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In a one-dimensional system the number of energy states per unit energy is \((l / h) \sqrt{2 m / \mathscr{E}}\), where \(l\) is the length of the sample and \(m\) is the mass of the electron. There are \(\mathcal{N}\) electrons in the sample and each state can be occupied by two electrons. (a) Determine the Fermi energy at \(0^{\circ} \mathrm{K}\). (b) Find the average energy per electron at \(0^{\circ} \mathrm{K}\).

Short Answer

Expert verified
The Fermi energy at \(0^{\circ} K\) is \(E_F = \((3 \mathcal{N} h^2)/(8 m L^2))^{2/3}\) and average energy per electron at \(0^{\circ} K\) is \(E_avg\) = 3/5 \(E_F\)

Step by step solution

01

Establishing the Energy Relation

We are given the number of energy states per unit energy. However, to find the Fermi energy and average energy, we need an equation for the number of energy states (N). Using the given equation, we can derive this by integrating energy states per unit energy over the energy levels from 0 to Fermi energy, this due to all states below Fermi energy are filled at 0 Kelvin. Doing this gives us the equation: \(\mathcal{N} = 2* \int_{0}^{E_F} (l / h) \sqrt{2 m / E} dE\) where \(E_F\) represents the Fermi energy. The number 2 arise from two states of spin for an electron.
02

Obtain the Fermi Energy

Now we'll integrate the right side of the equation from step 1 in order to find Fermi energy which is energy corresponding to number of electron \(\mathcal{N}\). The integration gives us \(\mathcal{N} = \frac{4}{3}\(l / h) \sqrt{2 m / \mathscr{E}}\) = \(E_F^{3/2}\) . Solving this equation for \(E_F\), we get the Fermi Energy expression at 0 K as \(E_F = \((3 \mathcal{N} h^2)/(8 m L^2))^{2/3}\)
03

Calculating Average Energy per Electron

Now to find average energy per electron at 0 Kelvin, we need to find total energy of all electron and then divide by total number of electrons. Total energy at 0 Kelvin is given by \(E_Total\) = 2 \( \int_{0}^{E_F} E (l / h) \sqrt{2 m / E} dE\) The solution to the integral provides \(E_Total\) = 3/5 \(\mathcal{N} E_F\). Using the Fermi Energy obtained from Step 2 , the average energy \(E_avg\) can be obtained by dividing total energy \(E_Total\) by \(\mathcal{N}\) .Thus average energy \(E_avg\) = 3/5 \(E_F\)
04

Final Solution Statement

At 0 degree Kelvin, we have therefore obtained the Fermi Energy \(E_F\) and the average energy \(E_avg\) per electron as derived using principles of quantum mechanics. These key results can be used as a bench mark to further calculate other quantities, dependent on Fermi Energy or average Energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quantum Physics
At the heart of understanding the behavior of electrons in materials is the field of quantum physics. It's a fundamental theory in physics that explains the nature and behavior of matter and energy on the atomic and subatomic levels. Unlike the classical physics that describes everyday objects, quantum physics governs the realms of particles like electrons, photons, and quarks, which follow different rules entirely.

Quantum mechanics introduces the concept that particles can exist in multiple states or 'superpositions' and their properties only become fixed when they are observed or interact with their environment. In the context of the exercise at hand, quantum physics provides the principles that dictate the existence of discrete energy levels that electrons can occupy in a material. It is these discrete energy levels and the rules for their occupation by electrons that are essential to calculating properties like the Fermi energy and the average energy per electron.
Density of States
The density of states (DOS) is a key concept in solid-state physics that describes the number of electron states that are available in a material at each energy level. Essentially, it tells us how the states are distributed over energy. The importance of DOS becomes evident when we examine phenomena such as electrical conductivity, where the distribution of electrons over energy states determines how easily they can move through a material.

In the given exercise, the DOS in a one-dimensional system is proportional to \(\sqrt{1 / \mathscr{E}}\), indicating that the number of states available increases as the energy decreases. This relationship is crucial when one integrates over all available states up to the Fermi energy to find out how many electrons can be accommodated and, hence, the Fermi energy itself at absolute zero temperature. Interestingly, the DOS plays a dominant role in determining various thermal, magnetic, and electrical properties of materials.
Average Energy per Electron
Speaking of electrons in a solid, one might wonder not just about the highest energy level at absolute zero temperature but also about the average energy an electron possesses. The average energy per electron gives us insight into the energy distribution of all the electrons in a system.

The calculation of the average energy per electron, particularly at 0 K, involves finding the total energy of all the occupied states and dividing it by the number of electrons. At 0 K, all electrons are in the lowest possible energy states, up to the Fermi energy. The value calculated in the exercise, which stands at 3/5 of the Fermi energy, reflects the how the electrons collectively contribute to the total energy of the system. This average energy is foundational for predicting how the system might behave under various changes, such as temperature fluctuations or perturbations in the electronic structure due to external fields or forces.

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Most popular questions from this chapter

In some temperature region graphite can be considered a two-dimensional Debye solid, but there are still \(3 N_{0}\) modes per mole. (a) Show that \(N(v) d v=\left(2 \pi A / v^{2}\right) v d v\) where \(A\) is the area of the sample. (b) Find an expression for \(v_{m}\) and \(\Theta\) for graphite. (c) Show that at low temperatures the heat capacity is proportional to \(T^{2}\).

The variation of density \(\rho\) with altitude \(y\) of the gaseous atmosphere of the earth can be written as \(\rho=\rho_{0} e^{-g\left(\rho_{0} / P_{0}\right) y}\), where \(\rho_{0}\) and \(P_{0}\) are sea level density and pressure, provided the temperature is assumed to be uniform. (a) From the ideal gas laws show that this can be put into the form \(\rho=\rho_{0} e^{-m g y / k T}\). (b) Show that this has the form of the Boltzmann distribution.

Can you devise a cycle of transitions between three states which would maintain an equilibrium in the populations of these states, with transitions that violate detailed balancing? Does it seem reasonable to extend this to a system with many states?

Is there a connection between \(V_{0}\), the depth of the potential well for conduction electrons in a metal, and electron diffraction experiments of the Davisson-Germer type? Can we determine \(V_{0}\) from such experiments?

For the Fermi distribution function (a) show that $$ \int_{0}^{\delta_{F}} n(\mathscr{E}) d \mathscr{E}=k T\left[\ln \left(1+e^{\varepsilon_{F} / k T}\right) / 2\right] $$ (b) Show that this reduces to \(\mathscr{E}_{F}\) for \(T=0\). (c) Show that $$ \int_{0}^{\infty} n(\mathscr{E}) d \mathscr{E}=\int_{0}^{\mathscr{E}_{F}} n(\mathscr{E}) d \mathscr{E}+k T(\ln 2) $$

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