Uncertainty Principle & Probability Density

[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 v_x=0 m/s passes through the hole. Afterward, is v_x still zero? If not, within what range is v_x likely to be?

My textbook says that the uncertainty principle is (delta x)(delta p_x) > 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 v_x 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 -E_o compared to a free electron, where E_o 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 "-E_o compared to a free electron", what is the energy of a free electron? why is there a negative sign in front of E_o?

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!

 

[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.

 

[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...

 

[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.

 

[kingwinner]

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

 

[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.

 

[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?

 

[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.