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.

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Most popular questions from this chapter

Consider a free Fermi gas in two dimensions, confined to a square area A=L2

(a) Find the Fermi energy (in terms of Nand A), and show that the average energy of the particles is F2.

(b) Derive a formula for the density of states. You should find that it is a constant, independent of .

(c) Explain how the chemical potential of this system should behave as a function of temperature, both when role="math" localid="1650186338941" kTFand when Tis much higher.

(d) Because gis a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for μas a function of N. Show that the resulting formula has the expected qualitative behavior.

(e) Show that in the high-temperature limit, kTF, the chemical potential of this system is the same as that of an ordinary ideal gas.

Starting from the formula for CV derived in Problem 7.70(b), calculate the entropy, Helmholtz free energy, and pressure of a Bose gas for T<Tc. Notice that the pressure is independent of volume; how can this be the case?

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.

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(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.)

Although the integrals (7.53and 7.54) forNand Ucannot be

carried out analytically for all T, it's not difficult to evaluate them numerically

using a computer. This calculation has little relevance for electrons in metals (for

which the limit kT<<EFis always sufficient), but it is needed for liquid H3eand

for astrophysical systems like the electrons at the center of the sun.

(a) As a warm-up exercise, evaluate theNintegral (7.53) for the casekT=εF

and μ=0, and check that your answer is consistent with the graph shown

above. (Hint: As always when solving a problem on a computer, it's best to

first put everything in terms of dimensionless variables. So let t=kTεFrole="math" localid="1649996205331" ,c=μεF

, and x=εkT. Rewrite everything in terms of these variables,

and then put it on the computer.)

(b) The next step is to varyμ holdingT fixed, until the integral works out to

the desired value,N. Do this for values of kTεFranging from 0.1 up to 2,

and plot the results to reproduce Figure7.16. (It's probably not a good idea

to try to use numerical methods when kTεF is much smaller than 0.1, since

you can start getting overflow errors from exponentiating large numbers.

But this is the region where we've already solved the problem analytically.)

(c) Plug your calculated values ofµ into the energy integral (7.54), and evaluate

that integral numerically to obtain the energy as a function of temperature

forkTup to 2εF Plot the results, and evaluate the slope to obtain the

heat capacity. Check that the heat capacity has the expected behavior at

both low and high temperatures.

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