Reading Quiz

Question 1:

Answer:

1. Not really. It's not the partition function that is really important, it's the probabilites you get by using it. And our probabilites will be about the same for the exercise.
2. It does change the numbers a bit, but it's still a 4 multiplying something times 10^-178, so it doesn't change the calculations of the partition function in any meaningful way.
3. It changes our answer but by a very small amount. Instead of 1.6 x10-5 we now get 0.4 x 10-5. This does make a difference when we are dealing with things as large as stars because of the great number of atoms in a star.
4. It does not change the partition function because only the ground state with degeneracy = 1 contributes significantly to Z
5. ?
6. No, because the first factor had a degeneracy of 1, and the following factors were negligable.
7. The way the question is written, it seems like the degeneracy is taken care of when the energy values are substituted into the expression for the partition function.
8. No because the first term (n=1) dominates the partition function anyway.
9. It will change it but not by much. The degenerecies increase as the exponential terms decrease.
10. No because exp[-1/n^2] --> 0 faster than n^2. Multiplying 6.42x10^-178 is still a really small (read: negligable) number.

Question 2:

Answer:

1. This is the degeneracy discussed above. The number of degenerate states for level j is 2j+1 so we need to multiply the normal partition function by this probability.
2. This factor of (2j + 1) came from the number of degenerate states available in each energy level.
3. This factor is the number of degenerate states for level j.
4. 2j+1 is the degeneracy at level, j.
5. The factor of (2j+1) is the number of states with energy E(j).
6. That's the number of degenerate states.
7. The (2j+1) factor accounts for the degeneracy, since energy level j has a degeneracy of (2j+1).
8. This is the number of degenerate energy levels
9. That's the degenerecy!
10. (2j+1) is the number of degernate states per level j.

Question 3:

Answer:

1. The equipartition theorem! Here f =2 so Average rotational energy is kT.
2. We may have been able to obtain the same result as eq. (6.32) after our first week of class from section 1.3 by multiplying the formula of the Equipartition Theorem, (1/2)kT, by two, in order to account for the two rotational states of the diatomic molecule.
3. We had the original equipartition theorem at this point. Eq. 6.32 tells us that we are only worried about rotational energy which means only rotational degrees of freedom for which there are 2. so 2/2 *kT= kT, the answer to eq 6.32.
4. For each degree of freedom, the ave. rotational energy is 1/2kT
5. It's just the partition theorem result with 2 degrees of freedom (the two rotational degrees of freedom).
6. Well, if you consider one molecule and the equipartition theorem, and think of a diatomic molecule having two rotational degrees of freedom (f), then Urotational=2/2* kT which is the same as(6.32)
7. It's the equipartition thereom, U = N(f/2)kT, for a diatomic gas where f = 2. so U = NkT, or if you divide both sides by N, the average energy = kT.
8. Well, in section 1.3 we said U=N*f/2, since rotational energies are f=2, and we are dealing with single molecules, so N=1. Since U is overall thermal energy, it is sort of like E(avg)
9. Because we said the average kinetic energy was directly related to the average temperature.
10. We could have used the old equipartition fn.

Question 4:

Answer:

1. It didn't derive the entire equipartion theorem did it? I don't see any generic result for number of particles (N) or degreees of freedom (f).
2. While the delta_q is turned into dq inside the integral, there is still a 1/delta_q outside it. How do we differentiate this delta_q from dq? How would be calculate delta_q if it is very small?
3. The step from 6.37 to 6.38 and 6.39 to 6.40.
4. I can follow this, but there's no way I could have derived it.
5. This derivation assumes that the coordinate q can have any arbitrary value. Should it be different for, say, a system that can only have 0 < x < 1?
6. "the clever trick" described in appendix B.
7. I think I understand the math, but I'm a little unsure of how you set up the problem, i.e., how you knew how to define the energy and why you make the assumptions you make, etc.
8. Everything seems to follow fairly logically, so not much, maybe more about the assumptions that go into the equipartition theorem.
9. THe later steps. Like starting at 6.39
10. Looks good to me.

Question 5:

Answer:

1. We get large gaps between energy levels, so the probabilities can't be turned into a smooth curve and integrated as a Gausian. That makes the 2nd half of the derivation bunk.
2. The equipartition theorem fails for quantum mechanical systems at low temperature because most of the energy states at low temperatures will be too small to appear on the graph. The gaussian will not approximate the curve of the bar graph well at all.
3. This equipartition theorem uses a Gaussian curve to approximate a bar graph. At low temperature quantum mechanical systems, there are two few distinct states that have significant probabilities.
4. At low temperatures, few states have probabilities of being occupied and the Gaussian curve is not a good fit for the bar graph.
5. In calculating the partition function, we approximate the sum over all states as an integral. Quantum mechanical systems at low temperatures have few enough possible states that the spacing becomes relevant and hence the integral is a bad approximation.
6. Because in quantum mechanical systems, the spacings between the states becomes more important, and the gaussian curve does not become a good approximation to the bar graph of q vs. Boltzmann constant.
7. If the temperature is too small, then the graph of the Boltzmann factor in the expression for the partition function gets too thin and the curve approximation is no longer accurate, so we can't just take the area under that curve (integrate).
8. The gaussian fails to be a good approximation to the bar graph.
9. At lower energy levels the spacing between adjacent energies is much less than kT, and the equipartition theorem works when many distinct states contribute and in the high temperature, the spacing is unimportant.
10. Because a Gaussian in not a good approximation for the Boltzmann factor.

Question 6:

Answer:

2. I have nothing at this time.
3. Aside from that cloudy derivation in question 4, i thought it was straight-forward.
7. Why do we take the standard deviation to be the root of the mean of the squares of the individual deviations, instead of just taking the mean of the absolute value of the deviations?
8. None
9. This new equipartition theorm stuff.