Explain in some detail why the three graphs in Figure 7.28 all intercept the vertical axis in about the same place, whereas their slopes differ considerably.

Short Answer

Expert verified

The conduction electrons in copper, silver, and gold is one. The fermi energy for these elements is approximately equal. Hence, the intercept for copper, silver, and gold is equal on vertical axis in the figure 7.28.

Basically , the slopes of the graphs depend on the speed of sound CSin each material of the three metals. The metal copper has the largest value Cs, and hence largest Debye temperature, and smallest slope. On the other hand, Gold has the smallest value of CSand hence the largest slope. Silver is somewhere in between copper and the gold metal slope.

So,the slope of the graph between CTand T2in the figure 7.28for copper, silver, and gold is not same.

Step by step solution

01

Step 1. Given information

The total heat capacity at low temperature is equal to the sum of the electronic heat capacity lattice vibrational heat capacity.

C=γT+αT3

02

Step 2. Putting the value of γ and αin above equation we get 

Here,γ=π2NkB22εF,α=12Nπ4kB5TD3andTis the temperature.

Firstly, rearranging the equationC=γT+αT3forCT.

CT=γ+αT2

in the above equation αis the slope onCTversusT2plot andγis the intercept.

03

Step 3. finding the intercept value for copper, silver, gold.

The intercept of the graph betweenCTandT2in the figure7.28for copper, silver, and gold is

role="math" localid="1647622591195" γ=π2NkB22εF

Here, Nis the number of the conduction electrons per mole of the metal, kBis the Boltzmann's constant, and εFis the fermi energy.

As the intercept in the graph betweenCTand T2is directly proportional to Nand indirectly proportional to the fermi energy.

So, the conduction electrons in copper, silver, and gold is one. The fermi energy for these elements is approximately equal . Hence, the intercept for copper, silver, and gold is equal on vertical axis in the figure 7.28.

04

Step 4. finding the slope relation for copper, silver, gold.

The slope of the graph betweenCTandT2in the figure7.28for copper, silver, and gold is as

α=12Nπ4kB5TD3

Here,TDis the Debye temperature.

So, the slope is indirectly proportional to the cube root of the Debye temperature.

TD=hcs2kB6NπV1/3

Here,his the Planck's constant, csis the speed of the sound in the liquid, Nis the Avogadro number, Vis the volume, and kis the Boltzmann's constant.

Hence, we can say that the slopes of the graphs depend on the speed of soundCsin each material of the three metals. The metal copper has the largest value CS, and so largest Debye temperature, and smallest slope.

Similarly, Gold has the smallest value of CSand hence the largest slope. Silver is somewhere in between copper and the gold metal slope.

So, basically the slope of the graph between CTand T2in the figure 7.28for copper, silver, and gold is not same.

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Most popular questions from this chapter

In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and antineutrinos (v-), currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarisation state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless

(a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of antineutrinos, so that their chemical potentials are equal: μν=μν¯. Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction

ν+ν¯2γ

(where y is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that u =0 for both the neutrinos and the antineutrinos.

(b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. differences between this "neutrino gas" and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a table or consult Appendix B. (Hint: There are very few

(c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K.

d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn't have affected the production of neutrinos in the early universe, when me would have been negligible compared to typical thermal energies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass m. What would mc2 have to be (in eV), in order for the total mass of neutrinos in the universe to be comparable to the total mass of ordinary matter?

A black hole is a blackbody if ever there was one, so it should emit blackbody radiation, called Hawking radiation. A black hole of mass M has a total energy of Mc2, a surface area of 16πG2M2/c4, and a temperature ofhc3/16π2kGM(as shown in Problem 3.7).

(a) Estimate the typical wavelength of the Hawking radiation emitted by a one-solar-mass (2 x 1030 kg) black hole. Compare your answer to the size of the black hole.

(b) Calculate the total power radiated by a one-solar-mass black hole.

(c) Imagine a black hole in empty space, where it emits radiation but absorbs nothing. As it loses energy, its mass must decrease; one could say it "evaporates." Derive a differential equation for the mass as a function of time, and solve this equation to obtain an expression for the lifetime of a black hole in terms of its initial mass.

(d) Calculate the lifetime of a one-solar-mass black hole, and compare to the estimated age of the known universe (1010 years).

(e) Suppose that a black hole that was created early in the history of the universe finishes evaporating today. What was its initial mass? In what part of the electromagnetic spectrum would most of its radiation have been emitted?

Consider any two internal states, s1 and s2, of an atom. Let s2 be the higher-energy state, so that Es2-Es1=ϵ for some positive constant. If the atom is currently in state s2, then there is a certain probability per unit time for it to spontaneously decay down to state s1, emitting a photon with energy e. This probability per unit time is called the Einstein A coefficient:

A = probability of spontaneous decay per unit time.

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B=probability of absorption per unit timeu(f)

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B'=probability of stimulated emission per unit timeu(f)

As Einstein showed in 1917, knowing any one of these three coefficients is as good as knowing them all.

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B'=BandAB=8πhf3c3

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