GREPhysics.NET: GR0177 #77

GR0177 #77

Problem

GREPhysics.NET Official Solution

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Statistical Mechanics $\Rightarrow$ }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).

Alternate Solutions

Ning Bao
2008-02-01 07:39:44

Quick elimination: higher states are less likely->D or E.

Ratio of given energy to kT mist be important: as Energy of A increases, likelihood in state A decreases ->E.

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Comments

gbenga 2014-10-18 22:35:18	semi-fast soln: rn $\frac{N_{A}}{N_{B}}=\frac{e^{\frac{-E_A}{kT}}}{e^{\frac{-E_B}{kT}}}=e^{\frac{-\triangle{E}}{kT}}=e^{\frac{-0.1}{0.025}}$ rnThis power is negative so A,B, & C are eliminated. The denominator is small so D is unlikely. Remaining is E Reply to this comment
gbenga 2014-10-18 22:32:03	semi-fast soln: $\frac{N_{A}}{N_{B}}=\frac{e^{\frac{-E_A}{kT}}}{e^{\frac{-E_B}{kT}}}=e^{\frac{-\triangle{E}}{kT}}=e^{\frac{-0.1}{0.025}}$ This power is negative so A,B, & C are eliminated. The denominator is small so D is unlikely. Remaining is E Reply to this comment
QuantumCat 2014-09-01 10:39:10	A quick way to solve this problem (knowing that the occupation number depends on the energy) is to say that state B is at zero energy so that the exponential for state B just becomes 1, which is infinitely easier to deal with. Reply to this comment
Ning Bao 2008-02-01 07:39:44	Quick elimination: higher states are less likely->D or E. Ratio of given energy to kT mist be important: as Energy of A increases, likelihood in state A decreases ->E. Reply to this comment

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Quick elimination: higher states are less likely->D or E. Ratio of given energy to kT mist be important: as Energy of A increases, likelihood in state A decreases ->E.

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