/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Metallic lithium has a bce cryst... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
(a) Number of atoms per unit volume is \( \frac{2}{a^3} \). (b) Fermi energy \( E_F \approx 7.49 \text{ eV} \).

Step by step solution

01

Find Atoms per Unit Cell in BCC Lattice

In a body-centered cubic (bcc) lattice, there are two atoms per unit cell: one atom at each corner of the cube (each shared by eight neighboring cells) and one atom at the center of the cube. This means that there are \( 1 \) corner atom and \( 1 \) center atom in each bcc unit cell.
02

Calculate Atoms per Unit Volume

Each corner atom is shared by \( 8 \) unit cells, so its contribution to a single unit cell is \( \frac{1}{8} \). With 8 corners, the total is \( 8 \times \frac{1}{8} = 1 \) atom from corners and \( 1 \) atom from the center, giving us \( 2 \) atoms per unit cell. The volume of the unit cell is \( a^3 \), hence the number of atoms per unit volume is \( \frac{2}{a^3} \).
03

Understand the Connection Between Fermi Energy and Atom Density

Fermi energy is related to the density of electrons, which equals the density of atoms when there is one free electron per atom in a metal. This allows us to use the result from Step 2 in computing the Fermi energy.
04

Calculate Fermi Energy

The Fermi energy for a free electron gas is given by:\[E_F = \frac{\hbar^2}{2m}(3\pi^2n)^{2/3}\]where \( \hbar \) is the reduced Planck's constant, \( m \) is the electron mass, and \( n \) is the number of electrons per unit volume which equals the atomic density from Step 2, \( n = \frac{2}{a^3} \). Substitute \( n \) into the equation to find the Fermi energy:\[ E_F = \frac{\hbar^2}{2m} \left( 3\pi^2 \left( \frac{2}{a^3} \right) \right)^{2/3} \]
05

Substitute Values

Given \( a = 0.35 \, \text{nm} = 0.35 \times 10^{-9} \, \text{m} \), substitute into the Fermi energy equation:\[ E_F = \frac{\hbar^2}{2m} \left( 3\pi^2 \left( \frac{2}{(0.35 \times 10^{-9})^3} \right) \right)^{2/3} \]
06

Simplify and Solve for \(E_F\)

Calculate numerically to find \( E_F \).Using known constants: \( \hbar = 1.0545718 \times 10^{-34} \text{ J s} \) and \( m = 9.10938356 \times 10^{-31} \text{ kg} \), compute:\[ E_F \approx 7.49 \text{ eV} \]

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Key Concepts

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

bcc lattice
In the world of crystal structures, the body-centered cubic (bcc) lattice is a fascinating topic. Imagine a cube, like a dice, but made up of atoms. A bcc lattice is different from other cubic structures because it has an extra atom right in the center of the cube, in addition to the atoms at the corners. Each corner atom is shared with 8 other cubes, which means that, practically, only 1/8 of each corner atom 'belongs' to one unit cell. When you add it all up across all 8 corners, you have one full atom from corners. Add in the center atom, and that gives you a total of 2 atoms in every bcc unit cell. This unique arrangement has significant implications for the material's properties. Since there are 2 atoms in a unit cell and the unit cell volume is given by the cube of its side length, or \(a^3\), the atomic density or the number of atoms per unit volume can be calculated as \(\frac{2}{a^3}\). This sets the stage for deeper explorations into electron behavior, such as calculating the Fermi energy.
electron density
Electron density is a crucial concept when exploring the properties of solids, particularly metals like lithium with their bcc lattice structure. In metals, electrons can move freely, and electron density gives insight into that availability of free electrons. In a bcc lattice, the electron density is directly tied to the structure's atomic density, especially since we assume one free electron per atom. That means the number of electrons in the unit volume is equal to the number of atoms per unit volume, \( n = \frac{2}{a^3} \). This electron density is then used to calculate important properties like the material's Fermi energy. Understanding electron density helps us grasp how conductive a metal might be, and how its electrons will behave at different energy levels.
crystal structure
Crystal structure gives a material its unique set of properties by dictating the arrangement of atoms within the material. In the context of the bcc lattice, lithium showcases a specific type of crystal structure, where its properties are greatly influenced by how its atoms are arranged.
  • At the most fundamental level, a crystal structure defines mechanical properties like strength and ductility.
  • It also affects electrical properties, making the study of crystal structures crucial for understanding how materials will perform in real-world applications.
  • For lithium, which is used in lightweight batteries, its crystal structure is a key factor behind its widespread use.
When we talk about Fermi energy, it's this exact atomic arrangement within the crystal structure that determines the density of states for the electrons, influencing how electrons are distributed across energy levels at absolute zero temperature.

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Most popular questions from this chapter

Light of wavelength 3.10 \(\mathrm{mm}\) strikes and is absorbed by a molecule. Is this process most likely to alter the rotational, vibrational, or atomic energy levels of the molecule? Explain your reasoning. (b) If the light in part (a) had a wavelength of 207 \(\mathrm{nm}\) , which energy levels would it most likely affect? Explain.

If the energy of the \(\mathrm{H}_{2}\) covalent bond is \(-4.48 \mathrm{eV},\) what wavelength of light is needed to break that molecule apart? In what part of the electromagnetic spectrum does this light lie?

When an OH molecule undergoes a transition from the \(n=0\) to the \(n=1\) vibrational level, its internal vibrational energy increases by 0.463 eV. Calculate the frequency of vibration and the force constant for the interatomic force. (The mass of an oxygen atom is \(2.66 \times 10^{-26} \mathrm{kg},\) and the mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg} .\) )

Hydrogen is found in two naturally occurring isotopes; normal hydrogen (containing a single proton in its nucleus) and deuterium (having a proton and a neurron). Assuming that both molecules are the same size and that the proton and neutron have the same mass (which is almost the case), find the ratio of (a) the energy of any given rotational state in a diatomic hydrogen molecule to the energy of the same state in a diatomic deuterium molecule and (b) the energy of any given vibrational state in hydrogen to the same state in deuterium (assuming that the force constant is the same for both molecules) Why is it physically reasonable that the force constant would be the same for hydrogen and deuterium molecules?

cp calt (a) Consider the hydrogen molecule $$\left(\mathrm{H}_{2}\right)$$ to be a simple harmonic oscillator with an equilibrium spacing of \(0.074 \mathrm{nm},\) and estimate the vibrational energy-level spacing for H. The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) . Hint: Estimate the force constant by equating the change in Coulomb repulsion of the protons, when the atoms move slightly closer together than \(r_{0}\) to the "spring" force. That is, assume that the chemical binding force remaing" approximately constant as \(r\) is decreased slightly from \(r_{0}\) ) ( b) Use the results of part (a) to calculate the vibrational energy-level spacing for the deuterium molecule, \(\mathrm{D}_{2}\) . Assume that the spring constant is the same for \(D_{2}\) as for \(\mathrm{H}_{2}\) . The mass of a deuterium atom is \(3.34 \times 10^{-27} \mathrm{kg}\)

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