Exercise 54 gives a rough lifetime for a particle trapped particle to escape an enclosure by tunneling.

(a) Consider an electron. Given thatW=100 nm,L=1 nm  and  U0=5 eV, first verify that theEGS<<U0assumption holds, then evaluate the lifetime.

(b) Repeat part (a), but for a0.1µgparticle, withW=1nm,L=1µm, and a barrier heightU0that equals the energy the particle would have if its speed were just1 mmperyear.

Short Answer

Expert verified
  1. The assumptionEGS<<U0 holds. The lifetime is τ33min.
  2. The assumption EGS<<U0holds.

The relaxation time is very large ( 1020σ2eσ) for such a huge classical particle.

Step by step solution

01

Concepts involved

Tunneling is a phenomena in which a particle is able to tunnel through a potential barrier when its kinetic energy

The ground state energy of infinite well is given by,

EGS=π222mL2(1)

Where, = Modified Plank’s Constant

m= mass of the particle

L = Barrier width

Lifetime of the particle is given by,

τmW4σ22000L2eσ(2)

Where, =Wavelength/2

σ=2L2mU0(3)

02

Step 2(a): Determine the lifetime of the particles

The infinite well ground state Energy from Equation (1):

EGS=π222mL2=π2(1.055×1034 J.s)22(9.11×1031 kg)(107 m)2=6×1024 JEGS=3.8×105eV<<U0

Hence, the assumption holds, the value of EGSis found to be much smaller than U0.

Now, if you put the values in equation (3), and then in equation (2) you get,

σ=(109)8(9.11×1031 kg)(8×1019 J)(1.055×1034 J.s)=22.9τ=(9.11×1031 kg)(107m)42000(1.055×1034J.s)(109 m)222.92e22.92000 s33min

Hence, lifetime of the particle is 33 min.

03

Step 3(b): verify that the  EGS<<U0

Now, for new parameters:

From Equation (1), you get,

E1=π2(1.055×1034 J.s)2(1010)(103)2E1=5×1052JU0=12(1010kg)102m/s3.16×107sU0=5×1032J

Now, if you put above obtained values in equation (3) and then in equation (2), you get,

  σ=(106 m)8(1010 kg)(5×1032 J)(1.055×1034 J.s)=6×107   τ=(1010kg)(103m)42000(1.05×1034J.s)(106m)2σ2eσ      1020σ2eσ

τwould be very large for such a huge, classical particle.

The assumption ofEGS<<U0holds good.

For such a huge classical particle, relaxation time would be very large.

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

The diagram below plots ω(k) versus wave number for a particular phenomenon. How do the phase and group velocities compare, and do the answer depend on the central value of k under consideration? Explain.

Verify that the reflection and transmission probabilities given in equation (6-7) add to 1.

For a general wave pulse neither E nor p (i.eneither ωnor k)areweIldefined. But they have approximate values E0andp0. Although it comprisemany plane waves, the general pulse has an overall phase velocity corresponding to these values.

vphase=ω0k0=E0/hp0/h=E0p0

If the pulse describes a large massive particle. The uncertaintiesarereasonably small, and the particle may be said to have energy E0and momentum p0.Usingtherelativisticallycorrect expressions for energy and momentum. Show that the overall phase velocity is greater thancand given by

vphase=c2uparticle

Note that the phase velocity is greatest for particles whose speed is least.

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?

As we learned in example 4.2, in a Gaussian function of the formψ(x)αe-(x2/2ε2)is the standard deviation or uncertainty in position.The probability density for gaussian wave function would be proportional toψ(x)squared:e-(x2/2ε2). Comparing with the timedependentGaussian probability of equation (6-35), we see that the uncertainty in position of the time-evolving Gaussian wave function of a free particle is given by

.Δx=ε1+h2t24m2ε4 That is, it starts atand increases with time. Suppose the wave function of an electron is initially determined to be a Gaussian ofuncertainty. How long will it take for the uncertainty in the electron's position to reach5m, the length of a typical automobile?

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