# Homework Help: Uncertainty Principle & Probability Density

1. Apr 12, 2007

### kingwinner

I find quantum mechanics to be very hard, and I am currently having trobule with the following 2 problems, can someone please help me out?

1) A thin solid barrier in the xy-plane has a 10-micrometer-diameter circular hole. An electron traveling in the z-direction with vx=0 m/s passes through the hole. Afterward, is vx still zero? If not, within what range is vx likely to be?

My textbook says that the uncertainty principle is (delta x)(delta px) > h/2
But I am not sure what delta x would be in this situation...and I got completely confused with the x,y,z direction business as well...if vx is zero originally, why would it possibly be different after going through the hole? I don't understand what is going on...

2) A typical electron in a piece of metallic sodium has energy -Eo compared to a free electron, where Eo is the 2.7 eV work function of sodium. At what distance beyond the surface of the metal is the electron's probability density 10% of its value at the surface?
First of all, I don't get what it means by saying that the elctron has energy "-Eo compared to a free electron", what is the energy of a free electron? why is there a negative sign in front of Eo?

Secondly, I don't even know which equation to use, can someone give me some hints on solving this problem? I am totally lost...

Any help is greatly appreciated!

Last edited: Apr 12, 2007
2. Apr 12, 2007

### malawi_glenn

1) Electron diffraction, gives you that you must have some v_x after passage of the hole.

2) Well the electron energy in a metal piece is negative because the electron is bound. Bound states have negative enery.

3. Apr 12, 2007

### kingwinner

1) The electron is moving in the z-direction only and we are given that v_x=0 m/s, so v_x should be 0 m/s always, right? Passing a hole shouldn't affect anything...there is no force acting on it...

4. Apr 12, 2007

### Dick

In quantum mechanics particles aren't points. They are represented by a wavefunction that can be extended in space. In squeezing itself through the hole, the uncertainty principle tells you have to expect the possibility there will be a change in momentum.

5. Apr 12, 2007

### kingwinner

So should I substitute delta x=5x10^-6 m into the formula (delta x)(delta px) > h/2 ?

6. Apr 12, 2007

### Dick

Sure. Though be warned if you are comparing with a printed answer they might decide to put delta x=10 microns. This is not an exact use of the uncertainty relation since we haven't clearly defined 'delta'. It's only an estimate.

7. Apr 12, 2007

### kingwinner

Yes, the only way I can match the answer is to substitute delta x=10x10^-6 m, why is that? Is it because it may be at one end?

8. Apr 12, 2007

### Dick

Essentially, yes. But realize that answer is not 'correct' either. The exact definitions of the deltas in the uncertainty principle are standard deviations of wavefunctions. In these sorts of problems you aren't given exact wavefunctions, you just 'estimate' the width from the parameters of the problem. It's a 'ball park' estimate. You shouldn't expect it to be exact.