Question

What is the number of occupied states in the energy range of 0.0300 eV that is centered at a height of 6.10 eV in the valence band if the sample volume is$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}{\mathbf{m}}^{\mathbf{3}}$, the Fermi level is 5.00 eV, and the temperature is 1500 K?

Short answer

The number of occupied states in the energy range of $0.0300eV$centered at $6.10\hspace{0.33em}eV$is $5.1\times {10}^{15}$ .

Step-by-step solution

The given data

a) The energy range is,$\Delta E=0.03\hspace{0.33em}eV$

b) The center of the energy range is,$E=6.10\hspace{0.33em}eV$

c) Volume of the sample,$V=5\times {10}^{-8}\hspace{0.33em}{m}^3$

d) Fermi energy of the sample,$E_F=5\hspace{0.33em}eV$

e) Temperature of the sample,$T=1500\hspace{0.33em}K$

f) The value of density of states at 6 eV from figure 41-6, $N\left(E\right)=1.7\times {10}^{28}\hspace{0.33em}/m^3.eV$

Understanding the concept of occupied states

the probability of an electron to occupy a particular energy state is known as the occupancy probability of the state. The occupancy probability of Fermi level is 0.5. At T=0 K, all the state below Fermi state are fully occupied whereas the ones above Fermi state are empty.

Formulae:

The occupancy probability for a state is given as-

$P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)/kT}+1}$ (i)

The density of occupied states,$N_0\left(E\right)=N\left(E\right)P\left(E\right)$ (ii)

The number of occupied states in the given energy range$∆E$

$n=N_0\left(E\right)V\Delta E$ (iii)

Here, $E_F$ is the Fermi energy, T is the absolute temperature and V is the volume of the

Sample.

Calculation of the number of occupied states

The occupied probability for the given energy level can be calculated using equation (i) as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{exp\left[{\displaystyle \frac{\left(6.10-5\right)eV\left(1.6\times {10}^{-19}J/eV\right)}{\left(1.38\times {10}^{-23}J/K\right)\left(1500K\right)}}\right]+1} \\ =2.01\times {10}^{-4} \end{gathered}$$

The density of occupied states using the above value in equation (ii) as follows:

$$\begin{gathered} N_0\left(E\right)=\left(1.7\times {10}^{28}/m^3.eV\right)\left(2.01\times {10}^{-4}\right) \\ =3.42\times {10}^{24}/m^3 \end{gathered}$$

The number of occupied states can be given using the above values in equation (iii) as follows:

$$\begin{gathered} n=\left(3.42\times {10}^{24}/m^3.eV\right)\left(5\times {10}^{-8}{m}^3\right)\left(0.03eV\right) \\ =5.1\times {10}^{15} \end{gathered}$$

Hence, the number of occupied states is $5.1\times {10}^{15}$.