How Do You Derive Particle Distributions Using the Boltzmann Factor?

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In a thermal equilibrium system with N particles and two energy states separated by energy Δε, the number of particles in each state can be derived using the Boltzmann factor. The equations for the particle distributions are N1 = N/(1 + exp(-Δε/kBT)) and N2 = N*exp(-Δε/kBT)/(1 + exp(-Δε/kBT)). The ratio of particles in the two states is given by N1/N2 = exp(-E1/kBT)/exp(-E2/kBT). It is suggested that Δε should be defined as E2 - E1, leading to the relation N2/N1 = exp(-Δε/kBT). The total number of particles N1 + N2 must equal N, allowing for the derivation of individual expressions for N1 and N2.
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Homework Statement



A system in thermal equilibrium at temperature T consists of N particles that have two
energy states separated by an energy Δε.

If the number of particles in the two states is N1 and N2, show that:

N{1} = N(\frac{1}{1+exp(-Δε/k{B}T})) and N{2} = N(\frac{exp(-Δε/k{B}T}{1+exp(-Δε/k{B}T}))

Homework Equations


\frac{N{1}}{N{2}} = \frac{exp(-E{1}/k{B}T}{exp(-E{2}/k{B}T}

Δε=E1 - E2


The Attempt at a Solution



Really struggling to see where to get started with this the lectures and the lecture notes we have are not helping.
 
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ferret123 said:

Homework Equations


\frac{N{1}}{N{2}} = \frac{exp(-E{1}/k{B}T)}{exp(-E{2}/k{B}T)}

Δε=E1 - E2

From the way the problem is worded, I think Δε should be Δε = E2 - E1

See if you can show \frac{N{2}}{N{1}}= {exp(-Δε/k_{B}T)}

Also, what must the sum N1+N2 equal?
 
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Well N1 + N2 must equal N?

So now that I have it in terms of Δε I can rearrange for expressions for N1 and N2 then add them for N?
 
You have two equations for N1 and N2. So you should be able to solve for N1 and N2 separately.
 
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