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The Fermi energy of aluminum is \(11.6 \mathrm{eV} ;\) its density and molar mass are \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\) and \(27.0 \mathrm{~g} / \mathrm{mol},\) respectively. From these data, determine the number of conduction electrons per atom.

Short Answer

Expert verified
There are 3 conduction electrons per aluminum atom.

Step by step solution

01

Calculate the number density of aluminum atoms

Given the density of aluminum \( \rho = 2.70 \mathrm{~g} / \mathrm{cm}^{3} \), and the molar mass \( M = 27.0 \mathrm{~g}/\mathrm{mol} \), we use Avogadro's number \( N_A = 6.022 \times 10^{23} \mathrm{atoms/mol} \) to find the number density \( n \) as follows:\[ n = \frac{\rho \times N_A}{M} = \frac{2.70 \times 6.022 \times 10^{23}}{27.0} = 6.022 \times 10^{22} \mathrm{~atoms/cm}^3. \]
02

Estimate the electron number density

The Fermi energy \( E_F \) for a free electron gas is related to the electron number density \( n_e \) by the formula:\[ E_F = \frac{\hbar^2}{2m_e}(3\pi^2 n_e)^{2/3}, \]where \( \hbar = 1.0545718 \times 10^{-34} \mathrm{~J}\cdot\mathrm{s} \) and \( m_e = 9.10938356 \times 10^{-31} \mathrm{~kg} \). Given that \( E_F = 11.6 \mathrm{eV} = 11.6 \times 1.602 \times 10^{-19} \mathrm{J} \), solve for \( n_e \):\[ n_e = \left(\frac{2m_e E_F}{\hbar^2}\right)^{3/2} \frac{1}{3\pi^2}. \]
03

Convert results to consistent units and solve

First, convert \( E_F \) to Joules:\[ E_F = 11.6 \times 1.602 \times 10^{-19} \mathrm{~J}. \]Insert known values into the expression for \( n_e \):\[ n_e = \left(\frac{2 \times 9.10938356 \times 10^{-31} \times 11.6 \times 1.602 \times 10^{-19}}{(1.0545718 \times 10^{-34})^2}\right)^{3/2} \frac{1}{3 \pi^2}. \]Calculate \( n_e \), which results in an approximate value of \( 18.1 \times 10^{28} \mathrm{~electrons/m}^3. \)
04

Determine the number of conduction electrons per atom

Using the number densities \( n \) (from Step 1) and \( n_e \) (from Step 3), calculate the number of electrons per atom as:\[ \text{Electrons per atom} = \frac{n_e}{n}, \]where \( n_e = 18.1 \times 10^{28} \mathrm{~electrons/m}^3 \) and \( n = 6.022 \times 10^{28} \mathrm{~atoms/m}^3 \) (converted to \( \mathrm{m}^3 \) for consistency).Calculate this ratio to get approximately 3 electrons per atom.

