What is the occupancy probability at an energy of 12.00 eV for a material with a Fermi energy (level) of 11.63 eV at a temperature of 500 K? (^mel = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s, Boltzmann constant = 1.38 × 10^-23, 1 eV = 1.60 × 10^-19 J)

Question

Quizplus · Accepted Answer

The occupancy probability at a certain energy can be found using the Fermi-Dirac distribution function:

f(E) = 1 / {exp[(E-E_F)/(kT)] + 1}

where E is the energy, E_F is the Fermi energy, k is the Boltzmann constant, and T is the temperature.

Plugging in the given values, we get:

f(12.00 eV) = 1 / {exp[(12.00-11.63)/(1.38×10^-23×500)] + 1}

= 0.019

So the occupancy probability at an energy of 12.00 eV is 0.019, or 0.019%, which corresponds to answer choice B.

What Is the Occupancy Probability at an Energy of 12