Final answer:
For a state with energy 0.10eV above the Fermi energy and a sample temperature of 800K, the probability of occupation is approximately 20.4%.
Step-by-step explanation:
The probability that a quantum state whose energy is 0.10eV above the Fermi energy will be occupied can be determined using the Fermi-Dirac distribution function. The Fermi distribution function describes the probability of finding a fermion in a particular energy state at a given temperature. The probability is given by:
P(E) = 1 / (1 + exp((E - EF) / (kB * T))),
where P(E) is the probability of occupation, E is the energy of the state, EF is the Fermi energy, kB is Boltzmann's constant, and T is the temperature in Kelvin.
For this problem, we can plug in the values:
E = 0.10eV, EF = 0eV (assuming the Fermi energy is at zero), kB = 8.617333262145x10-5 eV/K, and T = 800K.
Substituting these values into the equation:
P(E) = 1 / (1 + exp((0.10 - 0) / (8.617333262145x10-5 * 800)))
Simplifying the expression:
P(E) = 1 / (1 + exp(1.36))
P(E) ≈ 0.204,
Therefore, the probability that the quantum state will be occupied is approximately 0.204 or 20.4%.