Question

The Fermi energy for copper is 7.00eV. For copper at 1000K, (a) find the energy of the energy level whose probability of being occupied by an electron is 0.900. For this energy, evaluate (b) the density of states N(E) and (c) the density of occupied states ${\mathit{N}}_{\mathbf{0}}\left(E\right)$.

Short answer

a) The energy of the energy level whose occupied probability has value of 0.900 is 6.81 eV.

b) The density of the states N (E) for the energy is $1.77\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

b) The density of occupied states $N_0\left(E\right)\mathrm{is}1.59\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

Step-by-step solution

The given data

a) The Fermi energy of copper,$E_F=7\mathrm{eV}$

b) The temperature value of the energy level, T = 1000 K

c) The probability of being occupied by an electron, P(E) = 0.900

Understanding the concept of density and probability

Using the concept of probability of an energy level being occupied by an electron, we can get the value of the required energy level. Now, using this energy level, we can get the value of the density of states. At last, the product of the probability and the density of states will determine the value of the density of occupied states.

Formulae:

The density of states associated with the conductions electrons of a metal $\mathbf{N}\left(E\right)\mathbf{=}{\mathbf{CE}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{,}\mathbf{where}\mathbf{}\mathbf{C}\mathbf{=}\frac{\mathbf{8}\sqrt{\mathbf{2}}{\mathbf{\tau \tau m}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}{{\mathbf{h}}^{\mathbf{3}}}$(i)

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics,$\mathbf{P}\left(E\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\left(E-E\right)\mathbf{/}\mathbf{kT}}\mathbf{+}\mathbf{1}}\mathbf{,}\mathbf{where}\mathbf{}\mathbf{k}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{eV}$ (ii)

The density of states for the condition of occupancy by the conduction electrons,

${\mathbf{N}}_{\mathbf{0}}\left(E\right)\mathbf{=}\mathbf{N}\left(E\right)\mathbf{P}\left(E\right)$ (iii)

Calculation of the energy at the energy level

Using equation (ii) and the given data, we can get the energy of the energy level at which the probability of occupancy has value 0.900 is given as follows:

$$\begin{gathered} E=E_F+kT\mathrm{In}\left({\mathrm{P}}^{-1}+1\right) \\ =7\mathrm{Ev}+\left(8.62\times {10}^{-5}\mathrm{Ev}\right)\left(1000\mathrm{k}\right)\mathrm{In}\left[{\left(0.900\right)}^{-1}+1\right] \\ =6.81\mathrm{eV} \end{gathered}$$

Hence, the value of the energy is 6.81 eV.

b) Calculation of the density of states

At first, we know that

$$\begin{gathered} C=\frac{8\sqrt{2}\pi m^{3/2}}{h^3} \\ =\frac{8\sqrt{2}\pi {\left(9.1\times {10}^{-31}\mathrm{kg}\right)}^{3/2}}{{\left(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^3} \\ =1.062\times {10}^{56}{\mathrm{kg}}^{3/2}/{\mathrm{J}}^3.{\mathrm{s}}^3 \\ =6.81\times {10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3} \end{gathered}$$

Thus, using this value and the energy value in equation (i), we can get the density of the states as follows:

$$\begin{gathered} N\left(E\right)=(6.81\times {10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3})\left(6.81\mathrm{eV}\right) \\ =1.77\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1} \end{gathered}$$

Hence, the value of the density of states is $1.77\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

c) Calculation of the number of occupied states

Using the above value of density of states and occupied probability value in equation (iii), we can get the density of the occupied states as follows:

$$\begin{gathered} N_0\left(E\right)=\left(1.77\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}\right)\left(0.900\right) \\ =1.59\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1} \end{gathered}$$

Hence, the value of density of occupied states is $1.59\times {10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.