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Key Concepts

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

Conduction Electrons
Conduction electrons are a crucial part of understanding electrical conductivity in metals like aluminum. These are the electrons that are free to move through the metal lattice and carry electric current when a potential difference is applied. Essentially, they are not bound to any specific atom but can move freely throughout the material.
In metals, such as aluminum, conduction electrons are responsible for the metal's conductive properties. Their freedom to move results from the loosely bound nature of valence electrons in metallic bonds, providing pathways for easy electron flow.
Understanding how these electrons function is key to calculating values like the number of conduction electrons per atom. This knowledge helps in predicting and explaining the behavior of metals in different conditions, such as during electrical conductivity or thermal transport.
Electron Number Density
Electron number density refers to the number of conduction electrons per unit volume of a material. This density is a fundamental concept in solid-state physics, particularly when exploring the electronic properties of materials like aluminum.
To calculate the electron number density, you need to consider various factors, including the material's Fermi energy, which indicates the highest energy level occupied by electrons at absolute zero temperature. Using quantum mechanical principles and specific equations like the one given for a free electron gas, you can determine this density based on Fermi energy values and other constants like Planck's constant and the electron mass.
This concept is critical because it helps characterize how densely packed the free electrons are in a material. High electron number density can imply greater electrical and thermal conductivity, making it essential for applications in electronics and engineering.
Density of Aluminum
The density of aluminum is a measure of how much mass is contained in a given volume of the material. Aluminum, with a density of approximately 2.70 g/cm³, is considered a lightweight metal, making it ideal for applications where reduced weight is crucial, such as in aerospace and automotive industries.
Density is calculated based on the material's mass and volume, and in this example, it's expressed in grams per cubic centimeter (g/cm³). This value is intrinsic to aluminum and doesn't change without external factors like temperature or pressure. Knowing the density allows us to understand how tightly the aluminum atoms are packed within their crystalline structure, influencing many of its mechanical properties such as strength and malleability.
In calculations involving electron characteristics in aluminum, having an accurate density is fundamental. It helps convert between macroscopic properties (like mass) and microscopic behaviors (like electron density).
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For aluminum, the molar mass is about 27.0 g/mol, which means that one mole of aluminum atoms weighs 27 grams.
This value is critical for a range of applications, from simple chemical reactions to complex materials science calculations. It helps in determining the number density of a material, which is the number of atoms, ions, or molecules in a given volume.
Knowing the molar mass is essential for converting between different units of measure – such as converting from grams to moles, which is often necessary in chemical equations and reactions, or when deducing electron properties in materials. In the context of conduction electron calculations, molar mass combined with density allows for the determination of how many aluminum atoms—and therefore electrons—are present in a certain volume.

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Most popular questions from this chapter

Pure silicon at room temperature has an electron number density in the conduction band of about \(5 \times 10^{15} \mathrm{~m}^{-3}\) and an equal density of holes in the valence band. Suppose that one of every \(10^{7}\) silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, \(n\) or \(p ?\) (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and holes in the valence band) in the doped silicon to that in pure silicon?

A sample of a certain metal has a volume of \(4.0 \times 10^{-5} \mathrm{~m}^{3}\). The metal has a density of \(9.0 \mathrm{~g} / \mathrm{cm}^{3}\) and a molar mass of \(60 \mathrm{~g} / \mathrm{mol} .\) The atoms are bivalent. How many conduction electrons (or valence electrons) are in the sample?

Doping changes the Fermi energy of a semiconductor. Consider silicon, with a gap of \(1.11 \mathrm{eV}\) between the top of the valence band and the bottom of the conduction band. At \(300 \mathrm{~K}\) the Fermi level of the pure material is nearly at the midpoint of the gap. Suppose that silicon is doped with donor atoms, each of which has a state \(0.15 \mathrm{eV}\) below the bottom of the silicon conduction band, and suppose further that doping raises the Fermi level to \(0.11 \mathrm{eV}\) below the bottom of that band (Fig. \(41-22\) ). For (a) pure and (b) doped silicon, calculate the probability that a state at the bottom of the silicon conduction band is occupied. (c) Calculate the probability that a state in the doped material (at the donor level) is occupied.

The Fermi energy for silver is \(5.5 \mathrm{eV}\). At \(T=0^{\circ} \mathrm{C},\) what are the probabilities that states with the following energies are occupied: (a) \(4.4 \mathrm{eV},(\mathrm{b}) 5.4 \mathrm{eV},(\mathrm{c}) 5.5 \mathrm{eV},(\mathrm{d}) 5.6 \mathrm{eV},\) and (e) \(6.4 \mathrm{eV} ?\) (f) At what temperature is the probability 0.16 that a state with energy \(E=5.6 \mathrm{eV}\) is occupied?

At \(1000 \mathrm{~K},\) the fraction of the conduction electrons in a metal that have energies greater than the Fermi energy is equal to the area under the curve of Fig. \(41-8 b\) beyond \(E_{\mathrm{F}}\) divided by the area under the entire curve. It is difficult to find these areas by direct integration. However, an approximation to this fraction at any temperature \(T\) is frac\(=\frac{3 k T}{2 E_{\mathrm{F}}}\) Note that \(f r a c=0\) for \(T=0 \mathrm{~K},\) just as we would expect. What is this fraction for copper at (a) \(300 \mathrm{~K}\) and (b) \(1000 \mathrm{~K}\) ? For copper, \(E_{\mathrm{F}}=7.0 \mathrm{eV} .\) (c) Check your answers by numerical integration using Eq. \(41-7\)

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