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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 2×1010atoms was trapped and cooled until its peak density was1.8×1014atoms/cm3. Calculate the condensation temperature for this system, and compare to the measured value of50μK.

Short Answer

Expert verified

The condensation temperature for the atomic hydrogen system is50.96μK

Step by step solution

01

Step 1. Given information

Condensation temperature for the substance,

Tc=0.527h22πmkNV23

Planck's constant,h=6.63×10-34

mass of the particle,m=1.67×10-27kg

N=number of theparticles

V=volume of the substance.

Density of atomic hydrogen=NV=1.8×1014atoms/cm3=1.8×1020atoms/m3

02

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

we get,

Tc=(0.527)6.63×10-34J·s22π1.67×10-27kg1.381×10-23J/K1.8×1020atoms/m323

=50.96×10-6K

=50.96×10-6K1μK1×10-6K

=50.96μK

The value of condensate temperature for the atomic hydrogen system is 50.96μK,

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

width="220">=50.96×10-6K1μK1×10-6K

=50.96μK

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Most popular questions from this chapter

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.

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.

Consider a free Fermi gas in two dimensions, confined to a square area A=L2•

(a) Find the Fermi energy (in terms of Nand A), and show that the average energy of the particles is ∈F2.

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

(c) Explain how the chemical potential of this system should behave as a function of temperature, both when role="math" localid="1650186338941" kT≪∈Fand when Tis much higher.

(d) Because g∈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 μas a function of N. Show that the resulting formula has the expected qualitative behavior.

(e) Show that in the high-temperature limit, kT≫∈F, the chemical potential of this system is the same as that of an ordinary ideal gas.

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they "heated" the photon radiation but not the neutrino radiation.

(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have ben conserved in the early universe, assuming that no other species of particles interacted with these.

(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by

TTν32π445+u(T)+f(T)=constant

as the universe expands and cools. Evaluate the constant by assuming that T=Tv when the temperatures are very high.

(c) Calculate the ratio T/Tv, in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K.

(d) Use a computer to plot the ratio T/Tv, as a function of T, for kT/mc2ranging from 0 to 3.*

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is

(a) 1eVless than μ

(b) 0.01eVless than μ

(c) equal to μ

(d) 0.01eVgreater than μ

(e) 1eVgreater thanμ
See all solutions

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