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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

Exercise 67 calculates the minimum total energy in a system of spin fermions and is applicable to conduction electrons in a metal. The average particle energy is the total energy divided by the number of particlesN .Show that the average particle energy E→of a conduction electron at low temperature(T=0) is(3/5)EF . This form is convenient, being rather simple, and it can easily he put in terms of N,Vandm via equation .

According to Wien's law, the wavelengthλmaxat which the thermal emission of electromagnetic energy from a body of temperatureTis maximum obeysλmaxT=2.898×10−3m⋅K.Show that this law follows from equation (9-47). To do this. Usef=c/λto expressin terms ofλrather than f, then obtain an expression that, when solved, would yield the wavelength at which this function is maximum. The transcendental equation cannot be solved exactly, so it is enough to show thatλ=(2.898×10−3m⋅K)/T solves it to a reasonable degree of precision.

Defend or refuel the following claim: An energy distribution, such as the Boltzmann distribution. specifies the microstate of a thermodynamic system.

Consider a gas of atoms that might serve as a laser medium but that is inequilibrium, With no population inversions. A photon gas coexists with the atoms. Woulda photon whose energy is precisely the differencebetween two atomic energy states be more likely to be absorbed or to induce a stimulated emission or neither? We expect that inequilibrium the number. of atomsat different levels and the number of photons of a given energy should be stable. Is your answer compactible?

Suppose that in Figure 9.27, the level labelled E1, rather than the one labelledE2, were metastable. Might the material still function as a laser? Explain.
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