We can use the Fermi-Dirac distribution to solve this problem. The probability that an electron occupies an energy state $E$ at a given temperature $T$ is given by:

$f(E) = \frac{1}{1 + e^{(E - E_F) / (k_B T)}}$

Where:

• $f(E)$ is the probability that the energy state is occupied.
• $E - E_F = 0.1 \, \text{eV}$ is the energy difference between the state and the Fermi level.
• $k_B$ is the Boltzmann constant, $k_B = 8.617 \times 10^{-5} \, \text{eV/K}$.
• $T$ is the temperature in Kelvin.
• $f(E) = 0.01$.

We need to solve for $T$ when $f(E) = 0.01$.

Rearranging the Fermi-Dirac equation:

$0.01 = \frac{1}{1 + e^{(0.1 \, \text{eV}) / (k_B T)}}$

$1 + e^{(0.1 \, \text{eV}) / (k_B T)} = 100$

$e^{(0.1 \, \text{eV}) / (k_B T)} = 99$

Now, take the natural logarithm on both sides:

$\frac{0.1 \, \text{eV}}{k_B T} = \ln(99)$

$T = \frac{0.1 \, \text{eV}}{k_B \ln(99)}$

Now, we calculate $T$.

$T = \frac{0.1 \, \text{eV}}{(8.617 \times 10^{-5} \, \text{eV/K}) \ln(99)}$

Let me compute that for you.The temperature at which an electron occupies an energy state 0.1 eV above the Fermi level with a probability of 0.01 is approximately 253 K.

Let me know if you'd like more details or have any questions!

Here are 5 related questions you might explore:

1. What is the Fermi-Dirac distribution used for in solid-state physics?
2. How does the probability of electron occupation change with temperature?
3. How is the Fermi level defined in a metal at absolute zero?
4. What is the significance of the Boltzmann constant in statistical mechanics?
5. How does the energy band structure affect electron occupancy in semiconductors?

Tip: The Fermi-Dirac distribution is most relevant for systems where quantum effects are significant, especially at very low temperatures.