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

The heat capacity of liquid H4ebelow 0.6Kis proportional to T3, with the measured valueCV/Nk=(T/4.67K)3. This behavior suggests that the dominant excitations at low temperature are long-wavelength photons. The only important difference between photons in a liquid and photons in a solid is that a liquid cannot transmit transversely polarized waves-sound waves must be longitudinal. The speed of sound in liquid He4is 238m/s, and the density is 0.145g/cm3. From these numbers, calculate the photon contribution to the heat capacity ofHe4in the low-temperature limit, and compare to the measured value.

Figure 7.37 shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph.

(a) Write down an expression for the total energy of a gas of Nbosons confined to a volume V, in terms of an integral (analogous to equation 7.122).

(b) For T<Tcyou can set μ=0. Evaluate the integral numerically in this case, then differentiate the result with respect to Tto obtain the heat capacity. Compare to Figure 7.37.

(c) Explain why the heat capacity must approach 32Nkin the high- Tlimit.

(d) For T>Tcyou can evaluate the integral using the values of μcalculated in Problem 7.69. Do this to obtain the energy as a function of temperature, then numerically differentiate the result to obtain the heat capacity. Plot the heat capacity, and check that your graph agrees with Figure 7.37.

Figure 7.37. Heat capacity of an ideal Bose gas in a three-dimensional box.

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.

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?

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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