# Calculating Tunneling Probability

1. Sep 24, 2007

### E1982

1. The problem statement, all variables and given/known data

There's two parts.

Part A) Find the Probability than an electron will tunnel through a barrier if energy is 0.1 ev less than height of the barrier. Barrier is 1nm.

Part B) Find tunneling probability if the barrier is widened to 3 nm.

2. Relevant equations

I believe relevant equations are attenuation factor alpha = sqrt (2*m*(Uo-E)/h^2) and that Probability = psi^2. Psi = A*e^(-alpha * x). Probably should mention that h = planck constant and m = mass of electron, Uo-E = change in energy

3. The attempt at a solution

I don't know how to write my solutions out here. I don't have a reference textbook on quantum physics on me, so I'm doing this by memory as much as I can. I calculated attenuation factor (alpha) but did not know what to do with it. I don't know what A is and what x is. Again, this is coming from memory, so I don't know how far away from the solution I am.

Last edited: Sep 24, 2007
2. Sep 24, 2007

### malawi_glenn

You can easily find tutorials on the internet if you google a bit, and your library should have plenty of those books.

Also check out this page to learn how to write nice formulas on this formum:

3. Sep 24, 2007

### E1982

I have found some google references to tunneling probability, but nothing that gives you a step by step how to on solving problems. There is one site with a Javascript function that will calculate the probability for you but without telling you how they did it (I will need to show my work!). I don't know if I'm using the correct keywords when it comes to searching on the internet, for I can't find a detailed how to guide. I've never taken a Quantum Mechanics or Modern Physics course before, so I'm going on what I know.

4. Sep 24, 2007

### malawi_glenn

you may also search for old exams with solutions, there should be plenty of those too =)
My first exam on QM was calculating and deriving things like these.. but the text is in swedish so hehe..

I can give you the formula I have..

Transmission coefficent:
$$T = \left[ 1 + \dfrac{E^2_p \sinh ^2 (\alpha a)}{4E(E-E_p )} \right] ^{-1}$$

E is particle energy
Ep is barrier potential energy
a is width of barrier
Sinh is sinus -hyperbolicus; can be found in your mathbooks or internet...

$$\alpha = \dfrac{\sqrt{2m \vert E_p - E \vert}}{\hbar}$$

Now, good luck! remember to have right units, and tell us more if you don't get it.

And one question; why are you doing this without any QM or modern physics in your back pocket? =/

Last edited: Sep 24, 2007
5. Sep 24, 2007

### E1982

I'm doing this for an Intro to Nanotech class, we're going over Scanning Tunneling Microscopes. However there was no pre-requisite for a modern physics course, and my major does not require it!

But one question... how does Transmission coefficient play in trying to find tunneling probability? I understand the attenuation factor is needed for the Schrodinger equation, but transmission coefficient now?

6. Sep 24, 2007

### nrqed

Are you supposed to simply plug in the numbers in a formula that you look up or are you supposed to derive the formula itself? (in which case you have to know how to solve Schrodinger's equation for constant potentials)

EDIT: Ah, I see from your other post that you are meant to simply use the formula.
To answer your question, th etransmission coefficient *is* the probability of transmission.

7. Sep 24, 2007

### E1982

Ah thanks for the clear up. Now the only problem I have is that I've not been given a value for E, I've only been told that E is 0.1 eV less than the height of the barrier. I'll see if I can cancel something out.

8. Sep 24, 2007

### malawi_glenn

Sorry I forgot to tell you that T = prob to tunnel.

And if E is not given; then you should probably express T as a function of E and Ep, i.e calculate everything that is not E And you get: T = [1 + K* Ep^2 *E^(-1) ]^(-1)

K is the constant that you should calculate.

At least I would have done so

Last edited: Sep 24, 2007
9. Sep 24, 2007

### nrqed

Oh... Then you won'tget a result. What is needed is the ratio E_particle/E_barrier. Without that information you can't find the value of T. So there is information missing if you want a numerical value.

PS: I did not double check the formula given by the other poster.

10. Sep 24, 2007

### nrqed

You should not have to square, T is already a probability.

11. Sep 24, 2007

### nrqed

I have a different formula: (I will use V_0 for the barrier height to make it as clear as possible):

$$T = \left[ 1 + \dfrac{V_0^2 \sinh ^2 (\alpha a)}{4E_p(V_0-E_p )} \right] ^{-1}$$

12. Sep 24, 2007

### malawi_glenn

But that is the formula if particle energy is less then barrier height..
If E_p now is the energy of the particle..

If you all want, I can scan in from my book and post it here..

Last edited: Sep 24, 2007
13. Sep 24, 2007

### E1982

Yes exactly is it! My electron energy is 0.1 eV less than barrier height! If I ever figure out tex code I'll write out the equation.

14. Sep 24, 2007

### nrqed

Yes, but that is precisely the situation the OP is dealing with.

15. Sep 24, 2007

### E1982

Another attempt at the equation

|$$T^2$$| = $$e^{-2 \alpha x} dx$$ where alpha = $$\sqrt{2m/h^2 * (V-E)}$$

So final equation to use would be |$$T^2$$| = $$e^{-2 \int \sqrt{2m/h^2 * (V-E) }dx}$$

This is the one I think!

So now only question is what are my limits of integration? If I know that I'm set!

Last edited: Sep 24, 2007
16. Sep 24, 2007

### malawi_glenn

ahh LOL I am not so awake now ;) sorry!

The equation you have now is called Gamow Factor, and I know how to use that method in alpha decay, I think that this approximation comes when you have an exponential decaying barrier height... But Iam not so sure about this so probably nrqed
or someone else know this =)

But MY intuitive is that the integral should be just multiplying with the widht of the barrier...

Last edited: Sep 24, 2007
17. Sep 24, 2007

### E1982

Wait a minute, shouldn't my limits of integration be the length of the barrier?!?!? I think it must... lol makes sense.

18. Sep 24, 2007

### nrqed

Ah! The Gamow factor.

well, actually, One can show that this is equivalent to the form we gave you if one makes some approximations (in other words, our expression is more general, for a square barrier)...... but whowing that would be a long post ...

To answer your question, in your case V is a constant (I think that's what they want you to do otherwise you could not answer the question) so that you can pull out (V-E) from the integral. Therfore [tex] \int dx (V-E) = (V-E) \int dx = (V -E) \times a [/itex] where "a" is the width of the barrier.

And you are done.

Sorry for misleading you a bit. But again, the expression we gave is actually the exact one for a square barrier.

Patrick

19. Sep 24, 2007

### E1982

YAY!!! I think we got it!

Doesn't say V is constant, but mentions that V-E = 0.1 eV, so since that is a constant I'm taking that it's safe to assume that (V-E) = constant and therefore I can do the integral you showed above with no repercussions Nrged.

I'm set! Thank you!