Boltzmann distribution problem

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  • #1
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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
 

Answers and Replies

  • #2
Simon Bridge
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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?
What is your reasoning?
 
  • #3
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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?
What is your reasoning?

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.
 
  • #6
DrClaude
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You have
$$
\frac{f(x)}{1+x} \approx (1-x) f(x)
$$
Then do the multiplication, and keep only terms up to first order.
 
  • #7
Simon Bridge
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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?

Your answer was:
E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE)

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

The answer you want is:
E1+deltaE*e^(-B*deltaE) = E1 + deltaE x

compare.
 
  • #8
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hint: the "order" is determined by the power of x.


So x=e^-BdeltaE right?

Your answer was:
E1+deltaE*e^(-B*deltaE) + E2*e^(-2B*deltaE)

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

The answer you want is:
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
 
  • #9
Simon Bridge
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Yep: that's all there is to it.

The idea is that if x is small enough to make the binomial approximation (which is what you did), then x^2 is way wayy too small to have a noticeable effect. It is safe to just ignore it. i.e. imagine x~0.00001... at that sort of scale the equations x^2+x+1 and x+1 are hard to tell apart.
 

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