Question

The Fermi energy for copper is $7.00\,eV$ . For copper at $1000\,K$ , (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 $N\left( E \right)$ and $\left( c \right)$ the density of occupied states ${N_ \circ }\left( E \right)$.

Answer 1

Hint: Fermi energy is defined as the energy difference between the highest occupied state and the lowest occupied state. The density of the occupied states is the product of the energy level of the occupied state and the probability of being occupied by the electron.

Formula used:
(1) The formula of the energy is given by
$E = {E_F} + kT\ln \left( {{P^{ - 1}} - 1} \right)$
Where $E$ is the energy, ${E_F}$ is the fermi energy of the copper, $k$ is the Boltzmann constant and $P$ is the probability.
(2) The density of the state is given by
$N\left( E \right) = C{E^{\frac{1}{2}}}$
Where $N\left( E \right)$ is the density of the state and $C$ is the constant.
(3) The density occupied by the states
${N_0}\left( E \right) = P \times N\left( E \right)$
Where ${N_0}\left( E \right)$ is the density occupied by the state.

Complete step by step answer:
Given: Fermi energy for copper,${E_F} = 7.00\,eV$
The temperature, $T = 1000\,K$
Probability being occupied by electron, $P = 0.009$
(a) Using the formula of the energy,
$E = {E_F} + kT\ln \left( {{P^{ - 1}} - 1} \right)$
Substituting the known values in the above formula,
$E = 0.07 + \left( {8.62 \times {{10}^{ - 5}} \times 1000} \right)\ln \left( {\dfrac{1}{{0.009}} - 1} \right)$
By doing the basic arithmetic operations, we get
$E = 6.81\,eV$
(b) Let us use the formula of the density of states,
$N\left( E \right) = C{E^{\frac{1}{2}}}$
$N\left( E \right) = 6.81\; \times \;{10^{27}} \times {\left( {6.81} \right)^{\frac{1}{2}}}$
By the simplification of the above step, we get
$N\left( E \right) = 1.77 \times {10^{28}}\,{m^{ - 3}}\,e{V^{ - 1}}$
(c) Let us use the formula of the density of the state,
${N_0}\left( E \right) = P \times N\left( E \right)$
${N_0}\left( E \right) = 0.900 \times 1.77 \times {10^8}$
By doing multiplication in the above step, we get
${N_0}\left( E \right) = 1.59 \times {10^{28}}{m^{ - 3}}.eV$

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

Note: $k$ , substituted in the above calculation is the Boltzmann constant and its value is equal to $8.62 \times {10^{ - 5}}$ . Remember the formula of the density of the free state and the occupied state since it differs by energy value is used in free state and the free state value is substituted in the occupied state.