Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q39E (page 188)

A finite well always has at least one bound state. Why does the argument of Exercises 38 fail in the case of a finite well?

Short Answer

Expert verified

In the case of the finite well, the wave extends into the classically forbidden region on the outside of the barriers on both sides of the well, so its wavelength can be very large.

Step by step solution

01

Given information and Formula used

A finite well always has at least one bound state

Formula used

λ=hp

K.E=p22m

02

Calculation

In the case of the finite well, the wave extends into the classically forbidden region on the outside of the barriers on both sides of the well, so its wavelength can be very large.

λ=hp

Here,

λ= wavelengths = momentum = Planck's constant

As the wavelength is inversely proportional to momentum, so for a shorter wavelength, momentum will be larger.

Now,

K.E=p22m

Here,

E.K= kinetic Energy = momentum = mass of a particle

Kinetic energy is directly proportional to the momentum. Hence, kinetic energy will be larger.

03

Conclusion

Even a small potential barrier can hold the particle.

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

The particle has E=0.

(a) Show that the potential energy for x>0is given by

U(x)=-ℏ2am1x+ℏ2a22m

(b) What is the potential energy for x<0?

The deeper the finite well, the more state it holds. In fact, a new state, the, is added when the well’s depthU0reachesh2(n−1)2/8mL2. (a) Argue that this should be the case based only onk=2mE/h2, the shape of the wave inside, and the degree of penetration of the classically forbidden region expected for a state whose energy E is only negligibly belowU0. (b) How many states would be found up to this same “height” in an infinite well.

The product of uncertainties in particle's momentum and position.

A bound particle of massdescribed by the wave function

ψ(x)=Axe-x2/2b2

What is the most probable location at which to find the particle?

Simple models are very useful. Consider the twin finite wells shown in the figure, at First with a tiny separation. Then with increasingly distant separations, In all case, the four lowest allowed wave functions are planned on axes proportional to their energies. We see that they pass through the classically forbidden region between the wells, and we also see a trend. When the wells are very close, the four functions and energies are what we might expect of a single finite well, but as they move apart, pairs of functions converge to intermediate energies.

(a) The energies of the second and fourth states decrease. Based on changing wavelength alone, argue that is reasonable.

(b) The energies of the first and third states increase. Why? (Hint: Study bow the behaviour required in the classically forbidden region affects these two relative to the others.)

(c) The distant wells case might represent two distant atoms. If each atom had one electron, what advantage is there in bringing the atoms closer to form a molecule? (Note: Two electrons can have the same wave function.)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Rotational Dynamics

Read Explanation

Electricity

Read Explanation

Work Energy and Power

Read Explanation

Radiation

Read Explanation

Magnetism

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