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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

Show that when a system is in thermal and diffusive equilibrium with a reservoir, the average number of particles in the system is

N—=kTZ∂Z∂μ

where the partial derivative is taken at fixed temperature and volume. Show also that the mean square number of particles is

N2¯=(kT)2Z∂2Z∂μ2

Use these results to show that the standard deviation of Nis

σN=kT∂N—/∂μ,

in analogy with Problem6.18Finally, apply this formula to an ideal gas, to obtain a simple expression forσNin terms ofN¯Discuss your result briefly.

Repeat the previous problem, taking into account the two independent spin states of the electron. Now the system has two "occupied" states, one with the electron in each spin configuration. However, the chemical potential of the electron gas is also slightly different. Show that the ratio of probabilities is the same as before: The spin degeneracy cancels out of the Saha equation.

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.

The Sommerfeld expansion is an expansion in powers of kTεF, which is assumed to be small. In this section I kept all terms through order kTεF2, omitting higher-order terms. Show at each relevant step that the term proportional to localid="1650117451748" T3is zero, so that the next nonvanishing terms in the expansions forlocalid="1650117470867" μand localid="1650117476821" Uare proportional to localid="1650117458596" T4. (If you enjoy such things, you might try evaluating the localid="1650117464980" T4terms, possibly with the aid of a computer algebra program.)

Problem 7.69. If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate μforT>Tc.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define t=T/Tc,c=μ/kTc,andx=ϵ/kTc. Express the integral that defines , equation 7.22, in terms of these variables. You should obtain the equation

2.315=∫0∞xdxe(x-c)/t-1

(b) According to Figure

the correct value of cwhen T=2Tcis approximately -0.8. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary μ, holding Tfixed, to find the precise value of μfor T=2Tc. Repeat for values of T/Tcranging from 1.2up to 3.0, in increments of 0.2. Plot a graph of μas a function of temperature.
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