Suppose that in Figure 9.27, the level labelled E1, rather than the one labelledE2, were metastable. Might the material still function as a laser? Explain.

Short Answer

Expert verified

Yes the material still function as laser .

Step by step solution

01

The lasing transition.

The lasing transition occurs from '1' to the ground level '0. ' In this three-level type of laser operation the pump field is used to raise the atoms from the ground level '0' to the pump level '2,' from which there occurs a rapid transition

02

Find the downward transition.

Figure 27 depicts the subsequent transition of electrons in an atomic gas for a four-level laser. Here,E2 is designated as the metastable state. After a continuous round of optical pumping, all energy levels exceptE2 depopulate. Because of the slow decay fromE2 , the number of electrons in the metastable state is now larger than the E1state. This condition then initiates population inversion and the subsequent cascade of stimulated emissions. Four this four-level laser, the downward transition E2E1is the lasing transition.

03

Find if the material still function as laser.

Now suppose that E1is metastable. In this case, the system will STILL FUNCTION as a laser. In fact, such a configuration now corresponds to a three-level laser, where the lasing transition now occurs fromE1 toE0 . MakeE1 metastable, andE2non-metastable however, this reduces the efficiency of the lasing process since more optical pumping is required to completely exhaust the ground state.

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 that equation (9- 16) follows from (9-15) and (9- 10).

The electrons’ contribution to the heat capacity of a metal is small and goes to 0as T0. We might try to calculate it via the total internal energy, localid="1660131882505" U=EN(E)D(E)dE, but it is one of those integrals impossible to do in closed form, and localid="1660131274621" N(E)FDis the culprit. Still, we can explain why the heat capacity should go to zero and obtain a rough value.

(a) Starting withN(E)FDexpressed as in equation (34), show that the slope N(E)FDdEatE=EFis-1(4kBT).

(b) Based on part (a), the accompanying figure is a good approximation to N(E)FDwhen Tis small. In a normal gas, such as air, whenTis raised a little, all molecules, on average, gain a little energy, proportional to kBT. Thus, the internal energy Uincreases linearly with T, and the heat capacity, UT, is roughly constant. Argue on the basis of the figure that in this fermion gas, as the temperature increases from 0to a small value T, while some particles gain energy of roughly kBT, not all do, and the number doing so is also roughly proportional to localid="1660131824460" T. What effect does this have on the heat capacity?

(c)Viewing the total energy increase as simply U= (number of particles whose energy increases) (energy change per particle) and assuming the density of states is simply a constant Dover the entire range of particle energies, show that the heat capacity under these lowest-temperature conditions should be proportional to kBREFT. (Trying to be more precise is not really worthwhile, for the proportionality constant is subject to several corrections from effects we ignore).

By considering its constituents, determine the dimensions (e.g. length, distance over lime. etc.) of the denominator in equation(926). Why is the result sensible?

When a star has nearly bumped up its intimal fuel, it may become a white dwarf. It is crushed under its own enormous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle, "squeezing" the potential well necessarily increases the energies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is minimum total energy and corresponding equilibrium radius.

(a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67) is
Uclocimns=310(3π2h3me32V)23N53

Note that as the volume Vis decreased, the energy does increase. For a neutral star. the number of electrons, N, equals the number of protons. If protons account for half of the white dwarf's mass M (neutrons accounting for the other half). Show that the minimum electron energy may be written

Uelectrons=9h280me(3π2M5mp5)131R2

Where, R is the star's radius?

(b) The gravitational potential energy of a sphere of mass Mand radius Ris given by

Ugray=-35GM2R

Taking both factors into account, show that the minimum total energy occurs when

R=3h28G(3π2me3mp5M)13

(c) Evaluate this radius for a star whose mass is equal to that of our Sun 2x1030kg.

(d) White dwarfs are comparable to the size of Earth. Does the value in part (c) agree?

In a large system of distinguishable harmonic oscillator how high does the temperature have to be for the probable number of particles occupying the ground state to be less than 1 ?

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