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

Estimate the probability that a hydrogen atom at room temperature is in one of its first excited states (relative to the probability of being in the ground state). Don't forget to take degeneracy into account. Then repeat the calculation for a hydrogen atom in the atmosphere of the starγ UMa, whose surface temperature is approximately 9500 K.

Short Answer

Expert verified

As a result, no Hydrogen atoms can be found in the first excited state.

On this star, there is one atom in the first excited state out of 64500 hydrogen atoms.


Step by step solution

01

Given information

A hydrogen atom at room temperature is in one of its first excited states (relative to the probability of being in the ground state).

02

Explanation

With n = 2, assuming E1 is ground state energy and E2 is first excited state energy, the chance of finding the atom in either of the first excited states is:

Ps2=1Ze-E2/kT

The partition function is equal to the sum of the Boltzmann factors, but because the initial state has the greatest energy magnitude, we can approximate it as follows:

Z=se-E(s)/kTe-E1/kT

Substitute in the above equation

Ps2=eE1/kTe-E2/kTPs2=e-E2-E1/kT

There are four such states, therefore the probability of E2 equals Ps2multiplied by 4, so:

PE2=4e-E2-E1/kT

Substitute the energies and Boltzmann constant in eV (k = 8.617 x 10-5 eV/K) at room temperature T = 300 K for the energy of the ground state E1=-13.6 eV and the energy of the first excited state B2 =-3.4 eV.

PE2=4e-(-3.4eV-(-13.6eV))/8.617×10-5eV/K(300K)PE2=1.75×10-171

As a result, no Hydrogen atoms can be found in the first excited state.

03

Explanation

We must now calculate this probability for the star γUMa, which has a surface temperature of 9500 K, as follows:

PE2=4e-(-3.4eV-(-13.6eV))/8.617×10-5eV/K(9500K)PE2=1.55×10-5

On this star, there is one atom in the first excited state out of 64500 hydrogen atoms.

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

At very high temperatures (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the "nucleon." (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron's mass is higher than the proton's by 2.3 x 10-30 kg, its energy is higher by this amount times c2. Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 1011 K. What fraction of the nucleons at that time were protons, and what fraction were neutrons?

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.)

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 an O2molecule, the constant is approximately 0.00018eV. Estimate the rotational partition function for an O2molecule at room temperature.

Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy. Repeat the calculation given in the text for the relative probability of being in a first excited state, taking spin degeneracy into account. Show that the results are unaffected.

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