Question

The Fermi energy of a system is 5.5eV. At 500K, the energy of a level for which the probability of occupancy is 0.2, is ________eV. (Rounded off to two decimal places) (Boltzmann constant 𝑘_𝐵 = 8.62x10⁻⁵ eV/K)

Answer 1

Correct Answer: 5.55

Given Data 

• Fermi energy: \( E_F = 5.5 \) eV
• Probability of occupancy: \( f(E) = 0.2 \)
• Temperature: \( T = 500K \)
• Boltzmann constant: \( k_B = 8.62 \times 10^{-5} \) eV/K

Step 1: Use the Fermi-Dirac Distribution

The Fermi-Dirac distribution is given by:

\[ f(E) = \frac{1}{1 + \exp\left(\frac{E - E_F}{k_B T}\right)} \]

We are given \( f(E) = 0.2 \), so we need to solve for \( E \).

Step 2: Rearrange the Equation

Taking the reciprocal of both sides:

\[ \frac{1}{f(E)} = 1 + \exp\left(\frac{E - E_F}{k_B T}\right) \]

Subtracting 1 from both sides:

\[ \frac{1}{f(E)} - 1 = \exp\left(\frac{E - E_F}{k_B T}\right) \]

Taking the natural logarithm on both sides:

\[ \ln \left(\frac{1}{f(E)} - 1\right) = \frac{E - E_F}{k_B T} \]

Solving for \( E \):

\[ E = E_F + k_B T \ln \left(\frac{1}{f(E)} - 1\right) \]

Step 3: Substitute the Known Values

Substituting \( f(E) = 0.2 \), \( E_F = 5.5 \) eV, \( k_B = 8.62 \times 10^{-5} \) eV/K, and \( T = 500K \):

\[ E = 5.5 + (8.62 \times 10^{-5} \times 500) \ln \left(\frac{1}{0.2} - 1\right) \]

\[ E = 5.5 + (8.62 \times 10^{-5} \times 500) \ln(4) \]

\[ E = 5.5 + (0.0431) \times 1.386 \]

\[ E = 5.5 + 0.0598 \]

Step 4: Final Calculation

\[ E \approx 5.56 \text{ eV} \]

Final Answer

The energy of the level is approximately 5.56 eV.