Use the results of this section to estimate the contribution of conduction electrons to the heat capacity of one mole of copper at room temperature. How does this contribution compare to that of lattice vibrations, assuming that these are not frozen out? (The electronic contribution has been measured at low temperatures, and turns out to be about40% more than predicted by the free electron model used here.)

Short Answer

Expert verified

The electrons contribute less than 1% of the total heat capacity at room temperature.

Step by step solution

01

Step 1. Given information

The contribution of conduction electrons to the heat capacity of one mole of copper at room temperature is given as

cve=π2Nk2T2ϵF

where,

N=the number of atoms and is equal to Avogadro number of atoms per one mole .

k=Boltzmann constant

T=room Temperature

ϵF=the fermi energy.

02

Step 2. Calculating the value of cve

ϵF=7.05eV{The fermi energy of copper }

role="math" localid="1647886513817" Puttingthevalue6.022×1023forN,8.617×10-5eV/Kfork,300KforT, and7.05eVforϵF

CVe=π26.022×10238.617×10-5eV/K2(300K)2(7.05eV)

=9.389×1017

=9.389×1017eV/K1.6×10-19J/K

=0.15J/K

So, the contribution of conduction electrons to the heat capacity of one mole of copper at room temperature is0.15J/K.

03

Step 3.  Calculating the value of CVl which is the specific heat due to lattice vibration.

According to Debye theory of lattice vibrations, specific heat is given as

CVl=12π45TTD3Nk

TD=Debye temperature.

The above formula is applicable whenT<TD.

Duringthe higher temperature whereTTD, the specific heat is

CVI=3Nk

Puttingthevalue6.022×1023forNand8.617×10-5eV/Kfork

CV/=36.022×10231.381×10-23J/K

=25J/K

04

Step 4. Calculating the ratio of the contribution of electrons to the heat capacity of the lattice vibrations at room temperature.

So, the contribution of electrons is very small as compared to the heat capacity of the lattice vibrations at room temperature.

CVeCV1=0.15J/K25J/K

=0.006

<1

CVe<CVI

So, the electrons contribute less than1%of the total heat capacity at room temperature.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The argument given above for why CvTdoes not depend on the details of the energy levels available to the fermions, so it should also apply to the model considered in Problem 7.16: a gas of fermions trapped in such a way that the energy levels are evenly spaced and non-degenerate.

(a) Show that, in this model, the number of possible system states for a given value of q is equal to the number of distinct ways of writing q as a sum of positive integers. (For example, there are three system states for q = 3, corresponding to the sums 3, 2 + 1, and 1 + 1 + 1. Note that 2 + 1 and 1 + 2 are not counted separately.) This combinatorial function is called the number of unrestricted partitions of q, denoted p(q). For example, p(3) = 3.

(b) By enumerating the partitions explicitly, compute p(7) and p(8).

(c) Make a table of p(q) for values of q up to 100, by either looking up the values in a mathematical reference book, or using a software package that can compute them, or writing your own program to compute them. From this table, compute the entropy, temperature, and heat capacity of this system, using the same methods as in Section 3.3. Plot the heat capacity as a function of temperature, and note that it is approximately linear.

(d) Ramanujan and Hardy (two famous mathematicians) have shown that when q is large, the number of unrestricted partitions of q is given approximately by

p(q)eπ2q343q

Check the accuracy of this formula for q = 10 and for q = 100. Working in this approximation, calculate the entropy, temperature, and heat capacity of this system. Express the heat. capacity as a series in decreasing powers of kT/η, assuming that this ratio is large and keeping the two largest terms. Compare to the numerical results you obtained in part (c). Why is the heat capacity of this system independent of N, unlike that of the three dimensional box of fermions discussed in the text?

Near the cells where oxygen is used, its chemical potential is significantly lower than near the lungs. Even though there is no gaseous oxygen near these cells, it is customary to express the abundance of oxygen in terms of the partial pressure of gaseous oxygen that would be in equilibrium with the blood. Using the independent-site model just presented, with only oxygen present, calculate and plot the fraction of occupied heme sites as a function of the partial pressure of oxygen. This curve is called the Langmuir adsorption isotherm ("isotherm" because it's for a fixed temperature). Experiments show that adsorption by myosin follows the shape of this curve quite accurately.

A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At t=0the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are Natoms, the total magnetization is typically~2μeN, where µa is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure 7.30. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of h21rand therefore reduces the magnetization by ~2μe. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin-wave is proportional to the square of 1λ.. (in the limit of long wavelengths). Therefore, since=hfand p=hλ.. for any "particle," the energy of a magnon is proportional

In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy because the difference in direction between neighboring dipoles is very small.

to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write =p22pm*, wherem* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.24×1029kg, about14times the mass of an electron. Another difference between magnons and phonons is that each magnon ( or spin-wave mode) has only one possible polarization.

(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by

NmV=2π2m×kTh2320xex-1dx.

Evaluate the integral numerically.

(b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(O)-M(T))/M(O).Write your answer in the form (T/To)32, and estimate the constantT0for iron.

(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find Cv/Nk=(T/Ti)32, where Tidiffers from To only by a numerical constant. EstimateTifor iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470k.)

(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverge in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section 8.2we will consider a different two-dimensional model in which magnetization does occur.)

The speed of sound in copper is 3560m/s. Use this value to calculate its theoretical Debye temperature. Then determine the experimental Debye temperature from Figure 7.28, and compare.

In Section 6.5 I derived the useful relation F=-kTln(Z)between the Helmholtz free energy and the ordinary partition function. Use analogous argument to prove that ϕ=-kT×ln(Z^), where Z^ is the grand partition function and ϕis the grand free energy introduced in Problem 5.23.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free