# Boltzmann distribution problem

A system has two non-degenerate energy levels E1 and E2, where E2>E1>0. The system is at tempreture T. The Average energy of the system is = E1+E2e^(-B*deltaE) / 1+e^(-B*deltaE) where deltaE= E2 -E1 and B=1/kT (k=Boltzmann constant). show that for very low temperatures kT<<deltaE, average energy= E1+deltaE*e^(-B*deltaE).

hint: use the first order expansion (1+x)^-1=1-x for x<<1
keep terms up to first order only

Here is what I get:

I multiplied 1-e^(-B*deltaE) by E1+E2e^(-B*deltaE) based on (1+x)^-1=1-x and I get E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE), but it is not the answer

Help would be appreciated

Simon Bridge
Homework Helper
You are told: $$\bar E = \frac{E_1+E_2 e^{-(E_2-E_1)/kT}}{1-e^{-(E_2-E_1)/kT}}$$ ... and you need to find this in the limit that ##kT<<E_2-E_1##

You did: $$(1-e^{-(E_2-E_1)/kT})(E_1+E_2 e^{-(E_2-E_1)/kT})$$ because ##(1+x)^{-1}\simeq 1-x## for ##x<<1## ... ??!

What are you treating as "x" in that expression?

You are told: $$\bar E = \frac{E_1+E_2 e^{-(E_2-E_1)/kT}}{1-e^{-(E_2-E_1)/kT}}$$ ... and you need to find this in the limit that ##kT<<E_2-E_1##

You did: $$(1-e^{-(E_2-E_1)/kT})(E_1+E_2 e^{-(E_2-E_1)/kT})$$ because ##(1+x)^{-1}\simeq 1-x## for ##x<<1## ... ??!

What are you treating as "x" in that expression?

I did it based on the equation (1+x)^-1=1-x. Therefore, (1+e^-BdeltaE)^-1 is equal to 1-e^-BdeltaE. The denominator is (1+e^-BdeltaE) and not (1-e^-BdeltaE) . I made a mistake, sorry.

DrClaude
Mentor
keep terms up to first order only
This is the part you forgot.

This is the part you forgot.

DrClaude
Mentor
You have
$$\frac{f(x)}{1+x} \approx (1-x) f(x)$$
Then do the multiplication, and keep only terms up to first order.

Simon Bridge
Homework Helper
hint: the "order" is determined by the power of x.

I did it based on the equation (1+x)^-1=1-x. Therefore, (1+e^-BdeltaE)^-1 is equal to 1-e^-BdeltaE.
So x=e^-BdeltaE right?

E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE)

... in terms of x, that is: E1 + deltaE x + E2 x^2

E1+deltaE*e^(-B*deltaE) = E1 + deltaE x

compare.

hint: the "order" is determined by the power of x.

So x=e^-BdeltaE right?

E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE)

... in terms of x, that is: E1 + deltaE x + E2 x^2

E1+deltaE*e^(-B*deltaE) = E1 + deltaE x

compare.

So, I just take E2 x^2 from the equation? or there is more to it? Thank you for your help

Simon Bridge