Barrier Tunneling Probability (Proton)

[bayan]

Homework Statement

A proton's energy is 1.30 MeV below the top of a 14.0 fm-wide energy barrier, What is the probability that the proton will tunnel through the barrier?

Homework Equations

η=\frac{\hbar}{\sqrt{2m(U_{0}-E)}}

P_{tunneling}=e^{\frac{-2w}{η}}

The Attempt at a Solution

I have got a probability of using the following method. Just want to check if it is the wright answer.

w=1.4*10^{-14} ∴ 2w=2.8*10^{-14}

m_{proton}≈ 1.67*10^{-27} ∴ 2m_{proton}≈ 3.35*10^{-27}

(U_{0}-E)=1.3MeV≈2.08*10^{-10}J

{\sqrt{2m(U_{0}-E)}}=8.35*10^{-19}

η=\frac{\hbar}{\sqrt{2m(U_{0}-E)}}=1.26*10^{-16}

P_{tunneling}=e^{\frac{-2.8*10^{-14}}{1.26*10^{-16}}}

P_{tunneling}≈e^{-221.63}

P_{tunneling}≈5.6^{-97}

The answer is 0% chance of tunneling ( is this wright or have I made a mistake somewhere?)

Thanks in advance

 

[Harrisonized]

I actually just worked out a similar problem a few days ago. The probability flux is given by

j = -iħ/(2m) ∫ψ*(dψ/dx)-(dψ*/dx)ψ dx

where ψ* is the complex conjugate of ψ. When solving something like this, you should split up the potential into different regions. A barrier potential is a superposition of an upward step potential and a downward step potential. Therefore, we should just solve it for the upward step potential. We can split the upward step potential into two regions. One for V(x)=0 and one for V₀=0.

The general solution to this is

ψ_I(x) = Ae^ikx + Be^-ikx
ψ_II(x) = Ce^iqx + De^-iqx

Recall that ψ(x,t) = ψ(x)ψ(t), where ψ(t) = e^-iωt. Hence, Ae^ikx and Ce^ikx correspond to waves belonging to a particle going from left (-x) to right (+x), and Be^-ikx and De^-ikx correspond to waves belonging to particles going from the right (+x) to left (-x).

Anyways, if you set up the problem such that the particle comes from one side of the potential (for example, the particle is coming in from the left), then you can eliminate one of the variables (in the example case I gave, D=0). This is because a positive D belongs to a wave coming in from the +x direction, which is impossible if all the particles are coming in from the left, and reflection can only occur at the boundary.

Put this into the flux equation and solve. You'll end up with something like

1-|R|² = T²

As you can guess, |R|² is the reflection probability and T² is the transmission probability.

I'll leave it to you to solve the integral and obtain |R|² and T². After solving the integral, you'll find that whenever the energy is below the barrier, the reflection coefficient is 1 and transmission coefficient is 0.

If the particle cannot pass through the potential step, then there's no chance that it will reach the other end of the potential barrier. Although we must be careful, since the wave function exists on the other side of the potential barrier (as well as inside the potential barrier). There is a finite probability of detecting the particle on the other end of the barrier. There just isn't a probability that the particle will actually tunnel through.

 

[bayan]

I forgot to mention that this is a first year one dimensional problem, would I still need to use the method you mentioned?

Thanks

 

[bayan]

Worked it out using first method as it was only a one dimensional problem. For some reason I had 1.3MeV=1.3*10^9 eV instead of 1.3*10^6 eV

changing that gave me a probability of 0.09%