Problem 27 - University Physics with Modern Physics

Question

For a solid metal having a Fermi energy of 8.500 eV, what is the probability, at room temperature, that a state having an energy of 8.520 eV is occupied by an electron?

Step-by-Step Solution

Answer

The probability is approximately 31.6%.

Step 1: Understand the Fermi-Dirac Distribution Formula

The Fermi-Dirac distribution function helps us determine the probability that a given energy level is occupied by an electron. It is given by the formula: \( f(E) = \frac{1}{e^{(E - E_f) / kT} + 1} \), where \( E \) is the energy of the state, \( E_f \) is the Fermi energy, \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin.

Step 2: Identify the Given Variables

We have a Fermi energy \( E_f = 8.500 \) eV and an energy state \( E = 8.520 \) eV. The room temperature \( T \) is approximately 300 K. Also, \( k \), the Boltzmann constant, is about \( 8.617333262145 \times 10^{-5} \) eV/K.

Step 3: Calculate the Exponent in the Distribution Function

Calculate \( \frac{E - E_f}{kT} \). Here, \( E - E_f = 8.520 - 8.500 = 0.020 \) eV. Calculate \( kT = (8.617333262145 \times 10^{-5} \text{ eV/K})(300 \text{ K}) = 0.025851 \) eV. Thus, \( \frac{E - E_f}{kT} = \frac{0.020}{0.025851} \approx 0.774 \).

Step 4: Compute the Fermi-Dirac Distribution

Substitute the exponent \( 0.774 \) into the Fermi-Dirac distribution function: \( f(E) = \frac{1}{e^{0.774} + 1} \). Calculate \( e^{0.774} \approx 2.168 \), so \( f(E) = \frac{1}{2.168 + 1} \approx \frac{1}{3.168} \approx 0.316 \).

Step 5: Interpret the Result

The result from the previous step tells us the probability. Thus, the probability that the state at 8.520 eV is occupied at room temperature is approximately 0.316, or 31.6%.

Key Concepts

Fermi Energy

Fermi energy represents a key concept in quantum mechanics and solid state physics. It is the energy level at which the probability of finding an electron is exactly 50% when the temperature is at absolute zero.
In metallic solids, the Fermi energy is a crucial parameter that profoundly influences electronic properties. This is because it marks the highest occupied energy level of electrons within a solid.
Key points about Fermi energy include:

• Designated as \( E_f \).
• Varies for different materials. In our original problem, it is 8.500 eV.
• At absolute zero, all energy states below \( E_f \) are filled, and those above are empty.

It helps us to understand the distribution of electrons and their dynamics at various energy levels, playing a critical role in analyzing electrical conductivity and thermal properties.

Boltzmann Constant

The Boltzmann constant is an important physical constant that serves as a bridge between macro and micro worlds. It links temperature to energy on a particle level.
The constant is symbolized by \( k \) and has a value of approximately \( 8.617333262145 \times 10^{-5} \, \text{eV/K} \). This measurement is crucial for calculations involving the Fermi-Dirac distribution.
Highlighting its importance:

• Relates energy to temperature in physics equations.
• Plays a significant role in thermodynamics and statistical mechanics.
• Used in calculating thermal volts, which are key for understanding semiconductor behavior.

In our exercise, it allows us to convert the energy difference \( E - E_f \) in terms of temperature influence, determining the probability that a particular energy state is occupied by an electron.

Room Temperature

Room temperature is often a default reference point used in scientific calculations. It is generally taken as around 20°C to 25°C, which translates to approximately 293 K to 298 K in the Kelvin scale.
In our particular exercise, room temperature is approximated to 300 K. This small assumption helps in simplifying complex equations.
Things to note about room temperature in context:

• Commonly used to approximate conditions for substances under normal conditions.
• Enables theoretical predictions in physics and chemistry due to its constancy in settings.
• Affects electron dynamics as warmer temperatures mean higher thermal energy.

Because of its consistency, room temperature serves as a useful basis for comparing the behavior of materials and electronic probabilities, like in the probability calculations of energy states using the Fermi-Dirac distribution function.