Question

Evaluate the temperature at which there is 4 percent probability that a state with energy 2 eV is occupied. Given that Fermi energy = 1.5 eV​

Answer 1

Explanation:

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Answer 2

Answer:

The temperature for the given probability is 1804.53K.

Explanation:

Given the probability of occupancy of a state,

$f(E) = 4\%=\dfrac{4}{100}$

The occupied energy state above the fermi energy, $E = 2 eV$

Fermi energy, $E_F = 1.5 eV$

The probability of occupation of certain energy level at a given temperature is given by Fermi-Dirac statistics.

$f(E) = \dfrac{1}{1+\mbox{exp}\bigg(\dfrac{E-E_f}{K_BT}\bigg) }$

where E is the allowed state,

$E_F$ is the fermi energy,

T is the temperature

and Boltzmann's constant,

$K_B=1.38\times 10^{-23} J/K=1.38\times 10^{-23}\times 6.24\times 10^{18}eV/K$.

Substituting the values,  

$\dfrac{4}{100} = \dfrac{1}{1+\mbox{exp}\bigg(\dfrac{2-1.5}{1.38\times 10^{-23}\times 6.24\times 10^{18}\times T} \bigg)}$

$\dfrac{1}{25} = \dfrac{1}{1+\mbox{exp}\bigg(\dfrac{0.5}{1.38\times 10^{-5}\times 6.24 \times T} \bigg)}$

$\implies 25 = 1+\mbox{exp}\bigg(\dfrac{0.5}{1.38\times 10^{-5}\times 6.24 \times T} \bigg)}$

$\implies 25 = 1+\mbox{exp}\bigg(\dfrac{0.5\times 10^{5}}{8.61 \times T} \bigg)}$

$\implies 25 \approx\mbox{exp}\bigg(\dfrac{0.05807\times 10^{5}}{T} \bigg)}$

Taking $ln$ on both sides

$ln25 =\dfrac{0.05807\times 10^{5}}{T}$

$\implies T =\dfrac{0.05807\times 10^{5}}{3.218}$

          $=0.0180453\times 10^{5}=1804.53K$

Therefore, the temperature for the given probability is 1804.53K.

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