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Statistical Mechanics}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).


See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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.
Alternate Solution - Unverified
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 NEC
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
NEC
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. NEC
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.
Alternate Solution - Unverified

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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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