Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q 6CQ (page 186)

When is the temporal part of the wave function 0? Why is this important?

Short Answer

Expert verified

Answer:

The likelihood of discovering the probability would be 0if the temporal part of a wave function was zero, implying that the particle would have vanished.

Step by step solution

01

Step 1: Theory of wave function

The wave function ψitself has no physical significance. However, the amplitude of wave function corresponds to ψand the square of the wave function ψrelates to the photon density, the number of photons present in a region, it relates to electron density in a certain region.

02

Conclusion

Therefore, using the square of wave function, we can measure of the probability that the electron can be found within a particular tiny volume of the atom.

Hence, The likelihood of discovering the probability would be 0 if the temporal part of a wave function was zero, implying that the particle would have vanished.

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

Write out the total wave functionψ(x,t).For an electron in the n=3 state of a 10nm wide infinite well. Other than the symbols a and t, the function should include only numerical values?

Calculate the uncertainty in the particle’s position.

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

sin(a+b)=sinacosb+cosasinband

cos(a+b)=cosacosb-sinasinbon the

functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that ψ(z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (α/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as αk=tankL2 and αk=-cotkL2. Finally, plug these separately back into the region I solutions and show that

ψ(z)=C×{eα(z+L/2)          z<L/2coskzcoskL2          -L/2<z<L/2e-α(z-L/2)          z>L/2


or

ψ(z)=C×{eα(z+L/2)          z<L/2-sinkzsinkL2          -L/2<z<L/2e-α(z-L/2)          z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |ψ(z)|2 over all space to 1 would give it in terms of kand α , but because we can’t solve (22) exactly for k(or E), neither can we obtain an exact value for C.

Determine the expectation value of the momentum of the particle. Explain.

Given that the particle’s total energy is0, show that the potential energy is role="math" localid="1657529957489" U(x)=h22mb4x2-3h22mb2.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Wave Optics

Read Explanation

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation

Turning Points in Physics

Read Explanation

Torque and Rotational Motion

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