Question

The Fermi energy for silver is5.5eV. At $\mathit{T}\mathbf{=}\mathbf{0}\mathbf{°}\mathit{C}$, what are the probabilities that states with the following energies are occupied: (a)4.4eV, (b)5.4eV, (c)5.5eV, (d)5.6eV, and (e)6.4eV? (f) At what temperature is the probability 0.16 that a state with energy E = 5.6eV is occupied?

Short answer

a) The occupied probability of state at 4.4 eVis 1.0.

b) The occupied probability of state at 5.4 eV is 0.99.

c) The occupied probability of state at 5.5 eV is 0.5.

d) The occupied probability of state at 5.6 eV is 0.014.

e) The occupied probability of state at 6.4 eV is 2.4 x ${10}^{-17}$.

f) The temperature at which the probability value is 0.16 with energy state 5.6 eV as is $7\times {10}^2\mathrm{K}$.

Step-by-step solution

The given data

a) Fermi energy of silver,$E_F=5.5\mathrm{eV}$

b) The temperature of the energy level,$T=0°C\mathrm{or}273\mathrm{K}$

c) The given energy values,$E=4.4\mathrm{eV},5.4\mathrm{eV},5.5\mathrm{eV},5.6\mathrm{eV}\mathrm{or}6.4\mathrm{eV}$

d) The energy level E = 5.6 eV has probability value,P = 0.16

Understanding the concept of occupied probability

Using the given data in the equation of occupied probability of Fermi-Dirac statistics, we can get the probability value of each energy level. Again, using this same concept, we can get the value of the temperature for the known energy level and probability.

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)

a) Calculation of the probability for energy E = 4.4 eV

At first, we calculated the exponential value of equation (i) for energy E =4.4 eV as follows:

$\frac{\left(E-E_F\right)}{kT}=\frac{\left(4.4-5.5\right)\mathrm{eV}\times \left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)}$

Now, using this value in equation (i), we can get the occupied probability at energy level as E = 4.4 eV follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{-46.25}+1} \\ =1.0 \end{gathered}$$

Hence, the value of probability is 1.0.

b) Calculation of the probability for energy E = 5.4eV

At first, we calculated the exponential value of equation (i) for energy E = 5.4eV as follows:

$$\begin{gathered} \frac{\left(E-E_F\right)}{kT}=\frac{\left(5.4-5.5\right)\mathrm{eV}\times \left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)} \\ =-4.25 \end{gathered}$$

Now, using this value in equation (i), we can get the occupied probability at energy level E = 5.4 eV as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{-4.25}+1} \\ =0.986 \\ \approx 0.99 \end{gathered}$$

Hence, the value of probability is 0.99.

c) Calculation of the probability for energy E = 5.5 eV

At first, we calculated the exponential value of equation (i) for energy E = 5.5 eV as follows:

$$\begin{gathered} \frac{\left(E-E_F\right)}{kT}=\frac{\left(5.5-5.5\right)\mathrm{eV}\times \left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)} \\ =0 \end{gathered}$$

Now, using this value in equation (i), we can get the occupied probability at energy level E = 5.5 eV as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^0+1} \\ =0.5 \end{gathered}$$

Hence, the value of probability is 0.5.

d) Calculation of the probability for energy E = 5.6 eV

At first, we calculated the exponential value of equation (i) for energy E = 5.6 eV as follows:

$$\begin{gathered} \frac{\left(E-E_F\right)}{kT}=\frac{\left(5.6-5.5\right)\mathrm{eV}\times \left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)} \\ =4.25 \end{gathered}$$

Now, using this value in equation (i), we can get the occupied probability at energy level$E=5.6\mathrm{eV}$ as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{{\mathrm{e}}^0+1} \\ =0.5 \end{gathered}$$

Hence, the value of probability is 0.5.

e) Calculation of the probability for energy E = 6.4 eV

At first, we calculated the exponential value of equation (i) for energy E = 5.6 eV as follows:

$$\begin{gathered} \frac{\left(E-E_F\right)}{kT}=\frac{\left(6.4-5.5\right)\mathrm{eV}\times \left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)} \\ =38.25 \end{gathered}$$

Now, using this value in equation (i), we can get the occupied probability at energy level E = 5.6 eV as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{38.25}+1} \\ =2.447\times {10}^{-17} \\ =2.4\times {10}^{-17} \end{gathered}$$

Hence, the value of probability is 0.014.

f) Calculation of the temperature

Now, equation (i) can be written for the temperature value as follows:

$$\begin{gathered} e^{\left(E-E_F\right)/kT}=\frac{1}{P}-1 \\ T=\frac{\left(E-E_F\right)}{k\mathrm{In}\left({\displaystyle \frac{1}{\mathrm{P}}}-1\right)}.....................\left(a\right) \end{gathered}$$

Thus, the temperature for the given conditions can be calculated using the given data in equation (a) as follows:

$$\begin{gathered} T=\frac{\left(5.6\mathrm{eV}-5.5\mathrm{eV}\right)\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\right)\mathrm{In}\left({\displaystyle \frac{1}{0.16}}-1\right)} \\ =699\mathrm{K} \\ \approx 7\times {10}^2\mathrm{K} \end{gathered}$$

Hence, the value of the temperature is $7\times {10}^2\mathrm{K}$.