A silicon sample is doped with atoms having donor states 0.110eV below the bottom of the conduction band. (The energy gap in silicon is 1.11eV ) If each of these donor states is occupied with a probability of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$at $\mathit{T}\mathbf{=}\mathbf{300}\mathbf{}\mathit{K}$, (a) is the Fermi level above or below the top of the silicon valence band and (b) how far above or below? (c) What then is the probability that a state at the bottom of the silicon conduction band is occupied?

1. The silicon sample is doped with atoms of donor states at an energy$∆E=0.110\mathrm{eV}$ below the bottom of the conduction band.
2. Energy gap in silicon,$E_g=1.11\mathrm{eV}$
3. The occupancy probability of silicon,$P\left(E\right)=5\times {10}^{-5}$
4. Temperature value,data-custom-editor="chemistry" $T=300K$

Due to doping the Fermi level, which lies in the mid-range of the energy gap, is shifted to the bottom of the conduction band. Now using the probability equation and the given data, we can get the value of the Fermi energy of silicon. Now, for the given energy gap, we can get the occupancy probability using the energy value in the given equation.

Formula:

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics, $P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)/kT}+1}$ (i)

Due to doping the Fermi level is shifted to the bottom of the conduction band. Thus, it should be above the top of the valence band.

The value of the energy gap for the donor state when measured from the top of the valence band is given as follows:

$$\begin{gathered} E=1.11\mathrm{eV}-0.11\mathrm{eV} \\ =1.0\mathrm{eV} \end{gathered}$$

Now, using the probability equation, we can get the value of the Fermi energy as follows:

$$\begin{gathered} E_F=E-kTIn\left[P^{-1}-1\right] \\ =1.0\mathrm{eV}-\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(300\mathrm{K}\right)\mathrm{In}\left[{\left(5\times {10}^{-5}\right)}^{-1}-1\right] \\ =0.744\mathrm{eV} \end{gathered}$$

Hence, the value of the energy of the Fermi level is 0.744 eV .

Using the given energy at the bottom of the silicon band, we can get the occupancy probability as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{\left(1.11eV-0.744eV\right)/\left(8.62\times {10}^{-5}eV/K\right)\left(300K\right)}+1} \\ =7.13\times {10}^{-7} \end{gathered}$$

Hence, the occupancy probability for a state at the bottom of conduction band is $7.13\times {10}^{-7}$.