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

Exercise 39 gives a condition for resonant tunneling through two barriers separated by a space width of2s, expressed I terms of factorβgiven in exercise 30. Show that in the limit in which barrier widthL, this condition becomes exactly energy quantization condition (5.22) for finite well. Thus, resonant tunneling occurs at the quantized energies of intervening well.

Short Answer

Expert verified

In the limit in which barrier width L, this condition becomes exactly energy quantization condition (5-22) for finite well. Thus, resonant tunneling occurs at the quantized energies of intervening well.

Step by step solution

01

Concepts involved

Resonant tunneling is a phenomenon in which an electron enters from one side of a double barrier structure, and travels across it at or near the metastable levels in the quantum well.

02

Given parameters

The condition for resonant tunneling from the Exercise 39 is given by,

2s=β/k

Where, β=tan12αkk2α2coth(αL).

k= wave number

L = barrier width

α =constant

03

Show that in the limit in which barrier width L→∞, this condition becomes exactly energy quantization condition

In the limit Lcoth1 and so:

β=tan12αkk2α22sk=tan12αkk2α2tan(2sk)=2αkk2α2

Rearranging, 2cot(k.2s)=kααk

2s is the distance between the barriers; this is same as the expression (5.22) which is the energy quantization condition.

Here, the condition of resonant tunneling is proved to be equal to the energy quantization condition, whenL . Thus, resonant tunneling occurs at the quantized energies of intervening well.

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

A method for finding tunneling probability for a barrier that is "wide" but whose height varies in an arbitrary way is the so-called WKB approximation.

T=exp[2122m(U(x)E)dx]

Here U(x) is the height of the arbitrary potential energy barrier.Whicha particle first penetrates at x=0 and finally exits at x=L. Although not entirely rigorous, show that this can be obtained by treating the barrier as a series of rectangular slices, each of width dx (though each is still a "wide" barrier), and by assuming that the probability of tunneling through the total is the product of the probabilities for each slice.

Given the same particle energy and barrier height and width, which would tunnel more readily: a proton or an electron? Is this consistent with the usual rule of thumb governing whether classical or non-classical behavior should prevail?

From equations (6-23) and (6-29) obtain the dispersion coefficient for matter waves (in vacuum), then show that probability density (6-35) follows from (6-28)

Could the situation depicted in the following diagram represent a particle in a bound state? Explain.

Exercise 39 gives the condition for resonant tunneling through two barriers separated by a space of width 2 s, expressed in terms of a factor βgiven in Exercise 30. (a) Suppose that in some system of units, k and α are both2π. Find two values of 2s that give resonant tunneling. What are these distances in terms of wavelengths ofψ? Is the term resonant tunneling appropriate?(b) Show that the condition has no solution if s = 0 and explain why this must be so. (c) If a classical particle wants to surmount a barrier without gaining energy, is adding a second barrier a good solution?

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