The Fermi energy of rubidium at a temperature of 5 K is 1.85 eV. An electron state in rubidium is 0.007 eV above the Fermi level. What is the probability that this state is occupied at a temperature of 9K? (^mel = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s, Boltzmann constant = 1.38 ∙ 10^-23 J/K, 1 eV = 1.60 × 10^-19 J)

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Quizplus · Accepted Answer

The probability that an electron state is occupied is given by the Fermi-Dirac distribution function:$$f(E) = \frac{1}{e^{(E-E_F)/k_BT} + 1}$$where:* $E$ is the energy of the electron state* $E_F$ is the Fermi energy* $k_B$ is the Boltzmann constant* $T$ is the temperatureIn this problem, we are given that $E = 1.857$ eV, $E_F = 1.85$ eV, $k_B = 1.38 imes 10^{-23}$ J/K, and $T = 9$ K. Plugging these values into the Fermi-Dirac distribution function, we get:$$f(E) = \frac{1}{e^{(1.857-1.85)/(1.38 imes 10^{-23})(9)} + 1} = 0.0001$$Therefore, the probability that the electron state is occupied is 0.0001, or 1 × 10^-4.

The Fermi Energy of Rubidium at a Temperature of 5