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?

Short answer

The probability is approximately 0.461.

Step-by-step solution

Understanding the Problem

We are given a solid metal with a Fermi energy (\(E_F\)) of 8.500 eV. We need to find the probability that an electron occupies a state with energy of 8.520 eV at room temperature (approx. 300 K). This requires using the Fermi-Dirac distribution formula.

Fermi-Dirac Distribution Formula

The Fermi-Dirac distribution provides the probability (\(f(E)\)) that a particular energy state is occupied by an electron. It is given by:\[ f(E) = \frac{1}{\exp\left(\frac{E - E_F}{kT}\right) + 1} \]where \(E\) is the energy level we are interested in, \(E_F\) is the Fermi energy, \(k\) is the Boltzmann constant (\(8.617 \, \times 10^{-5} \, \text{eV/K}\)), and \(T\) is the temperature in Kelvin.

Substitute Known Values

Substituting the given values into the formula, we have:\[ E = 8.520 \, \text{eV}, \quad E_F = 8.500 \, \text{eV}, \quad k = 8.617 \, \times 10^{-5} \, \text{eV/K}, \quad T = 300 \, \text{K} \]We need to compute the exponent in the formula:\[ \frac{E - E_F}{kT} = \frac{8.520 - 8.500}{8.617 \, \times 10^{-5} \, \times 300} \].

Calculate the Exponent

First, calculate the difference in energy: \(8.520 - 8.500 = 0.020 \, \text{eV}\).Then calculate the denominator:\(8.617 \, \times 10^{-5} \, \times 300 = 0.025851 \, \text{eV}\).Thus:\[ \frac{0.020}{0.025851} \approx 0.7738 \].

Compute the Probability

Plug the computed exponent back into the Fermi-Dirac formula:\[ f(8.520) = \frac{1}{\exp(0.7738) + 1} \approx \frac{1}{2.1685} \approx 0.461 \].This means the probability that a state with energy 8.520 eV is occupied by an electron is approximately 0.461.

Key concepts

Fermi Energy

Fermi energy is a fundamental concept in solid state physics, especially when studying the behavior of electrons in solids. It is the highest energy level that is filled with electrons at absolute zero temperature. Just imagine a well-packed concert, with the Fermi energy representing the topmost row of occupied seats. Knowing the Fermi energy allows us to predict electron distributions in metals and semiconductors.
A solid metal, for example, can be characterized by its Fermi energy, which is critical in determining electrical conductivity.
In our given problem, the Fermi energy is 8.500 eV. At room temperature, not all states below this energy are filled, and some above are occupied. This occupancy distribution is modeled by the Fermi-Dirac statistics and is essential to understanding energy level occupation at non-zero temperatures.

Probability Calculation

Probability calculation in the context of the Fermi-Dirac distribution helps determine how likely it is for electrons to occupy specific energy states above the Fermi energy.The Fermi-Dirac distribution formula, as used in this exercise, gives us the probability that a state is occupied by an electron.
The formula is expressed as:

• \( f(E) = \frac{1}{\exp\left(\frac{E - E_F}{kT}\right) + 1} \)

Here, \(f(E)\) is the probability, \(E\) is the given energy level, \(E_F\) is the Fermi energy, and \(kT\) represents the thermal energy.
In our example, this allows us to calculate a probability of 0.461 for an energy state of 8.520 eV. This means that there is about a 46.1% chance this state is occupied by an electron.

Boltzmann Constant

The Boltzmann constant \(k\) is a bridge between macroscopic and microscopic physics, playing a key role in statistical mechanics.It provides a way to relate temperature to energy, making it possible to calculate thermal effects on particles like electrons in solids.
The value of the Boltzmann constant is \(8.617 \times 10^{-5} \text{ eV/K}\), a crucial parameter in the context of the Fermi-Dirac distribution.This constant appears in the denominator of the exponent in the Fermi-Dirac formula, where it modulates how sharply energy levels transition from being occupied to unoccupied at a given temperature.
Understanding \(k\) helps us comprehend how temperature influences electron occupancy in materials. In essence, it affects how electrons distribute across different energy levels at various temperatures.

Solid State Physics

Solid state physics focuses on understanding the properties and behavior of solid materials, like metals and semiconductors. It deals with the interesting phenomena such as electrical conduction, magnetism, and optical properties that arise from atomic arrangements in solids.
In the realm of solid state physics, the Fermi-Dirac distribution is a fundamental concept. It describes how electrons populate energy states and is pivotal for explaining phenomena like electrical conductivity in metals.
The exercise involving Fermi energy, probability calculations, and the use of the Boltzmann constant reflects how these principles help us predict and understand real-world behaviors of materials. By applying these concepts, we can determine the likelihood of electrons filling certain energy levels at given temperatures, providing insights into the material's electronic properties.