
#1
Sep2407, 04:12 AM

P: 8

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*(UoE)/h^2) and that Probability = psi^2. Psi = A*e^(alpha * x). Probably should mention that h = planck constant and m = mass of electron, UoE = 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. 



#2
Sep2407, 08:06 AM

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You can easily find tutorials on the internet if you google a bit, and your library should have plenty of those books.
If you cant find any references, we'll help you. Also check out this page to learn how to write nice formulas on this formum: http://www.physicsforums.com/showthread.php?t=8997 



#3
Sep2407, 10:55 AM

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#4
Sep2407, 11:04 AM

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Calculating Tunneling Probability
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: [tex] T = \left[ 1 + \dfrac{E^2_p \sinh ^2 (\alpha a)}{4E(EE_p )} \right] ^{1} [/tex] 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... [tex] \alpha = \dfrac{\sqrt{2m \vert E_p  E \vert}}{\hbar} [/tex] 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? =/ 



#5
Sep2407, 11:15 AM

P: 8

I'm doing this for an Intro to Nanotech class, we're going over Scanning Tunneling Microscopes. However there was no prerequisite 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
Sep2407, 11:22 AM

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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
Sep2407, 11:46 AM

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#8
Sep2407, 11:51 AM

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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 



#9
Sep2407, 11:53 AM

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



#10
Sep2407, 12:00 PM

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#11
Sep2407, 12:02 PM

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[tex] T = \left[ 1 + \dfrac{V_0^2 \sinh ^2 (\alpha a)}{4E_p(V_0E_p )} \right] ^{1} [/tex] 



#12
Sep2407, 12:12 PM

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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.. 



#13
Sep2407, 12:15 PM

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#14
Sep2407, 12:17 PM

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#15
Sep2407, 12:19 PM

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Another attempt at the equation
[tex]T^2[/tex] = [tex] e^{2 \alpha x} dx [/tex] where alpha = [tex] \sqrt{2m/h^2 * (VE)}[/tex] So final equation to use would be [tex]T^2[/tex] = [tex] e^{2 \int \sqrt{2m/h^2 * (VE) }dx} [/tex] This is the one I think! So now only question is what are my limits of integration? If I know that I'm set! 



#16
Sep2407, 12:38 PM

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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... 



#17
Sep2407, 12:49 PM

P: 8

Thanks Malawi for all the help you offered!
Wait a minute, shouldn't my limits of integration be the length of the barrier?!?!? I think it must... lol makes sense. 



#18
Sep2407, 12:54 PM

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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 (VE) from the integral. Therfore [tex] \int dx (VE) = (VE) \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 


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