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An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 7: Q. 7.58 (page 313)

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

Short Answer

Expert verified

Hence, the Debye Temperature from experimental Data (graph) $T_{D} = 359.9176 K$ .

Step by step solution

01

Step 1. Given information

We have, $the speed of sound in copper is 3560 m / s$

$The Debye temperature is T_{D} = \frac{h C_{s}}{2 k_{B}} 路 {(\frac{6 N}{蟺 V})}^{\frac{1}{3}}$ .

02

Step 2. Putting the value of h , V , N ,CS ,kB.

$V = 7.11 {cm}^{3} / mol$

$C_{s} = 3560 m / s = 3560 脳 100 cm / s$

$N = 6.022 脳 10^{23} atoms / mol$

$k_{B} = 1.381 脳 10^{- 23} J / K$

$h = 6.626 脳 10^{- 34} J 路 s$

substituting all the above value in Debye Temperature we get

$T_{D} = \frac{6.626 脳 10^{- 34} J 路 s 脳 3560 脳 100 cm / s}{(2 脳 1.381 脳 10^{- 23} J / K)} {[\frac{6 脳 6.022 脳 10^{23} / mol}{蟺脳 7.11 {cm}^{3} / mol}]}^{\frac{1}{3}}$

$鈭� T_{D} = 465.3381 K$

03

Step 3. Calculating the experimental Debye Temperature.

Take any two points on experimental data calculate the slope [Approximately] Approximately take two points

Approximately take two points

$A (0, 0.75) B (18, 1.5)$

$\frac{螖 (\frac{C_{V}}{T})}{螖 (T^{2})} = \frac{1.5 - 0.75}{18 - 0}$

$= 0.0417$

04

Step 4. Finding the slope of the graph.

$Slope: \frac{螖 (\frac{C_{V}}{T})}{螖 (T^{2})} = \frac{12 蟺^{4} N k_{B}}{5 T_{D}^{3}}$

$\frac{12 蟺^{4} N k_{B}}{5 T_{D}^{3}} = 0.0417 mJ / K^{4}$

then , the Debye Temperature

$T_{D} = {[\frac{12 蟺^{4} N k_{B}}{5 脳 0.0417 脳 10^{- 3} J / K^{4}}]}^{\frac{1}{3}}$

$= {(\frac{12 蟺^{4} 脳 6.022 脳 10^{23} 脳 1.381 脳 10^{- 23}}{5 脳 0.0417 脳 10^{- 3}})}^{\frac{1}{3}}$

$= 359.9176 K$

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Most popular questions from this chapter

Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic $(系鈮� m c^{2})$ so that their energies are $系 = p c$ (where p is the magnitude of the momentum vector).

(a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by =

$渭 = h c (3 N / 8 蟺 V)^{1 / 3}$

(b) Find a formula for the total energy of this system in terms of N and $渭$ .

Consider a free Fermi gas in two dimensions, confined to a square area $A = L^{2}$ 鈥�

(a) Find the Fermi energy (in terms of $N$ and $A$ ), and show that the average energy of the particles is $\frac{{鈭�}_{F}}{2}$ .

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

(c) Explain how the chemical potential of this system should behave as a function of temperature, both when role="math" localid="1650186338941" $k T 鈮� {鈭�}_{F}$ and when $T$ is much higher.

(d) Because $g (鈭�)$ is 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 $渭$ as a function of $N$ . Show that the resulting formula has the expected qualitative behavior.

(e) Show that in the high-temperature limit, $k T 鈮� {鈭�}_{F}$ , the chemical potential of this system is the same as that of an ordinary ideal gas.

Change variables in equation 7.83 to $位 = h c / 系$ and thus derive a formula for the photon spectrum as a function of wavelength. Plot this spectrum, and find a numerical formula for the wavelength where the spectrum peaks, in terms of hc/kT. Explain why the peak does not occur at hc/(2.82kT).

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 ( $\bar{v}$ ), 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: $渭_{谓} = 渭_{\overset{炉}{谓}}$ . Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction

$谓 + \overset{炉}{谓} 鈫� 2 纬$

(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?

Consider a gas of $n$ identical spin-0 bosons confined by an isotropic three-dimensional harmonic oscillator potential. (In the rubidium experiment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are $蔚 = n h f$ , where $n$ is any nonnegative integer and $f$ is the classical oscillation frequency. The degeneracy of level $n is (n + 1) (n + 2) / 2$ .

(a) Find a formula for the density of states, $g (蔚)$ , for an atom confined by this potential. (You may assume $n > > 1$ .)

(b) Find a formula for the condensation temperature of this system, in terms of the oscillation frequency $f$ .

(c) This potential effectively confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by setting the particle's total energy (of order $k T$ ) equal to the potential energy of the "spring." Making these associations, and neglecting all factors of 2 and $蟺$ and so on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.

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