Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Q7CQ (page 223)

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?

Short Answer

Expert verified

The answer is electron.

Step by step solution

01

Definition of quantum tunneling

The quantum phenomenon, in which a particle can penetrate a barrier and pass through it even though it is forbidden classically, is known as quantum tunneling.

02

Explanation and conclusion

An electron would tunnel more readily because it has less mass as compared to the proton, provided all other things are kept the same.

Less mass would imply less momentum and quicker exponential decay and which in turn would result in a larger wavelength. The larger the wavelength, the more classical the behavior of 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

Fusion in the Sun: Without tunnelling. our Sun would fail us. The source of its energy is nuclear fusion. and a crucial step is the fusion of a light-hydrogen nucleus, which is just a proton, and a heavy-hydrogen nucleus. which is of the same charge but twice the mass. When these nuclei get close enough. their short-range attraction via the strong force overcomes their Coulomb repulsion. This allows them to stick together, resulting in a reduced total mass/internal energy and a consequent release of kinetic energy. However, the Sun's temperature is simply too low to ensure that nuclei move fast enough to overcome their repulsion.

a) By equating the average thermal kinetic energy that the nuclei would have when distant,32KBT. and the Coulomb potential energy they would have when 2fm apart, roughly the separation at which they stick, show that a temperature of about 1019K would be needed.

b) The Sun's core is only about 10k. If nuclei can’t make it "over the top." they must tunnel. Consider the following model, illustrated in the figure: One nucleus is fixed at the origin, while the other approaches from far away with energyE. As rdecreases, the Coulomb potential energy increases, until the separation ris roughly the nuclear radius rnuc. Whereupon the potential energy is Umaxand then quickly drops down into a very deep "hole" as the strong-force attraction takes over. Given then E≪Umax, the point b, where tunnelling must begin. will be very large compared with rnuc, so we approximate the barrier's width Las simply b. Its height, U0, we approximate by the Coulomb potential evaluated at b2. Finally. for the energy Ewhich fixes b, let us use 4×32KBT. which is a reasonable limit, given the natural range of speeds in a thermodynamic system.Combining these approximations, show that the exponential factor in the wide-barrier tunnelling probability is

exp[-e24πε0h4m3kBT]

c)Using the proton mass for , evaluate this factor for a temperature of107K. Then evaluate it at3000K. about that of an incandescent filament or hot flame. and rather high by Earth standards. Discuss the consequences.

Question: An electron bound in an atom can be modeled as residing in a finite well. Despite the walls. When many regularly spaced atoms are relatively close together as they are in a solid-all electrons occupy alltheatoms. Make a sketch of a plausible multi-atom potential energy and electron wave function.

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)

Solving the potential barrier smoothness conditions for relationships among the coefficients A,B and Fgiving the reflection and transmission probabilities, usually involves rather messy algebra. However, there is a special case than can be done fairly easily, through requiring a slight departure from the standard solutions used in the chapter. Suppose the incident particles’ energyEis preciselyU0.

(a) Write down solutions to the Schrodinger Equation in the three regions. Be especially carefull in the region0<x<L. It should have two arbitrary constants and it isn’t difficult – just different.

(b) Obtain the smoothness conditions, and from these findR and T.

(c) Do the results make sense in the limitL→∞?

In the wide-barrier transmission probability of equation (6-18), the coefficient multiplying the exponential is often omitted. When is this justified, and why?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physics of Motion

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Space Physics

Read Explanation

Measurements

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