Complete Question
Evaluate the Fermi function for an energy kT above the Fermi energy. Find the temperature at which there is a 1% probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron.
Answer:
a
The Fermi function for the energy KT is [tex]F(E_o) = 0.2689[/tex]
b
The temperature is [tex]T_k = 1261 \ K[/tex]
Step-by-step explanation:
From the question we are told that
The energy considered is [tex]E = 0.5 eV[/tex]
Generally the Fermi function is mathematically represented as
[tex]F(E_o) = \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }[/tex]
Here K is the Boltzmann constant with value [tex]k = 1.380649 *10^{-23} J/K[/tex]
[tex]E_F[/tex] is the Fermi energy
[tex]E_o[/tex] is the initial energy level which is mathematically represented as
[tex]E_o = E_F + KT[/tex]
So
[tex]F(E_o) = \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}[/tex]
=> [tex]F(E_o) = \frac{1}{e^{\frac{KT}{KT} } + 1}[/tex]
=> [tex]F(E_o) = \frac{1}{e^{ 1 } + 1}[/tex]
=> [tex]F(E_o) = 0.2689[/tex]
Generally the probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron is mathematically represented by the Fermi function as
[tex]F(E_k) = \frac{1}{e^{\frac{[E_k - E_F]}{KT_k} } + 1 } = 0.01[/tex]
Here[tex]E_k[/tex] is that energy level that is 0.5 ev above the Fermi energy [tex]E_k = 0.5 eV + E_F[/tex]
=> [tex]F(E_k) = \frac{1}{e^{\frac{[[0.50 eV + E_F] - E_F]}{KT_k} } + 1 } = 0.01[/tex]
=> [tex]\frac{1}{e^{\frac{0.50 eV ]}{KT_k} } + 1 } = 0.01[/tex]
=> [tex]1 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} } + 0.01[/tex]
=> [tex]0.99 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} }[/tex]
=> [tex]e^{\frac{0.50 eV ]}{KT_k} } = 99[/tex]
Taking natural log of both sides
=> [tex]\frac{0.50 eV }{KT_k} } =4.5951[/tex]
=> [tex]0.50 eV =4.5951 * K * T_k[/tex]
Note eV is electron volt and the equivalence in Joule is [tex]eV = 1.60 *10^{-19} \ J[/tex]
So
[tex]0.50 * 1.60 *10^{-19 } =4.5951 * 1.380649 *10^{-23} * T_k[/tex]
=> [tex]T_k = 1261 \ K[/tex]
