Question

Calculate the probability for an electron to be found at an energy of (EF +2kBT) in a
metal.

Answer 1

Answer on Question #51573, Physics, Solid State Physics

Calculate the probability for an electron to be found at an energy of $(\mathrm{E_F} + 2\mathrm{k_B T})$ in a metal.

Solution

Electrons are particles with half-integer spin. And they obey Fermi-Dirac statistics. The function of Fermi-Dirac distribution is written as follows:

$$P (E, T) = \frac {1}{1 + \exp \left(\frac {E - E _ {F}}{k T}\right)}$$

where $P(E,T)$ and $P(E,T)$ - the probability that the electron occupies an energy level $E$, above or below the Fermi level $E_{F}$.

Then

$$P \left(E _ {F} + 2 k _ {B} T, T\right) = \frac {1}{1 + \exp \left(\frac {E _ {F} + 2 k _ {B} T - E _ {F}}{k _ {B} T}\right)} = \frac {1}{1 + e ^ {2}} \approx 0.119 = 11.9\%$$

Answer: $P(E_{F} + 2k_{B}T, T) = \frac{1}{1 + e^{2}} \approx 11.9\%$

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