Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states.

(a) Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly in Figure 6.2. (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies according!y.)

(b) Show that if all bound states are included in the sum, then the partition function of a hydrogen atom is infinite, at any nonzero temperature. (See Appendix A for the full energy level structure of a hydrogen atom.)

(c) When a hydrogen atom is in energy level n, the approximate radius of the electron wavefunction is a0n2, where ao is the Bohr radius, about 5 x 10-11 m. Going back to equation 6.3, argue that the PdV term is Tot negligible for the very high-n states, and therefore that the result of part (a), not that of part (b), gives the physically relevant partition function for this problem. Discuss.

Short Answer

Expert verified

Therefore, the partition function for hydrogen atom isZ=1.000000013

Step by step solution

01

Given information

In the numerical example in the text, only the ratio of the probabilities of a hydrogen atom being in two different states is calculated. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states.

02

Explanation

(a)The partition function is equal to the total of all Boltzmann factors.

Z=se-E(s)/kT

Consider a hydrogen atom; in figure 6.2, the ground state energy is zero, therefore the energies must be shifted by 15.6; the three energies are thus ϵ1=0eV,ϵ2=-3.4+13.6=10.2eVandϵ3=-1.5+13.6=12.1eV; the partition function is:

Z=e-ϵ1/kT+4e-ϵ2/kT+9e-ϵ3/kT

There are 9 degenerates in role="math" localid="1647326001026" ϵ3and four degenerates in ϵ2. Replace the energies and Boltzmann constant in eV (k = 8.617 x 10 eV/K) at T = 5800 K, as follows:

role="math" localid="1647326952362" Z=e0+4e-12.1eV/0.50eV+9e-10.2eV/0.50eVZ=1.000000013

(b) Because the nthenergy level has n2states, the partition function is as follows:

Z=nn2e-E(s)/kT

E=13.6eVis the maximum energy, and since the exponential is negative, we can deduce that:

Z>nn2e-E/kTZ>e-E/kTnn2Z>e-E/kT()=Z>

03

Explanation

(c) If we keep the PdV term in equation 6.3, the Boltzmann factor will be:

Boltzmann Factor=e-(E+PV)/kT

Where,

Pis the pressure of reservoir

V is the volume

Consider the volume of a hydrogen atom in its ground state, V=1.0×10-10m3=1.0×10-30m3at atmospheric pressureP=1.0×105Pa, the PV term is:

PV=1.0×10-30m31.0×105Pa=1.0×10-25J

using1.0eV=1.6×10-19J, we get:

PV=6.25×10-7eV

Now, because the radius is proportional to n, the volume will be 1003 greater than the volume of one atom for n = 10, hence the PV term is:

PV=1.0×1061.0×10-30m31.0×105Pa=1.0×10-19JPV=0.625eV

At T = 5800 K, kT = 0.5 eV, the term PV reduces the Boltzmann factor by e-PV/kT=e-0.625eV/0.50eV=0.2865, which is a significant factor; as n increases, this component grows exponentially, forcing the Boltzmann factor to approach l.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show explicitly from the results of this section thatG=Nμfor an ideal gas.

A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the "flexing' mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of 4.8 x 1013Hz. As for any quantum harmonic oscillator, the energy levels are 12hf,32hf,52hf, and so on. None of these levels are degenerate.

(a) state and in each of the first two excited states, assuming that it is in equilibrium with a reservoir (say the atmosphere) at 300 K. (Hint: Calculate 2 by adding up the first few Boltzmann factors, until the rest are negligible.) Calculate the probability of a water molecule being in its flexing ground

(b) Repeat the calculation for a water molecule in equilibrium with a reservoir at 700 K (perhaps in a steam turbine).

Cold interstellar molecular clouds often contain the molecule cyanogen (CN), whose first rotational excited states have an energy of 4.7x 10-4 eV (above the ground state). There are actually three such excited states, all with the same energy. In 1941, studies of the absorption spectrum of starlight that passes | through these molecular clouds showed that for every ten CN molecules that are in the ground state, approximately three others are in the three first excited states (that is, an average of one in each of these states). To account for this data, astronomers suggested that the molecules might be in thermal equilibrium with some "reservoir" with a well-defined temperature. What is that temperature?

Equations 6.92 and 6.93 for the entropy and chemical potential involve the logarithm of the quantity VZintNvQ. Is this logarithm normally positive or negative? Plug in some numbers for an ordinary gas and discuss.

In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H2 molecule, for example, are more accurately described by the approximate formula

Enϵ1.03n-0.03n2,n=0,1,2,

where ϵ is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for slightly higher n it would say that En decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n 15.) Use a computer to calculate the partition function, average energy, and heat capacity of a system with this set of energy levels. Include all levels through n = 15, but check to see how the results change when you include fewer levels Plot the heat capacity as a function of kT/ϵ. Compare to the case of a perfectly harmonic oscillator with evenly spaced levels, and also to the vibrational portion of the graph in Figure 1.13.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free