Answer:
See attached
Step-by-step explanation:
<u>Given</u>
The graph for the function is attached
Answer:
1000000
Step-by-step explanation:
10x10x10x10x10x10=1000000
Answer:
1) 2(t - 7)(t + 1)
2) $32,000
Step-by-step explanation:
v(t) = 2t² - 12t - 14
v(t) = 2(t² - 6t - 7) = 2(t - 7)(t + 1)
x = -b/2a = 6/2 = 3
y = 2(3)² - 12(3) - 14 = -32
(3, -32) means at 3 months, -32 thousand dollars
Value + value = 14 dollars
0.65x + 10-0.40x = 14
0.25x = 4
x = 16
25-x =9
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 
b
The temperature is 
Step-by-step explanation:
From the question we are told that
The energy considered is 
Generally the Fermi function is mathematically represented as
![F(E_o) = \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }](https://tex.z-dn.net/?f=F%28E_o%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5BE_o%20-%20E_F%5D%7D%7BKT%7D%20%7D%20%2B%201%20%7D)
Here K is the Boltzmann constant with value 
is the Fermi energy
is the initial energy level which is mathematically represented as

So
![F(E_o) = \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}](https://tex.z-dn.net/?f=F%28E_o%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5B%5BE_F%20%2B%20KT%5D%20-%20E_F%5D%7D%7BKT%7D%20%7D%20%2B%201%7D)
=> 
=> 
=> 
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](https://tex.z-dn.net/?f=F%28E_k%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5BE_k%20-%20E_F%5D%7D%7BKT_k%7D%20%7D%20%2B%201%20%7D%20%20%3D%200.01)
Here
is that energy level that is 0.5 ev above the Fermi energy 
=> ![F(E_k) = \frac{1}{e^{\frac{[[0.50 eV + E_F] - E_F]}{KT_k} } + 1 } = 0.01](https://tex.z-dn.net/?f=F%28E_k%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5B%5B0.50%20eV%20%2B%20E_F%5D%20-%20E_F%5D%7D%7BKT_k%7D%20%7D%20%2B%201%20%7D%20%20%3D%200.01)
=> ![\frac{1}{e^{\frac{0.50 eV ]}{KT_k} } + 1 } = 0.01](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D%20%2B%201%20%7D%20%20%3D%200.01)
=> ![1 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} } + 0.01](https://tex.z-dn.net/?f=1%20%3D%200.01%20%2A%20e%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D%20%2B%200.01)
=> ![0.99 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} }](https://tex.z-dn.net/?f=0.99%20%3D%200.01%20%2A%20e%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D)
=> ![e^{\frac{0.50 eV ]}{KT_k} } = 99](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D%20%20%3D%2099)
Taking natural log of both sides
=> 
=> 
Note eV is electron volt and the equivalence in Joule is 
So

=> 