Potential Barrier (quantum mechanics)

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Homework Help Overview

The discussion revolves around a quantum mechanics problem involving potential barriers and transmission coefficients. Participants are examining equations related to the transmission of particles through a potential barrier, specifically focusing on cases where the energy of the particles is less than, greater than, and equal to the potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the calculation of the transmission coefficient T(E) for different energy conditions (EU, and E=U). There is confusion regarding the interpretation of T(E) and whether it should be multiplied by the total number of particles. Questions are raised about the implications of setting E equal to U and the resulting behavior of the equations.

Discussion Status

Some participants have provided guidance on the interpretation of T(E) and its application in calculating the number of transmitted particles. There is ongoing exploration of the mathematical limits and simplifications as E approaches U, with varying levels of agreement on the correctness of the algebraic manipulations presented.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available and the assumptions that can be made. The discussion reflects uncertainty regarding the behavior of the equations at the boundary condition where E equals U.

roam
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Homework Statement



I need help with part (iii) and (ii) of the follwoing problem:

http://img228.imageshack.us/img228/516/79620850.jpg

Homework Equations



From my notes transmission coefficient for E<U is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) sinh^2 (\alpha L) \right]^{-1}[/tex]

And for E>U it is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(E-U)} \right) sin^2 (k' L) \right]^{-1}[/tex]

Where [tex]k'=\frac{\sqrt{2m(E-U)}}{\hbar}[/tex] and [tex]\alpha =\frac{\sqrt{2m(U-E)}}{\hbar}[/tex].

The Attempt at a Solution



So for part (ii). I used the first equation and got a value of 4.59x10-4. This doesn't seem like a correct number for the number of neutrons. So do I have to multiply this with the total number of neutrons? :confused:

For (iii) I am confused and I don't know which equation to use. We know the equation when E<U and when E>U, but what exactly is the equation for when E=U?

I don't quite understand the hint, does this mean we need the following equation:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) \alpha^2 L^2 \right]^{-1}[/tex]

Is that right?
 
Last edited by a moderator:
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roam said:

Homework Statement



I need help with part (iii) and (ii) of the follwoing problem:

http://img228.imageshack.us/img228/516/79620850.jpg

Homework Equations



From my notes transmission coefficient for E<U is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) sinh^2 (\alpha L) \right]^{-1}[/tex]

And for E>U it is:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(E-U)} \right) sin^2 (k' L) \right]^{-1}[/tex]

Where [tex]k'=\frac{\sqrt{2m(E-U)}}{\hbar}[/tex] and [tex]\alpha =\frac{\sqrt{2m(U-E)}}{\hbar}[/tex].

The Attempt at a Solution



So for part (ii). I used the first equation and got a value of 4.59x10-4. This doesn't seem like a correct number for the number of neutrons. So do I have to multiply this with the total number of neutrons? :confused:
What does T(E) represent? If you understand that, it should be clear how to proceed.

For (iii) I am confused and I don't know which equation to use. We know the equation when E<U and when E>U, but what exactly is the equation for when E=U?

I don't quite understand the hint, does this mean we need the following equation:

[tex]T(E)= \left[ 1+\frac{1}{4} \left( \frac{U^2}{E(U-E)} \right) \alpha^2 L^2 \right]^{-1}[/tex]

Is that right?
Yes. Substitute in for ##\alpha## now.
 
Last edited by a moderator:
vela said:
What does T(E) represent? If you understand that, it should be clear how to proceed.

Thank you for your reply. I believe T(E) represents the probability of the particles being transmitted relative to the total incident particles. So isn't it correct then to multiply it by the total number of particles to see how many are transmitted?

In that case I will have (106)x(4.59x10-4)=459

Yes. Substitute in for ##\alpha## now.

But doesn't ##\alpha## equal to 0 when U=E?

##\alpha = \frac{\sqrt{2(1.675\times 10^{-27})(4MeV-4MeV)}}{1.054 \times 10^{-34}} = 0##

This will make the whole T(E) expression equal to 1, so that means that all the particles get transmitted? Is that right or did I make a mistake?
 
roam said:
Thank you for your reply. I believe T(E) represents the probability of the particles being transmitted relative to the total incident particles. So is isn't it correct then to multiply it by the total number of particles to see how many are transmitted?

In that case I will have (106)x(4.59x10-4)=459
Exactly, T(E) is the fraction that gets transmitted.



But doesn't ##\alpha## equal to 0 when U=E?

##\alpha = \frac{\sqrt{2(1.675\times 10^{-27})(4MeV-4MeV)}}{1.054 \times 10^{-34}} = 0##

This will make the whole expression equal to 1, so that means that all the particles get transmitted? Is that right or did I make a mistake?
No, remember the expression for T(E) is valid for E<U, so U-E isn't 0. Do the substitution and simplify, and then take the limit as E approaches U.
 
vela said:
Exactly, T(E) is the fraction that gets transmitted.

Thanks, so I guess it is correct to multiply T(E) by the total number of particles.

No, remember the expression for T(E) is valid for E<U, so U-E isn't 0. Do the substitution and simplify, and then take the limit as E approaches U.

I see. So here is what I did:

##\lim_{E \to U} \left( 1+ \frac{1}{4}\frac{U^2}{E(U-E)} . \frac{2m(U-E)}{\hbar^2} L^2 \right)^{-1}##

##\lim_{E \to U} \left( 1+ \frac{mU^2}{2 \hbar E} L^2 \right)^{-1}##

And since for small x we have (1+x)-1=1-x,

##\lim_{E \to U} \left( 1- \frac{mU^2}{2 \hbar E} L^2 \right)^{-1} = \frac{mU^2}{2 \hbar U} L^2##

Is this correct now?
 
No, you made algebra mistakes, and you shouldn't approximate using a series.
 
Oops that was a typo, here is what I've got

[tex]\lim_{E \to U} \left( 1+ \frac{mU^2}{2 \hbar^2 E} L^2 \right)^{-1}[/tex]

I have substituted the limits first and it is possible to take the inverse later:

##T(x)=\left( 1+ \frac{mU}{2 \hbar^2} L^2 \right)^{-1}##

Is that fine now?
 
Looks fine.
 
Alright, thank you so much for the help.
 

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