Final answer:
The energy at which the probability of non-occupancy is 75% for copper with a Fermi energy of 7eV at room temperature, the Fermi-Dirac distribution function is used along with the values for Boltzmann's constant and typical room temperature.
Step-by-step explanation:
To find the energy for which the probability of non-occupancy at room temperature is 75%, we will use the Fermi-Dirac distribution function. This states that the probability F(E) that an electron state at energy E is occupied at absolute temperature T is given by:
F(E) = 1 / (exp[(E - EF) / (kBT)] + 1)
Where:
EF is the Fermi energy,
kb is the Boltzmann's constant (8.617 x 10^-5 eV/K),
T is the absolute temperature in Kelvin.
For non-occupancy, we are interested in the probability of a state being empty, which is (1 - F(E)). If we want this to be 75%, it means that F(E) should be 25% or 0.25. Therefore, we can write:
0.25 = 1 / (exp[(E - EF) / (kBT)] + 1)
Rearranging and solving for E, we get:
E = EF + kBT * ln[(1/F(E)) - 1]
Here EF = 7 eV (Fermi energy for copper) and room temperature is typically taken as 300 K. Substituting these values in we get:
E = 7 eV + (8.617 x 10^-5 eV/K * 300 K) * ln[(1/0.25) - 1]
After calculating the above expression, we obtain the value of E which is the energy at which the probability of non-occupancy is 75%.