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Title: GR8677.067
Post by: conantee on January 06, 2010, 12:08:00 PM
67. A large isolated system of N weakly interacting particles is in thermal equilibrium. Each particle has only 3 possible nondegenerate states of energies 0, \epsilon, and 3 \epsilon. When the system is at absolute temperature T >> \epsilon /k, where k is Boltzmann's constant, the average energy of each particle is
(A) 0
(B) \epsilon
(C) \frac{4}{3} \epsilon
(D) 2 \epsilon
(E) 3 \epsilon


Title: Re: GR8677.067
Post by: conantee on October 11, 2010, 12:28:41 PM
If you have taken statistical mechanics long time ago, I hope you still remember that
(1) at T=0, all particles goes to ground state energy.
(2) particles tend to be at lower energy than higher energy.

By this two facts, you can eliminate choices (A), (D) and (E) by reasoning as follows:
Since T is greater than some number greater than zero, average energy is not zero since some particles will not be at zero-energy level. (A) is wrong.
Next, we can set the upper bound for average energy by saying that the actual average energy is always no greater than the average energy assuming that particles prefer lower energy level and higher energy level equally. This just comes from (2).
Well, if particles are equally likely to be in any levels, the average energy is just  (0 + \epsilon + 3 \epsilon)/3 = \frac{4}{3} \epsilon. So, (D) and (E) are wrong.

The choices for guess are now (B) and (C).
Now, you need Boltzmann distribution to pick the right. According to Boltzmann distribution, the number of particles in each level is proportional Boltzmann factor \exp(-E/kT), where E is energy of the level.

If \epsilon >> kT, then Boltzmann factor for every level will go to 1. That is, every energy level is approximately likely to have particles in. The average energy goes to the one we have calculated.

The correct answer is (C). 22 of 100 people answer correctly.