Free-electron theory of metals: 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? (m_el = 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^-19J) A) 1 × 10^-4 B) 2 × 10^-5 C) 5 × 10^-6 D) 1 × 10^-6 E) 2 × 10^-7

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The Answer of Free-electron theory of metals: 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? (m_el = 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^-19J) A) 1 × 10^-4 B) 2 × 10^-5 C) 5 × 10^-6 D) 1 × 10^-6 E) 2 × 10^-7

Free-Electron Theory of Metals: the Fermi Energy of Rubidium at a Temperature