1. At what temperature we can expect a 15% probability that electrons in silver have an energy which is 2% of Fermi energy and above the Fermi energy? The Fermi energy of silver is 5.5 eV.
Solution: \( E_F = 5.5 \, \text{eV} = 5.5 \times 1.6 \times 10^{-19} \, \text{J} \), electron occupation probability, \( F(E) = 0.15 \),
Energy level, \( E = E_F + 2\% \, \text{of} \, E_F = E_F + 0.02 \, E_F = 1.02 \, E_F \), so, \( E - E_F = 0.02 \, E_F \).
Find Temperature \( T = ? \)
The electron occupation probability is given by
\[ F(E) = \frac{1}{1 + e^{\frac{E - E_F}{k_B T}}} \]
\[\Rightarrow F(E) = \frac{1}{1 + e^{\frac{0.02 \times E_F}{k_B T}}} = \frac{1}{1 + e^{\frac{0.02 \times 5.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23} \times T}}} \]
\[ \Rightarrow 0.15 = \frac{1}{1 + e^{\frac{0.02 \times 5.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23} \times T}}} \]
\[ \Rightarrow 1 + e^{\frac{0.02 \times 5.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23} \times T}} = \frac{1}{0.15} = 6.666 \]
\[ \Rightarrow e^{\frac{0.02 \times 5.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23} \times T}} = 5.666 \]
\[ \Rightarrow \frac{0.02 \times 5.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23} \times T} = \ln(5.666) = 1.7334 \]
\[ \Rightarrow \frac{1275.362}{T} = 1.7334 \]
\[ \Rightarrow T = \frac{1275.362}{1.7334} = 735.75 \, \text{K} \]
The temperature is \( 735.75 \, \text{K} \).
2. At what temperature there is 1% probability in an energy level having energy 0.5 eV above Fermi energy.
Solution:
Energy level, \( E = E_F + 0.5 \, \text{eV}; E - E_F = 0.5 \, \text{eV} \times 1.602 \times 10^{-19} = 8.01 \times 10^{-20} \, \text{J} \)
Electron occupation probability, \( f(E) = 0.01 \, (1\%) \).
The electron occupation probability is given by \[ f(E) = \frac{1}{1 + e^{(E - E_F)/kT}} \] \[ 0.01 = \frac{1}{1 + e^{(8.01 \times 10^{-20})/kT}} \] \[ \frac{1}{0.01} = 1 + e^{(8.01 \times 10^{-20})/kT} \] \[ 100 = 1 + e^{(8.01 \times 10^{-20})/kT} \] \[ e^{(8.01 \times 10^{-20})/kT} = 99 \] \[ \frac{8.01 \times 10^{-20}}{kT} = \ln(99) \] \[ \ln(99) \approx 4.595 \] \[ \frac{8.01 \times 10^{-20}}{kT} = 4.595 \] \[ kT = \frac{8.01 \times 10^{-20}}{4.595} \approx 1.743 \times 10^{-20} \, \text{J} \]Substitute \( k = 1.38 \times 10^{-23} \, \text{J/K} \):
\[ T = \frac{1.743 \times 10^{-20}}{1.38 \times 10^{-23}} \approx 1263 \, \text{K} \]The temperature is approximately 1263 K.
3. Evaluate the Fermi function for an energy kT above the Fermi level.
Solution:The Fermi function is given by:
\[ f(E) = \frac{1}{1 + \exp\left(\frac{E - E_F}{kT}\right)} \]
For an energy \( E = E_F + kT \), calculate the Fermi function.
Substitute \( E = E_F + kT \) into the Fermi function:
\[ f(E) = \frac{1}{1 + \exp\left(\frac{(E_F + kT) - E_F}{kT}\right)} \]
Simplify the exponent:
\[ f(E) = \frac{1}{1 + \exp(1)} = \frac{1}{1 + e} \]
Since \( e \approx 2.718 \), we have:
\[ f(E) = \frac{1}{1 + 2.718} = \frac{1}{3.718} \]
\[ f(E) = 0.269 \]
The value of the Fermi function for an energy \( kT \) above the Fermi level is approximately \( 0.269 \).
For a comprehensive understanding, refer to a article on Fermi-Dirac Function.
c) 0.5
b) States below \(E_f\) filled, above \(E_f\) empty
b) 0.0259 eV
b) 0.268
c) \(4 \times 10^{-9}\)