/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q7CQ Given the same particle energy a... [FREE SOLUTION] | 91影视

91影视

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

What fraction of a beam of 50eVelectrons would get through a 200V1nm wide electrostatic barrier?

The potential energy barrier in field emission is not rectangular, but resembles a ramp, as shown in Figure 6.16. Here we compare tunnelling probability calculated by the crudest approximation to that calculated by a better one. In method 1, calculate T by treating the barrier as an actual ramp in which U - E is initially, but falls off with a slop of M. Use the formula given in Exercise 37. In method 2, the cruder one, assume a barrier whose height exceeds E by a constant /2(the same as the average excess for the ramp) and whose width is the same as the distance the particle tunnels through the ramp. (a) Show that the ratio T1/T2 is e8m33hM . (b) Do the methods differ more when tunnelling probability is relatively high or relatively low?

For the E>U0 potential barrier, the reflection, and transmission probabilities are the ratios:

R=B*BA*AT=F*FA*A

Where A, B, and F are multiplicative coefficients of the incident, reflected, and transmitted waves. From the four smoothness conditions, solve for B and F in terms of A, insert them in R and T ratios, and thus derive equations (6-12).

A beam of particles of energy incident upon a potential step ofU0=(3/4),is described by wave function:inc(x)=eikx

The amplitude of the wave (related to the number of the incident per unit distance) is arbitrarily chosen as 1.

  1. Determine the reflected wave and wave inside the step by enforcing the required continuity conditions to obtain their (possibly complex) amplitudes.
  2. Verify the explicit calculation of the ratio of reflected probability density to the incident probability density agrees with equation (6-7).

The matter wave dispersion relation given in equation (6-23) is correct only at low speed and when mass/internal energy is ignored.

(a) Using the relativistically correct relationship among energy, momentum and mass, show that the correct dispersion relation is

=k2c2+m2c42

(b) Show that in the limit of low speed (small p and k) and ignoring mass/internal energy, this expression aggress with that of equation (6-23).

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.