Barrier Tunneling Probability (Proton)

In summary, the probability that a proton will tunnel through a 14.0 fm-wide energy barrier with an energy of 1.30 MeV below the top is 0.09%. This was found by solving the integral for the probability flux and using the fact that the reflection coefficient is 1 and the transmission coefficient is 0 when the energy is below the barrier.
  • #1
bayan
203
0

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



[itex]η=\frac{\hbar}{\sqrt{2m(U_{0}-E)}}[/itex]

[itex]P_{tunneling}=e^{\frac{-2w}{η}}[/itex]

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.

[itex]w=1.4*10^{-14}[/itex] ∴ [itex]2w=2.8*10^{-14}[/itex]

[itex]m_{proton}≈ 1.67*10^{-27}[/itex] ∴ [itex]2m_{proton}≈ 3.35*10^{-27}[/itex]

[itex](U_{0}-E)=1.3MeV≈2.08*10^{-10}J[/itex]

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

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

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

[itex]P_{tunneling}≈e^{-221.63}[/itex]

[itex]P_{tunneling}≈5.6^{-97}[/itex]

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

Thanks in advance
 
Last edited:
Physics news on Phys.org
  • #2
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

ψI(x) = Aeikx + Be-ikx
ψII(x) = Ceiqx + De-iqx

Recall that ψ(x,t) = ψ(x)ψ(t), where ψ(t) = e-iωt. Hence, Aeikx and Ceikx 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.
 
  • #3
I forgot to mention that this is a first year one dimensional problem, would I still need to use the method you mentioned?

Thanks
 
  • #4
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%
 
  • #5
.Your solution seems to be correct. The probability of tunneling for a proton with energy 1.30 MeV below the top of a 14.0 fm-wide barrier is indeed very close to 0%. This is because the barrier is relatively wide and the energy of the proton is relatively low, making it difficult for the proton to tunnel through.
 

Related to Barrier Tunneling Probability (Proton)

1. What is barrier tunneling probability?

Barrier tunneling probability is the probability of a particle, such as a proton, passing through a potential barrier that it does not have enough energy to overcome classically. This phenomenon is described by quantum mechanics and is also known as quantum tunneling.

2. How is barrier tunneling probability calculated?

Barrier tunneling probability is calculated using the Schrödinger equation, which describes the wave function of a particle. The wave function is used to determine the probability of finding the particle at a certain position. The probability of tunneling through a barrier is then calculated by comparing the wave function on both sides of the barrier.

3. What factors affect barrier tunneling probability for protons?

The main factor that affects barrier tunneling probability for protons is the height and width of the potential barrier. A higher and wider barrier will result in a lower probability of tunneling. The mass and energy of the proton also play a role, with higher energies and lighter masses resulting in a higher probability of tunneling.

4. What is the significance of barrier tunneling probability in particle physics?

Barrier tunneling probability is significant in particle physics as it allows particles to pass through barriers that they would not be able to overcome classically. This phenomenon is important in understanding the behavior of particles at the quantum level and is essential in many areas of research, including quantum computing and nuclear physics.

5. How is barrier tunneling probability experimentally observed?

Barrier tunneling probability can be observed experimentally by measuring the transmission of particles through a potential barrier. This can be done using techniques such as scanning tunneling microscopy or electron tunneling spectroscopy. Additionally, theoretical calculations can be compared to experimental results to verify the accuracy of the predictions.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
271
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
892
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
2K
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
24
Views
2K
Back
Top