Units of the given potential box

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SUMMARY

The discussion focuses on calculating the tunneling probability of a car (mass m = 1 t) attempting to overcome a barrier with a height of 1 m and thickness of 3 m. The relevant equation for tunneling probability is T = (|A'|/|A|)^2, where α = √[2m(Vo − E)]/(h/2π). The user initially considers using Vo = 1 J and E = 0.5 J, leading to a calculation of bα ≈ 10^36. However, discrepancies arise in the calculations of T, particularly when approximating sinh(bα), resulting in conflicting conclusions about the tunneling probability.

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Homework Statement


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?

Homework 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.
 
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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.
 
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

T=1/(sinh(bα))^2.

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 probability 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.
 
Hint: mgh
 

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