- #1

viviane363

- 17

- 0

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?