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Boltzmann distribution problem

  1. Mar 19, 2014 #1
    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
     
  2. jcsd
  3. Mar 20, 2014 #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?
     
  4. Mar 20, 2014 #3
    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.
     
  5. Mar 20, 2014 #4

    DrClaude

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    This is the part you forgot.
     
  6. Mar 20, 2014 #5
    what should I do? could you please help?
     
  7. Mar 20, 2014 #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.
     
  8. Mar 20, 2014 #7

    Simon Bridge

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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.
     
  9. Mar 20, 2014 #8
    So, I just take E2 x^2 from the equation? or there is more to it? Thank you for your help
     
  10. Mar 20, 2014 #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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