Free-electron theory of metals: 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? (m_el = 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) A) 0.00030% B) 0.019% C) 2.0% D) 6.0% E) 12%

Question

Quizplus · Accepted Answer

According to the free-electron theory of metals, the occupancy probability of an electron at a certain energy is given by the Fermi-Dirac distribution. The equation for the Fermi-Dirac distribution is:

f(E) = 1 / [exp((E-Ef)/(k_BT)) + 1]

where f(E) is the occupancy probability, E is the energy, Ef is the Fermi energy (level), k_B is the Boltzmann constant, and T is the temperature.

Substituting the given values into the equation and solving for f(12.00 eV):

f(12.00 eV) = 1 / [exp((12.00-11.63)/(1.38*10^-23*500*1.60*10^-19)) + 1]
f(12.00 eV) = 0.000190

Converting to a percentage:

f(12.00 eV) = 0.019%

Therefore, the answer is B.

Free-Electron Theory of Metals: What Is the Occupancy Probability at an Energy