91Ó°ÊÓ

• Number of conduction electrons per unit volume,$n=1.70×{10}^{28}{m}^{-3}$
• Volume of metal,$V=6×{10}^{-6}{m}^3$
• Temperature,$T=200K$
• Given energy range,$∆E=3.20×{10}^{-20}J$
• Energy in the center,$E=4x{10}^{-19}J$

At first, using the number of conduction electrons per unit volume, we can calculate the Fermi energy of the material. Then, we calculate the occupancy probability. Then, using Fermi-Dirac statistics we can calculate the value of the density of occupied states. Now, the value of the density of occupied states is calculated using the density of states at the energy level and the occupancy probability value is used in the formula of the number of occupied density states to get the required value.

Formulae:

The equation of Fermi energy $E_F=\frac{0.121h^2}{m_e}{n}^{2/3}$ (i)

The occupancy probability is , $P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)/kT}+1}$ (ii)

The density of states associated with the conduction electrons of a material,

$N\left(E\right)=\frac{8\sqrt{2}{\mathrm{Ï€³¾}}^{3/2}}{h^3}{E}^{1/2}$ (iii)

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

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

$N\text{'}=N_0\left(E\right)V∆E$ (v)

Using the value of conduction electrons per unit volume in equation (i), we can calculate the Fermi energy of the material as follows:

$$\begin{gathered} E_F=\frac{0.121{(6.26×{10}^{-34}J.s)}^2}{\left(9.11×{10}^{-31}kg\right)}{\left(1.70×{10}^{28}{m}^{-3}\right)}^{2/3} \\ ==3.86×{10}^{-19}J \end{gathered}$$

Now,

$$\begin{gathered} E-E_F=4.00x{10}^{-19}J-3.86×{10}^{-19}J \\ =1.4×{10}^{-20}J \end{gathered}$$

and

role="math" localid="1661573166752" $$\begin{gathered} \frac{E-E_F}{kT}=\frac{1.4×{10}^{-20}J}{\left(1.38×{10}^{-23}J/K\right)\left(200K\right)} \\ =5.07 \end{gathered}$$

Thus, using the above value in equation (ii), we can get the occupancy probability of the material as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{5.07}+1} \\ =6.20×{10}^{-3} \end{gathered}$$

Now, the density of the states associated with the conduction electrons for the given energy level can be calculated using equation (iii) as follows:

$$\begin{gathered} N\left(E\right)=\frac{8\sqrt{2}\mathrm{π}{\left(9.11×{10}^{-31}\mathrm{kg}\right)}^{3/2}}{{\left(6.626×{10}^{-34}J.s\right)}^3}{\left(4.00×{10}^{-19}J\right)}^{1/2} \\ =6.717×{10}^{46}/m^{3}J \end{gathered}$$

The density of the occupied states is given using equation (iv) as follows:

$$\begin{gathered} {\mathrm{N}}_0\left(\mathrm{E}\right)=\left(6.717×{10}^{46}/{\mathrm{m}}^3.\mathrm{J}\right)\left(6.20×{10}^{-3}\right) \\ =4.16×{10}^{44}/{\mathrm{m}}^3\mathrm{J} \end{gathered}$$

With the given energy range of $∆E=3.20×{10}^{-20}$and the given volume of material, we can get the number of occupied states using equation (v) as follows:

$$\begin{gathered} N\text{'}=\left(4.16×{10}^{44}/m^{3}J\right)\left(6×{10}^{-6}/m^{-3}\right)\left(3.20×{10}^{-20}J\right) \\ =7.98×{10}^{19} \end{gathered}$$

Hence, the value of the occupied states is $7.98×{10}^{19}$.