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Potential Barrier (quantum mechanics)

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data

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

    http://img228.imageshack.us/img228/516/79620850.jpg [Broken]


    2. Relevant 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].

    3. 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: May 6, 2017
  2. jcsd
  3. Sep 19, 2012 #2

    vela

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    What does T(E) represent? If you understand that, it should be clear how to proceed.

    Yes. Substitute in for ##\alpha## now.
     
    Last edited by a moderator: May 6, 2017
  4. Sep 19, 2012 #3
    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

    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?
     
  5. Sep 19, 2012 #4

    vela

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    Exactly, T(E) is the fraction that gets transmitted.



    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.
     
  6. Sep 20, 2012 #5
    Thanks, so I guess it is correct to multiply T(E) by the total number of particles.

    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?
     
  7. Sep 20, 2012 #6

    vela

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    No, you made algebra mistakes, and you shouldn't approximate using a series.
     
  8. Sep 20, 2012 #7
    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?
     
  9. Sep 20, 2012 #8

    vela

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    Looks fine.
     
  10. Sep 20, 2012 #9
    Alright, thank you so much for the help.
     
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