How Does Temperature Affect Particle Distribution in Two-State Systems?

[silence98]

Homework Statement

Consider a system of N particles, with two available energy states, 0 and E. What is the ratio of particles occupying the first state, n₀, to particles occupying the second state n₁?

Homework Equations

single particle partition function Z=[tex]\Sigma[/tex]exp(-e_i/kt)
system partition function Z_sys=Z^N

The Attempt at a Solution

Z = exp(0)+exp(-E/kt) = 1+exp(-E/kt)

I then looked at the probability of each state being occupied:

P(0)=1/(1+exp(-E/kt))
P(1)=exp(-E/kt)/(1+exp(-E/kt))

I assumed that the ratio of probabilities P(0)/P(1) was equal to the ratio of particles occupying the first state to the number occupying the second.

P(0)/P(1) = exp(E/kt)

I'm unsure if this is the right method? I don't see how to incorporate the system partition function..

 

[CompuChip]

I assumed that the ratio of probabilities P(0)/P(1) was equal to the ratio of particles occupying the first state to the number occupying the second.

That's actually quite simple to see.
If you have N particles, and for each one the probability of being in state n is P(n) independent of the other particles, then basic statistics tells you that on average there are
E(n) = N P(n)
particles in state n (it's a binomial distribution).
If you calculate their ratio,
E(1) / E(0) = (N P(1)) / (N P(0)) = P(1) / P(0)
(or E(0) / E(1), is fine with me) you will see that N drops out and you only get the ratio of the probabilities.

 

[silence98]

That's actually quite simple to see.
If you have N particles, and for each one the probability of being in state n is P(n) independent of the other particles, then basic statistics tells you that on average there are
E(n) = N P(n)
particles in state n (it's a binomial distribution).
If you calculate their ratio,
E(1) / E(0) = (N P(1)) / (N P(0)) = P(1) / P(0)
(or E(0) / E(1), is fine with me) you will see that N drops out and you only get the ratio of the probabilities.

thanks for the complete and concise response!