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Compute the Fermi energy of potassium by making the simple approximation that each atom contributes one free electron. The density of potassium is 851 \(\mathrm{kg} / \mathrm{m}^{3}\) , and the mass of a single potassium atom is \(6.49 \times 10^{-26} \mathrm{kg}\) .

Short Answer

Expert verified
The Fermi energy of potassium is approximately 2.12 eV.

Step by step solution

01

Calculate the Number Density of Atoms

First, we need to determine how many potassium atoms there are per cubic meter, which is known as the number density \( n \). This is calculated by dividing the density of potassium \( 851 \, \text{kg/m}^3 \) by the mass of a single potassium atom \( 6.49 \times 10^{-26} \, \text{kg} \). \[ n = \frac{851}{6.49 \times 10^{-26}} \, \text{atoms/m}^3 \]
02

Compute the Fermi Energy

The Fermi energy \( E_F \) is given by:\[ E_F = \frac{\hbar^2}{2m_e} (3\pi^2 n)^{2/3} \]Where \( \hbar \) is the reduced Planck's constant \( \hbar = 1.05\times 10^{-34} \, \text{J s} \) and \( m_e \) is the electron mass \( m_e = 9.11 \times 10^{-31} \, \text{kg} \). Substitute the number density from Step 1 into this expression to compute \( E_F \).
03

Substitute Values and Solve

Substitute the calculated number density \( n \) from Step 1 and constants into the Fermi energy formula:\[ E_F = \frac{(1.05 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31}} (3\pi^2 \times \frac{851}{6.49\times10^{-26}})^{2/3} \]Calculate the expression inside the parenthesis, raise it to the power \( \frac{2}{3} \), and then compute \( E_F \).

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Key Concepts

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

Number Density
Number density refers to the quantity of a certain entity within a given volume. It's a crucial concept in physics when discussing atoms in a material. For potassium, the number density tells us how many potassium atoms exist in one cubic meter of the element. This is important for calculating properties like the Fermi energy.To calculate the number density of potassium, you need to use the formula:\[ n = \frac{\text{Density of potassium}}{\text{Mass of single potassium atom}} \]Given the density of potassium as 851 kg/m extsuperscript{3} and the mass of a single potassium atom as \(6.49 \times 10^{-26}\) kg, the number density calculation becomes:\[ n = \frac{851}{6.49 \times 10^{-26}} \]This calculation results in a large number because there are many atoms in a cubic meter. Understanding how to calculate number density allows you to comprehend how densely packed atoms are in a material.
Reduced Planck's Constant
The reduced Planck's constant, denoted by \( \hbar \), plays a vital role in quantum mechanics, especially in the calculation of the Fermi energy. It's a fundamental constant derived from the Planck's constant \( h \) and is expressed as:\[ \hbar = \frac{h}{2\pi} \]In many quantum formulas, such as the computation of the Fermi energy, \( \hbar \) simplifies the equations by including the factor of \( 2\pi \). The value of \( \hbar \) is approximately \( 1.05 \times 10^{-34} \) Js.Using \( \hbar \) in physics helps to describe the properties at atomic scales where classical calculations wouldn't apply. For example, in the Fermi energy equation:\[ E_F = \frac{\hbar^2}{2m_e} (3\pi^2 n)^{2/3} \]\( \hbar \) helps to determine how electron energies distribute in a material at absolute zero, showing its relevance.
Electron Mass
The mass of an electron is a fundamental property critical in many physics calculations, including those involving quantum mechanics. The electron mass, denoted as \( m_e \), is approximately \( 9.11 \times 10^{-31} \) kg.When calculating the Fermi energy, the electron mass is part of the denominator in the equation:\[ E_F = \frac{\hbar^2}{2m_e} (3\pi^2 n)^{2/3} \]It helps to provide the scale necessary for accurate energy-level comparisons between materials. Although extremely small, each electron contributes significantly to the total mass in atomic-level computations.Understanding the importance of electron mass allows a better grasp of energy distributions in a solid. Being aware of this constant solidifies your understanding of how electron behavior affects material properties.
Potassium Density
The density of potassium, noted as 851 kg/m extsuperscript{3}, is the mass of potassium per unit volume. It's crucial in determining the number density of potassium atoms, which then leads to the calculation of the Fermi energy.Density helps to understand how much matter exists in a certain volume of space. For metals like potassium, density can indicate how atoms are packed tight within its structure.In calculations, when you know the density, as given here:\[ \text{Density of Potassium} = 851 \text{ kg/m}^3 \]you can determine how many atoms are occupying that space by using the mass of a single atom. This conversion from macroscopic density to microscopic number density is the foundation of many material science analyses, showing how essential understanding density is in practical scenarios.

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Most popular questions from this chapter

If a sodium chloride (NaC) molecule could undergo an \(n \rightarrow n-1\) vibrational transition with no change in rotational quantum number, a photon with wavelength 20.0\(\mu \mathrm{m}\) would be emitted. The mass of a sodium atom is \(3.82 \times 10^{-26} \mathrm{kg}\) , and the mass of a chlorine atom is \(5.81 \times 10^{-26} \mathrm{kg}\) . Calculate the force constant \(k^{\prime}\) for the interatomic force in \(\mathrm{NaCl}\) .

Two atoms of cesium (Cs) can form a \(\mathrm{Cs}_{2}\) molecule. The equilibrium distance between the nuclei in a \(\mathrm{Cs}_{2}\) . molecule is 0.447 \(\mathrm{nm}\) . Calculate the moment of inertia about an axis through the center of mass of the two nuclei and perpendicular to the line joining them. The mass of a cesium atom is \(2.21 \times 10^{-25} \mathrm{kg}\) .

When a hypothetical diatomic molecule having atoms 0.8860 \(\mathrm{nm}\) apart undergoes a rotational transition from the \(I=2\) state to the next lower state, it gives up a photon having energy \(8.841 \times 10^{-4} \mathrm{eV}\) . When the molecule undergoes a vibrational transition from one energy state to the next lower energy state, it gives up 0.2560 \(\mathrm{eV}\) . Find the force constant of this molecule.

Calculate the wavelengths of (a) a \(6.20-\mathrm{keV} \times\) ray; (b) a 37.6 -eV electron; (c) a 0.0205 -eV neutron.

Metallic lithium has a bee crystal structure. Each unit cell is a cube of side length \(a=0.35 \mathrm{nm}\) . (a) For a bcc lattice, what is the number of atoms per unit volume? Give your answer in terms of \(a\) . (Hint: How many atoms are there per unit cell?) (b) Use the result of part (a) to calculate the zero-temperature Fermi energy \(E_{\mathrm{FO}}\) for metallic lithium. Assume there is one free electron per atom.

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