What Temperature Gives a 25% Population Probability at 7.00 eV in Copper?

[viviane363]

Homework Statement

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)?

Homework Equations

The formula for the fermi-Dirac Distribution is f(E) = 1/(1+e^((E-EF)/kT))

The Attempt at a Solution

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?

 

[nickjer]

Hmmm... That seems a bit much. What Boltzmann constant are you using? And double check your work, because I get a different answer than you.

 

[motoroller]

That could well be the right answer (not sure of the exact answer, but you will get a big number). Metals have very high fermi temperatures - you can look at it in the following way. Fermi energy levels cannot be multiply-occupied. If that metal was made of bosons, it would have a temperature of 10^21 K because of where the highest energy electrons are.

 

[nickjer]

That is way too large of a temperature. Hah, that is hotter than 1 second after the big bang. Also, the temperature of the Fermi energy is not even close to that. A Fermi energy of 6.95 eV has a Fermi temperature of 80,654 K.

viviane363 must have made a mistake somewhere in the calculation. Because I used the exact same formula and I got a completely different answer. But I think she forgot about this thread.