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An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 7: Q. 7.67 (page 323)

Problem 7.67. In the first achievement of Bose-Einstein condensation with atomic hydrogen, a gas of approximately $2 脳 10^{10}$ atoms was trapped and cooled until its peak density was $1.8 脳 10^{14} a t o m s / c m^{3}$ . Calculate the condensation temperature for this system, and compare to the measured value of $50 渭 K$ .

Short Answer

Expert verified

The condensation temperature for the atomic hydrogen system is $50.96 渭 K$

Step by step solution

01

Step 1. Given information

Condensation temperature for the substance,

$T_{c} = 0.527 (\frac{h^{2}}{2 蟺 m k}) {(\frac{N}{V})}^{\frac{2}{3}}$

${Planck's constant,h=6.63脳10}^{- 34}$

${mass of the particle,m=1.67脳10}^{- 27} kg$

$N =number of theparticles$

$V =volume of the substance.$

Density of atomic hydrogen $=$ $\frac{N}{V} = 1.8 脳 10^{14} atoms / {cm}^{3}$ $= 1.8 脳 10^{20} a t o m s / m^{3}$

02

Step 2. Substituting the values of h, N/V, m, k in the equation Tc=0.527h22πmkNV23

we get,

$T_{c} = (0.527) \frac{{(6.63 脳 10^{- 34} J 路 s)}^{2}}{(2 蟺 (1.67 脳 10^{- 27} kg)) (1.381 脳 10^{- 23} J / K)} {(1.8 脳 10^{20} atoms / m^{3})}^{\frac{2}{3}}$

$= 50.96 脳 10^{- 6} K$

$= 50.96 脳 10^{- 6} K (\frac{1 渭 K}{1 脳 10^{- 6} K})$

$= 50.96 渭 K$

The value of condensate temperature for the atomic hydrogen system is $50.96 渭 K$ ,

which is approximately equal to the measured value of Bose-Einstein condensation with atomic hydrogen $(50 渭 K)$

width="220">=50.96脳10-6K1渭K1脳10-6K

$= 50.96 渭 K$

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Most popular questions from this chapter

Fill in the steps to derive equations $7.112$ and $7.117$ .

Sketch the heat capacity of copper as a function of temperature from $0$ to $5 K$ , showing the contributions of lattice vibrations and conduction electrons separately. At what temperature are these two contributions equal?

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, $\frac{N_{c}}{V}$ .

(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 $N_{c}$ in terms of the number of donor atoms $(N_{d})$ , eliminating碌. Solve for $N_{c}$ using the quadratic formula. (Hint: It's helpful to introduce some abbreviations for dimensionless quantities. Try $x = \frac{N_{c}}{N_{d}}$ , $t = \frac{k T}{l}$ and so on.)

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

Most spin-1/2 fermions, including electrons and helium-3 atoms, have nonzero magnetic moments. A gas of such particles is therefore paramagnetic. Consider, for example, a gas of free electrons, confined inside a three-dimensional box. The z component of the magnetic moment of each electron is 卤碌a. In the presence of a magnetic field B pointing in the z direction, each "up" state acquires an additional energy of $- 渭_{B} B$ , while each "down" state acquires an additional energy of $+ 渭_{B} B$

(a) Explain why you would expect the magnetization of a degenerate electron gas to be substantially less than that of the electronic paramagnets studied in Chapters 3 and 6, for a given number of particles at a given field strength.

(b) Write down a formula for the density of states of this system in the presence of a magnetic field B, and interpret your formula graphically.

(c) The magnetization of this system is $渭_{B} (N_{鈫�} - N_{鈫�}$ , where Nr and N1 are the numbers of electrons with up and down magnetic moments, respectively. Find a formula for the magnetization of this system at $T = 0$ , in terms of N, 碌a, B, and the Fermi energy.

(d) Find the first temperature-dependent correction to your answer to part (c), in the limit $T 鈮� T_{F}$ . You may assume that $渭_{B} B 鈮� k T$ ; this implies that the presence of the magnetic field has negligible effect on the chemical potential $渭$ . (To avoid confusing 碌B with 碌, I suggest using an abbreviation such as o for the quantity 碌aB.)

For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of "background radiation," in addition to the photons and neutrinos (see Figure 7.21). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (ea.ch with the same mass), and each has two independent polarization states. During the time period of interest, the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in Figure 7.21. When the temperature was greater than the electron mass times $\frac{c^{2}}{k}$ , the universe was filled with three types of radiation: electrons and positrons (solid arrows); neutrinos (dashed); and photons (wavy). Bathed in this radiation were a few protons and neutrons, roughly one for every billion radiation particles. the previous problem. Recall from special relativity that the energy of a massive particle is $系 = \sqrt{(p c)^{2} + {(m c^{2})}^{2}}$ .

(a) Show that the energy density of electrons and positrons at temperature $T$ is given by

$u (T) = {鈭�}_{0}^{鈭�} \frac{x^{2} \sqrt{x^{2} + {(m c^{2} / k T)}^{2}}}{e^{\sqrt{x^{2} + {(m c^{2} / k T)}^{2}}} + 1} d x; w h e r e u (T) = {鈭�}_{0}^{鈭�} \frac{x^{2} \sqrt{x^{2} + {(m c^{2} / k T)}^{2}}}{e^{\sqrt{x^{2} + {(m c^{2} / k T)}^{2}}} + 1} d x$

(b) Show that $u (T)$ goes to zero when $k T 鈮� m c^{2}$ , and explain why this is a

reasonable result.

( c) Evaluate $u (T)$ in the limit $k T 鈮� m c^{2}$ , and compare to the result of the

the previous problem for the neutrino radiation.

(d) Use a computer to calculate and plot $u (T)$ at intermediate temperatures.

(e) Use the method of Problem 7.46, part (d), to show that the free energy

density of the electron-positron radiation is

$\frac{F}{V} = - \frac{16 蟺 (k T)^{4}}{(h c)^{3}} f (T); w h e r e f (T) = {鈭�}_{0}^{鈭�} x^{2} \ln (1 + e^{- \sqrt{x^{2} + {(m c^{2} / k T)}^{2}}}) d x$

Evaluate $f (T)$ in both limits, and use a computer to calculate and plot $f (T)$ at intermediate

temperatures.

(f) Write the entropy of the electron-positron radiation in terms of the functions

$u (T)$ and $f (T)$ . Evaluate the entropy explicitly in the high-T limit.

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