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Homework Help: Units of the given potential box

  1. Nov 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A car (particle) with mass m = 1 t drives into a barrier with a height of 1 m and a thickness of 3 m. The kinetic energy shall be sufficient to get classically over a barrier with half of the height. Derive an equation for the tunnelling probability. What is the probability to tunnel through the barrier?

    2. Relevant equations
    Tunneling probability is given by

    (i) T = (|A'|/|A|)^2 = (1−E/Vo)/[1 − E/Vo + (Vo/4E) sinh2(bα] ; where

    α = √[2m(Vo − E)]/(h/2π) , b=3 m the barrier thickness, Vo: barrier potentail and E: particle's (car's) energy.

    This is an experimental physics homework. So I should be getting some numbers at the end. I have the equation (i) and I know how to use it. But, I'm not sure what "the potential barrrier to having a height of 1 m" mean unit-wise. In other words, I don't know which units I should use for the energy while calculating α? Feels like I should be using Joules but it's a just an intiution. If so, why?

    Apologies for not being able to use LaTeX.
  2. jcsd
  3. Nov 27, 2016 #2


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    Hello pheno, :welcome:

    In SI energy is in Joules. Doesn't give a very high speed, but I think that's what the composer of the exercise means.
  4. Nov 27, 2016 #3
    So you suggest I use Vo = 1 J? If I do that, I will get

    bα = b*√[2m(Vo − E)]/(h/2π) = 3√(2*1000*(1-1/2)/(1.054*10^-34)) ≈ 10^36. With Vo = 1 J, E = 0.5 J the equation (i) becomes


    New problems arise:
    1) WolframAlpha didn't calculate (sinh(10^36))^-2 so I did a Taylor expansion (just to get an idea about the value I should have):

    (sinh(bα))^-2 = (bα + (bα)^3/3! +o(5))^-2 ≈ 10^-72 ≈ 0.

    This could have been okay I guess, saying the probabilty of the car to tunnel is nearly zero.

    2) However when I do the following to check my conclusion, I get something completely different.

    For bα>10, we can approximate sinh(bα) = (1/2)(e^bα - e^-bα) ≈ (1/2)e^bα. Then 1/(sinh(bα))^2 ≈ 1/((1/4)e^(2bα)) = 4*e^(-2bα) = 4* e^(-2*10^36) ≈ 4.

    This is clearly wrong, since T cannot be bigger then one. Hence, I'm stuck again.
  5. Nov 27, 2016 #4


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    Hint: mgh
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