Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Q42E (page 281)

In Appendix G. the operator for the square of the angular momentum is shown to be

L^2=-h2[cscθ∂∂θsinθ∂∂θ+csc2θ∂2∂ϕ2]

Use this to rewrite equation (7-19) asL^2⚇Φ=-Ch2⚇Φ

Short Answer

Expert verified

L^2θΦ=-Ch2θΦ

Step by step solution

01

 Given data

The angular momentum operator is given as:

L^2=-h2[cscθ∂∂θ(sinθ∂∂θ)+csc2θ∂2∂ϕ2]

Where, h is the reduced Planck’s constant.

02

 Calculation

The separation of the variable of the above equation can be written as:

cscθ∂∂θ(sinθ∂∂θ)+csc2θ∂2∂ϕ2ΘΦ=CΘΦcscθ∂∂θ(sinθ∂ΘΦ∂θ)+csc2θ∂2ΘΦ∂ϕ2=CΘΦ1Θcscθ∂∂θ(sinθ∂Θ∂θ)+csc2θ1Φ∂2Φ∂ϕ2=C

Now we can write the operator as:

L^2=-h2cscθ∂∂θsinθ∂∂θcsc2θ∂2∂ϕ2L^2⊝Φ=-h2cscθ∂∂θsinθ∂∂θcsc2θ∂2∂ϕ2⊝Φ=-h2Φcscθ∂∂θsinθ∂∂θ-h2Φcsc2θ∂2Φ∂ϕ2=-h2⊝Φ1⊝cscθ∂∂θsinθ∂Φ∂θcsc2θ1Φ∂2Φ∂ϕ2L^2⊝Φ=-Ch2⊝Φ

03

 Conclusion

Thus, it can be written as L^2⊝Φ=-Ch2⊝Φ.

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.

Start your free trial

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

Classically, what happens when a moving object has a head-on elastic collision with a stationary object of exactly equal mass? What if it strikes an object of smaller mass? Of larger mass? How do these ideas relate to Rutherford’s conclusion about the nature of the atom?

Classically, an orbiting charged particle radiates electromagnetic energy, and for an electron in atomic dimensions, it would lead to collapse in considerably less than the wink of an eye.

(a) By equating the centripetal and Coulomb forces, show that for a classical charge -e of mass m held in a circular orbit by its attraction to a fixed charge +e, the following relationship holds

ω=er-3/24πε0m.

(b) Electromagnetism tells us that a charge whose acceleration is a radiates power P=e2a2/6ε0c3. Show that this can also be expressed in terms of the orbit radius as P=e696π2ε03m2c3r4. Then calculate the energy lost per orbit in terms of r by multiplying the power by the period T=2π/ωand using the formula from part (a) to eliminate .

(c) In such a classical orbit, the total mechanical energy is half the potential energy, or Eorbit=-e28πε0r. Calculate the change in energy per change in r : dEorbit/dr. From this and the energy lost per obit from part (b), determine the change in per orbit and evaluate it for a typical orbit radius of 10-10m. Would the electron's radius change much in a single orbit?

(d) Argue that dividing dEorbit/dr by P and multiplying by dr gives the time required to change r by dr . Then, sum these times for all radii from rinitial to a final radius of 0. Evaluate your result for rinitial=10-10m. (One limitation of this estimate is that the electron would eventually be moving relativistically).

In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,λ=2L/n , where is the length of its home. Given that λ=h/p, we would have p=nh/2L, and the kinetic energy, p2/2m, would thus be n2h2/8mL2. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: E=U. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?

At heart, momentum conservation is related to the universe being "translationally invariant," meaning that it is the same if you shift your coordinates to the right or left. Angular momentum relates to rotational invariance. Use these ideas to explain at least some of the differences between the physical properties quantized in the cubic three-dimensional box versus the hydrogen atom.

A wave function with a non-infinite wavelength-however approximate it might be- has nonzero momentum and thus nonzero kinetic energy. Even a single "bump" has kinetic energy. In either case, we can say that the function has kinetic energy because it has curvature- a second derivative. Indeed, the kinetic energy operator in any coordinate system involves a second derivative. The only function without kinetic energy would be a straight line. As a special case, this includes a constant, which may be thought of as a function with an infinite wavelength. By looking at the curvature in the appropriate dimension(s). answer the following: For a givenn,isthe kinetic energy solely

(a) radial in the state of lowest l- that is, l=0; and

(b) rotational in the state of highest l-that is, l=n-1?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Particle Model of Matter

Read Explanation

Wave Optics

Read Explanation

Waves Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
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