\nu/kT. Since the whole quantity of the denominator is squared, the top (h\nu/kT)^2 term is canceled.

In the numerator, one has e^{h\nu/kT}\rightarrow 1, since e^{1/\infty} \rightarrow e^0 \rightarrow 1.

Thus, one arrives at choice (D).

One either remembers this fact about solid or one derives it, as shown above. (Also, for low temperatures, at say around 20K for most solids, the Debye T^3 law applies, and specific heat is proportional to T^3. One can't find this result from the Einstein formula, which is why Debye's theory is more accurate for solids.)

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GR8677 #77

Statistical Mechanics}Maxwell-Boltzmann Distributions

The Maxwell-Boltzmann distribution is N\propto g e^{-\epsilon/(kT)}, where g is the degeneracy.

Given \epsilon_a = 0.1+\epsilon_b, one finds the ratio of distributions (thus ratio of numbers) to be e^{-(0.1+\epsilon_b)/kT}/e^{-\epsilon_b/kT}= e^{-0.1/kT}.

The problem gives kT = 0.025 eV, and thus the above ratio becomes e^{-.1/.025}=e^{-4}, as in choice (E).


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GR8677 #98

Statistical Mechanics}Average Energy

The average energy is given by \sum_i E_i e^{-E_i/kT}/Z, where Z=\sum_i e^{-E_i/kT} is the partition function.

The above is just the stat mech generalization of the baby-averaging \bar{u}=\sum_i U_i/N, where N is the number of energies summed over. Z is like a generalized N.

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GR8677 #14

Statistical Mechanics}Blackbody Radiation Formula

Recall

where P is the power and u the energy and T the temperature.

So, initially, the blackbody radiation emits P_1=kT^4. When its temperature is doubled, it emits P_2=k(2T)^4=16kT^4.

Recall that water heats according to Q=mc\Delta T= \kappa \Delta T. So, initially, the heat gain in the water is Q_1=\kappa (0.5^\circ). Finally, Q_2=\kappa x, where x is the unknown change in temperature.

Conservation of energy in each step requires that kT^4t=\kappa/2 and 16kT^4t=\kappa x, i.e., that P_i t = Q_i. Divide the two to get \frac{1}{16}=\frac{2}{x}\Rightarrow x=\Delta T = 8^\circ. Assuming the experiment is repeated from the same initial temperature, this would bring the initial 20^\circ to 28^\circ, as in choice (C).

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GR8677 #15

Statistical Mechanics}Heat Capacity

Note that this problem wants the regime of high temperatures, and thus the answer is not \frac{5}{2}R from classical thermodynamics, but rather \frac{7}{2} R.

The problem suggests that a quantized linear oscillator is used. From the energy relation \epsilon = \left(j+\frac{1}{2}\right)\hbar \nu, one can write a partition function and do the usual Stat Mech jig. Since one is probably too lazy to calculate entropy, one can find the specific heat (at constant volume) from c_v=\left.\frac{\partial U}{\partial T}\right|_v, where U=NkT^2\left(\frac{\partial Z}{\partial T}\right)_V, where N is the number of particles, k is the Boltzmann constant.

There are actually three contributions to the specific heat at constant volume. c_v=c_{translational}+c_{rotational}+c_{vibrational}. Chunk out the math and take the limit of high temperature to find that c_v=\frac{7}{2}R.

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GR8677 #23

Statistical Mechanics}Fermi Temperature

(Much of the stuff I classified as Stat Mech might also be considered Condensed Matter or Solid State Physics. They are classified as thus because the Stat Mech book I mentioned in the booklist on the site http://grephysics.yosunism.com is perhaps the best intro to all this.)

The Fermi velocity is related by \epsilon_F=k T_F =\frac{1}{2}m v^2\Rightarrow v=\sqrt{\frac{2kT_F}{m}}, where \epsilon_F is the fermi energy, and T_F is the Fermi temperature.

One should know by heart the following quantities, k=1.381E-23 and m=9.11E-31 (but then again, they are also given in the table of constants included with the exam). Plug these numbers into the expression above to find v,

v=\sqrt{\frac{2kT_F}{m}}\approx \sqrt{\frac{2\times 1.4 E-23 \times 8E5}{9E-31}}=\sqrt{\frac{2.8 E-23 \times 8E5}{9E-31}}\approx\sqrt{0.3 E8 \times 8E5}=\sqrt{24/10 E13}\approx \sqrt{10^{12}}=10^6,
the choice that comes closest to this order is choice (E).

Note that the hardest part of this problem is the approximation bit. No calculators allowed. Sadness.

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GR8677 #63

Statistical Mechanics}Maximal Probability

According to statistical mechanics, maximal probability is the sate of highest entropy---it's the peak of a Gaussian curve, the average score on a normally-curved test.

Spontaneous change to lower probability thus does not occur since maximal probability is the most stable state--one of highest entropy. Boltzmann's constant never approaches 0, however in the third law of thermodynamics, one has the entropy approaching 0 for T\rightarrow 0.

Eliminating choices, one has choice (D).

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GR8677 #71

Statistical Mechanics}Distributions

The Fermi-Dirac distribution, in general, gives the number of states in E_i to be N_{FD}=N_0\frac{1}{1+e^{-E_i/kT}}, where N_0 is the total number of states. (The Fermi-Dirac distribution is used since there are only two states.)

Define E_1=\epsilon and E_2=2\epsilon.

The number of states in 1 is just N_{1}=N_0\frac{1}{1+e^{-E_1/kT}}= N_0\frac{1}{1+e^{-\epsilon/kT}}, which is choice (B).

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GR8677 #72

Statistical Mechanics}Heat Capacity

The heat capacity is just dU/dT, where ETS generously supplies U, the internal energy. Since E_1 and N_0 are constants, the first term is trivial.

The temperature-derivative of the second term is N_0\epsilon^2/(kT^2)e^{\epsilon/kT} /(1+e^{\epsilon/kT})^2=, as in choice (A).

(The temperature derivative is easily done if one applies the chain-rule \frac{df}{dy}\frac{dy}{du}\frac{du}{dT} where f=1/y, y=1+e^u, u=\epsilon/(kT).)

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GR8677 #73

Statistical Mechanics}Entropy

The third law of thermodynamics says that S(T\rightarrow 0) \rightarrow 0.

Also, the statistical definition of entropy is just S = Nk ln Z, where Z is the partition function. For this problem, one has Z=e^{-\epsilon/kT}+e^{-2\epsilon/kT}. For high temperatures, one has Z \rightarrow 1 + 1 = 2, since e^{x}\approx 1+x for small x (and then 1+x \approx 1 for very small x).

Thus the entropy behaves as in choice (C).

 
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