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

For a system obeying Boltzmann statistics, we know what μis from Chapter 6. Suppose, though, that you knew the distribution function (equation 7.31) but didn't know μ. You could still determine μ by requiring that the total number of particles, summed over all single-particle states, equal N. Carry out this calculation, to rederive the formula μ=-kTlnZ1/N. (This is normally how μ is determined in quantum statistics, although the math is usually more difficult.)


Fill in the steps to derive equations 7.112and7.117.

Consider a system consisting of a single impurity atom/ion in a semiconductor. Suppose that the impurity atom has one "extra" electron compared to the neighboring atoms, as would a phosphorus atom occupying a lattice site in a silicon crystal. The extra electron is then easily removed, leaving behind a positively charged ion. The ionized electron is called a conduction electron because it is free to move through the material; the impurity atom is called a donor, because it can "donate" a conduction electron. This system is analogous to the hydrogen atom considered in the previous two problems except that the ionization energy is much less, mainly due to the screening of the ionic charge by the dielectric behavior of the medium.

(a) Write down a formula for the probability of a single donor atom being ionized. Do not neglect the fact that the electron, if present, can have two independent spin states. Express your formula in terms of the temperature, the ionization energy I, and the chemical potential of the "gas" of ionized electrons.

(b) Assuming that the conduction electrons behave like an ordinary ideal gas (with two spin states per particle), write their chemical potential in terms of the number of conduction electrons per unit volume,NcV.

(c) Now assume that every conduction electron comes from an ionized donor atom. In this case the number of conduction electrons is equal to the number of donors that are ionized. Use this condition to derive a quadratic equation for Ncin terms of the number of donor atoms Nd, eliminatingµ. Solve for Ncusing the quadratic formula. (Hint: It's helpful to introduce some abbreviations for dimensionless quantities. Tryx=NcNd,t=kTland so on.)

(d) For phosphorus in silicon, the ionization energy is localid="1650039340485" 0.044eV. Suppose that there are 1017patoms per cubic centimeter. Using these numbers, calculate and plot the fraction of ionized donors as a function of temperature. Discuss the results.

Carry out the Sommerfeld expansion for the energy integral (7.54), to obtain equation 7.67. Then plug in the expansion for μto obtain the final answer, equation 7.68.

In this problem you will model helium-3 as a non-interacting Fermi gas. Although He3liquefies at low temperatures, the liquid has an unusually low density and behaves in many ways like a gas because the forces between the atoms are so weak. Helium-3 atoms are spin-1/2 fermions, because of the unpaired neutron in the nucleus.

(a) Pretending that liquid 3He is a non-interacting Fermi gas, calculate the Fermi energy and the Fermi temperature. The molar volume (at low pressures) is 37cm3

(b)Calculate the heat capacity for T<<Tf, and compare to the experimental result CV=(2.8K-1)NkT(in the low-temperature limit). (Don't expect perfect agreement.)

(c)The entropy of solid H3ebelow 1 K is almost entirely due to its multiplicity of nuclear spin alignments. Sketch a graph S vs. T for liquid and solid H3eat low temperature, and estimate the temperature at which the liquid and solid have the same entropy. Discuss the shape of the solid-liquid phase boundary shown in Figure 5.13.

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