Fermi-Dirac Distribution

  • Thread starter viviane363
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  • #1
viviane363
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Pleas can you help me figure out what I do wrong?
At what temperature is the probability that an energy state at 7.00 eV will be populated equal to 25 percent for copper (EF = 6.95 eV)?
The formula for the fermi-Dirac Distribution is f(E) = 1/(1+e^((E-EF)/kT)) and looking at the problem I figured that f(E) = 25%=0.25 and E-EF=7.00 - 6.95 = 0.05eV
solving for T and found that T=3.2979e21 K, but it doesn't seem to be the right answer, why?
 

Answers and Replies

  • #2
f95toli
Science Advisor
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I think this thread should be move to the homework forum.
Anyway, f(E) is DISTRIBUTION function; not a probability so setting f(E)=0.25 doesn't make sense.

Consider this: How would you calculate the probability that the system is in ANY state?
You already know that this probability is one, but in problems like this it neverthless help to write down the expression.
 

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