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Chapter 9: Q10CQ (page 403)

There are more permutations of particle labels when two particles have energy0 and two have energy1 than when three particles have energy 0and one has energy . 2(The total energiesarethe same.) From this observation alone argue that the Boltzmann distribution should be lower than the Bose-Einstein at the lower energy level.

Short Answer

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The Boltzmann suppresses the possibility of discovering a particle with the lowest energy. And it’s encouraged for Bose- Einstein, demonstrating the Boltzmann probability is smaller.

Step by step solution

01

Boltzmann Statistics:

In the classical limit, Boltzmann statistics describes the energy distribution of an ensemble of identifiable particles. Bose - Einstein statistics, on the other hand, describes the energy distribution associated with a group of identical particles. A Boltzmann distribution can approximate the Bose-Einstein distribution in the case where temperature is effectively high or particle concentration is low.

02

Einstein Distribution:

Because there are more permutations for four particles occupying 0and ,1 where both states are doubly-filled, this suggests that at very low energies, these particles tend to cluster towards the same state.

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Most popular questions from this chapter

Heat capacity (at constant volume) is defined as∂U/∂T. (a) Using a result derived in Example 9.6. obtain an expression for the heat capacity per unit volume, inJ/K⋅mi3, of a photon gas. (b) What is its value at300K?

Show that equation (9- 16) follows from (9-15) and (9- 10).

(a) Show that the number of photons per unit volume in a photon gas of temperature Tis approximately(2×107K−3m−3)T3⋅(Note:∫0∞x2(ex−1)−1dx≅2.40.)

(b)Combine this with a result derived in Example 9.6 to show that the average photon energy in a cavity at temperatureTis given byE¯≅2.7kBT.

We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy ! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of11oscillators sharing a total energy of just5hω0 . In the symbols of Section 9.3. N=11andM=5 .

  1. Using equation(9-9) , calculate the probabilities ofn , being0,1,2, and3 .
  2. How many particlesNn , would be expected in each level? Round each to the nearest integer. (Happily. the number is still 11. and the energy still5hω0 .) What you have is a distribution of the energy that is as close to expectations is possible. given that numbers at each level in a real case are integers.
  3. Entropy is related to the number of microscopic ways the macro state can be obtained. and the number of ways of permuting particle labels withN0 ,N1,N2 , and N3fixed and totaling11 is11!(N0!N1!N2!N3!) . (See Appendix J for the proof.) Calculate the number of ways for your distribution.
  4. Calculate the number of ways if there were6 particles inn=0.5 inn=1 and none higher. Note that this also has the same total energy.
  5. Find at least one other distribution in which the 11 oscillators share the same energy, and calculate the number of ways.

  6. What do your finding suggests?

Given an arbitrary thermodynamic system, which is larger. the number of possible macro-states. or the numberof possible microstates, or is it impossible to say? Explain your answer. (For most systems, both are infinite, but il is still possible to answer the question,)
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