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Ann [662]
3 years ago
6

The question is :

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
7 0
I believe the answer is 42 ft
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A shopper paid $2.52 for 4.5 pounds of potatoes, $7.75 for 2.5 pounds of broccoli and $2.45 f0r 2.5 pounds of pears. What is the
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Unit rate refers to how much the buyer paid per unit. Let’s do the math:

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2 years ago
evaluate the fermi function for an energy KT above the fermi energy. find the temperature at which there is a 1% probability tha
dybincka [34]

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  F(E_o) =  0.2689

b

The temperature is  T_k  =  1261 \  K

Step-by-step explanation:

From the question we are told that

   The energy considered is  E = 0.5 eV

Generally the Fermi  function is mathematically represented as

       F(E_o) =  \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }

    Here K is the Boltzmann constant with value k = 1.380649 *10^{-23} J/K

            E_F  is the Fermi energy

            E_o  is the initial energy level which is mathematically represented as

     E_o = E_F + KT

So

     F(E_o) =  \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}

=>   F(E_o) =  \frac{1}{e^{\frac{KT}{KT} } + 1}

=>   F(E_o) =  \frac{1}{e^{ 1 } + 1}

=>   F(E_o) =  0.2689

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

     F(E_k) =  \frac{1}{e^{\frac{[E_k - E_F]}{KT_k} } + 1 }  = 0.01

HereE_k is that energy level that is  0.5 ev above the Fermi energy  E_k = 0.5 eV  + E_F

=>   F(E_k) =  \frac{1}{e^{\frac{[[0.50 eV + E_F] - E_F]}{KT_k} } + 1 }  = 0.01

=>   \frac{1}{e^{\frac{0.50 eV ]}{KT_k} } + 1 }  = 0.01

=>   1 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} } + 0.01

=>   0.99 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} }

=>   e^{\frac{0.50 eV ]}{KT_k} }  = 99

Taking natural  log of both sides

=>   \frac{0.50 eV }{KT_k} }  =4.5951

=>    0.50 eV   =4.5951 *  K *  T_k

Note eV is electron volt and the equivalence in Joule is     eV  =  1.60 *10^{-19} \  J

So

     0.50 * 1.60 *10^{-19 }   =4.5951 *  1.380649 *10^{-23} *  T_k

=>   T_k  =  1261 \  K

7 0
3 years ago
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