How do calculate the prob of being in a state in therm eq?

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SUMMARY

The discussion focuses on calculating the probabilities of a pore being in three energy states while in thermal equilibrium at temperature T. The energy levels are defined as state 1 with energy 0, state 2 with energy U0, and state 3 with energy 2U0. Using the Boltzmann factor, the probabilities are derived as p1 = p2 = p3 = 1/3, confirming that each state is equally likely due to the equal number of microstates associated with each energy level. The application of the Boltzmann factor reinforces this conclusion by demonstrating that the probabilities are consistent across the defined energy states.

PREREQUISITES
  • Understanding of Boltzmann statistics
  • Familiarity with thermal equilibrium concepts
  • Knowledge of energy levels and microstates
  • Basic proficiency in statistical mechanics
NEXT STEPS
  • Study the derivation of the Boltzmann distribution in detail
  • Explore the implications of microstate counting in statistical mechanics
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This discussion is beneficial for students of statistical mechanics, physicists exploring thermodynamic systems, and anyone interested in the principles of probability in energy states.

eprparadox
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Homework Statement


A pore has three configurations with the energy levels shown. The pore is in thermal equilibrium with the surroundings at temperature T . Find the probabilities ##p_1##, ##p_2##, ##p_3## of being in each level. Each level has only one microstate associated with it.

Also, state 1 corresponds to energy of 0, state 2 to energy of ##U_0## and state 3 to energy of ##2U_0##.

Homework Equations


The Boltzmann factor = ##e^{-(U_s - U_r)/kT}##

The probability of being in a state is equal to $$p_i = \frac{e^{\frac{-E_i}{kT}}}{\sum{\frac{-E_i}{kT}}}$$

where the denominator is the sum over all states.

The Attempt at a Solution


This is for self study and not for a course and so I'm really trying to explain these solutions to myself.

In terms of an attempt, I know that

$$\frac{p_2}{p_1} = \frac{\Omega(2)}{\Omega(1)}$$

But $$\Omega(2) = \Omega(1) = 1$$ so the probability of being in any state is the same. And since the we have 3 energy states, I get that $$p_1 = p_2 = p_3 = \frac{1}{3}$$.

Is that correct?

And if so, how can I use the Boltzmann factor to arrive at the same answer.
 
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The principle of equally likely microstates only applies where those microstates are sufficiently alike. In particular, they must have the same energy. That is not the case here.
Use the relevant equations you quote (Boltzmann) to find the probabilities from the energies.
 

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