Question

At T = 300K, how far above the Fermi energy is a state for which the probability of occupation by a conduction electron is 0.10?

Short answer

The value of the energy of the state above the Fermi energy is $9.1\times {10}^{-21}\mathrm{J}$9.1 .

Step-by-step solution

The given data

a) The temperature value, T = 300 K

b) The probability value, P (E) = 0.10

Understanding the concept of energy state

Using the equation of occupancy probability for an energy state, we can get the value of the energy that is above the level of the Fermi energy.

Formula:

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics, $P\left(E\right)=\frac{1}{e^{\left(E_1-E_F\right)/kT}+1}\mathrm{where}k=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$ (i)

Calculation of the value of energy above Fermi level

Let the energy of the state in the problem be an amount $∆E$above the Fermi level$E_F$.

Then, the equation of the required energy can be given using the occupancy probability of equation (i) and the given data as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{∆E/kT}+1} \\ ∆E=\mathrm{kT}\mathrm{In}\left(\frac{1}{\mathrm{P}\left(\mathrm{E}\right)}-1\right) \\ =\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right)\mathrm{In}\left(\frac{1}{0.1}-1\right) \\ =9.1\times {10}^{-21}\mathrm{J} \end{gathered}$$

Hence, the value of the energy is $9.1\times {10}^{-21}\mathrm{J}$.