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

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).

Short Answer

Expert verified

The probability of water molecule is:

P1=0.9997P2=4.618×10-4P3=2.133×10-7

Step by step solution

01

Given information

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.

02

Explanation

(a) The partition function is: because none of the levels are degenerate.

Z=se-E(s)/kT

Where,

E(s)=s+12hfs=0,1,2,

So,

Z=e-hf/2kT+e-3hf/2kT+e-5hf/2kT+

Let x=hf/kT, so

Z=e-x/2+e-3x/2+e-5x/2+(1)

As a result, the value of z is (at T = 300 K, we substitute the Boltzmann constant in eV, k = 8.617x 10-5 eV/K, and the Planck constant in eV, h = 4.136 x 10-15 eV s):

x=4.136×10-15eV·s4.8×1013Hz8.617×10-5eV(300K)=7.68

Substitute x into (1)

Z=e-(7.68)/2+e-3(7.68)/2+e-5(7.68)/2Z=0.0215

Probability of first state is:

P=1Ze-x/2

Substitute with x and z:

role="math" localid="1647366550518" P1=10.0215e-7.68/2P1=0.9997

Probability of second state is:

P2=1Ze-3x/2

Substitute x and z:

role="math" localid="1647366573852" P2=10.0215e-3(7.68)/2P2=4.618×10-4

Probability of third state:

P3=1Ze-3x/2

Substitute x and z:

role="math" localid="1647366591263" P3=10.0215e-5(7.68)/2P3=2.133×10-7

03

Explanation

We have temperature value, T=300k, so

x=4.136×10-15eV·s4.8×1013Hz8.617×10-5eV(700K)=3.2913

Substitute x into (1):

role="math" localid="1647366281815" Z=e-(3.2913)/2+e-3(3.2913)/2+e-5(3.2913)/2Z=0.20033

Substitute the value of x and z to get the probabilities:

P1=10.20033e-3.2913/2P1=0.962847P2=10.20033e-3(3.2913)/2P2=0.035823P3=10.20033e-5(3.2913)/2P3=0.001333

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

Prove that the probability of finding an atom in any particular energy level is P(E)=(1/Z)e-F/kT, whereF=E-TS and the "'entropy" of a level is k times the logarithm of the number of degenerate states for that level.

For a mole nitrogen (N2)gas at room temperature and atmospheric pressure, compute the following U,H,F,G,Sand μ. (The electronic ground state of nitrogen is not degenerate.)

Consider a hypothetical atom that has just two states: a ground state with energy zero and an excited state with energy 2 eV. Draw a graph of the partition function for this system as a function of temperature, and evaluate the partition function numerically at T = 300 K, 3000 K, 30,000 K, and 300,000 K.

The dissociation of molecular hydrogen into atomic hydrogen, H22Hcan be treated as an ideal gas reaction using the techniques of Section 5.6. The equilibrium constant K for this reaction is defined as

K=PH2P0PH2

whereP0is a reference pressure conventionally taken to be1bar,and the other P's are the partial pressures of the two species at equilibrium. Now, using the methods of Boltzmann statistics developed in this chapter, you are ready to calculate K from first principles. Do so. That is, derive a formula for K in terms of more basic quantities such as the energy needed to dissociate one molecule (see Problem 1.53) and the internal partition function for molecular hydrogen. This internal partition function is a product of rotational and vibrational contributions, which you can estimate using the methods and data in Section 6.2. (AnH2 molecule doesn't have any electronic spin degeneracy, but an H atom does-the electron can be in two different spin states. Neglect electronic excited states, which are important only at very high temperatures. The degeneracy due to nuclear spin alignments cancels, but include it if you wish.) Calculate K numerically atT=300K,1000K,3000K,and6000K. Discuss the implications, working out a couple of numerical examples to show when hydrogen is mostly dissociated and when it is not.

A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = mμB, where μ the constant is 1.03 x 10-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show that, if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T =-300 K.

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