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I realize I'm trying to get a simple answer on a complicated subject, but here goes anyway. I've done some research and I understand (well, mostly) about how Vc is the amount of energy required to get two nucleons past the electric repulsive force so the strong force can take over. I know about the Maxwell-Boltzmann distribution, quantum tunneling, and the Gamow peak.

But all that is about how the Coulomb barrier is overcome by ambient temperatures in stars. The texts always go from how Vc is calculated, to how hot a star would have to be to overcome it, and how they aren't actually that hot, and why quantum tunneling comes to the rescue, yada yada, yada.

I want a more concrete understanding of what Vc really represents. If I had two protons and was using my patented Nucleon Energizer 3000, does Vc represent the amount of energy I'd have to set the MeV dial to so that I could force the two protons together past the Coulomb barrier and get them to fuse? And how much would my electric bill be afterwards?

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  • $\begingroup$ Which texts are you reading, to name a few? $\endgroup$
    – Qmechanic
    Commented Mar 6, 2014 at 22:02
  • $\begingroup$ I've been a lot of places. Mostly wikipedia and college websites. When I hit something new, I open a new tab and look it up. So I've been on a lot of pages. $\endgroup$
    – BLAMM
    Commented Mar 6, 2014 at 22:49
  • $\begingroup$ You may want to check out this website: hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html $\endgroup$
    – Tim D
    Commented Mar 6, 2014 at 23:11
  • $\begingroup$ Yes, I had found that site too. Lots of good information there. $\endgroup$
    – BLAMM
    Commented Mar 7, 2014 at 15:11

1 Answer 1

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To recap: a classic (not as in "classic versus quantum") picture is the one-dimensional model of $\alpha$ decay by Gamow and Gurney/Condon.

enter image description here

$Q$ represents the energy of the particle within the well, which in this example will be the "disintegration energy" of the system, i.e. the energy the escaping particles are seen to have after a decay.

$r<a$ represents the area within the nucleus, where the strong force is dominant. The attraction therein is represented by the negative potential $-V_0$ (where $V_0>0 \in \mathcal{R}$).

$a<r<b$ is the "classically forbidden region"; if we think of macroscopic objects such as a ball rolling up a hill, we intuitively know that a ball with total energy (potential+kinetic) $Q$ coming from e.g. the right will have depleted all its kinetical energy when it reaches the point $b$ on the hill, and will never be able to "jump" over the crest at $a$.

However, particles in the quantum mechanical region will "leak" some of their probability density through such barriers. In practice it means that in a fraction of the many attempts a particle bounces against such wall, it will actually penetrate and be measurably found on the other side.

The potential at $r>a$ is modeled on the Coulomb potential, which falls off as $1/r$. The "Coulomb barrier" is the hill that a particle (from either direction) faces due to this effect, that originates from the electromagnetic force.


"What does $V_\mathrm{C}$ really represent?" — this nomenclature most probably represents the height of the potential at $a$ in the picture above, which in the model is given by the electrostatic potential energy emanating from a point charge at $r=0$. What it implies is different in classical (CM) and quantum mechanics (QM).

In CM, it is an actual "hard" potential barrier that needs to be overcome to pass the point in space. The ball will never roll over the hill no matter how many times we try unless we give it energy to overcome the potential at $a$. Consequentially, if the ball has such an energy coming up against the hill, it will always roll over (if we go too far into the analogy we need to worry about friction and geometry and how that can be represented by a potential, so leave it at a theoretical stage for clarity).

In QM, it enters as part of the Schrödinger equation for the particle, from which we can decide the tunneling probability through the barrier. A higher (and wider) barrier means that the probability is (drastically) lowered and vice versa, but even if the particle has an energy above the barrier, there is always a non-zero chance that it will still not have passed it when we later measure the system.

For the case of $\alpha$ decay, there will be many "attempts" of the particle inside the nucleus to escape, so the lifetime will simply put be decided by the (inverse) product of the frequency with which the particle presents itself at the barrier, and the probability that it will tunnel through said barrier each time. I have an order of magnitude calculation available for $^{238}\mathrm{U}$, which puts the average probability of $\alpha$ tunneling at ${\sim}10^{-38}$, with "attempts" happening at ${\sim}10^{21}\,/\mathrm{s}$, to yield a lifetime of ${\sim}10^9\,\mathrm{y}$ (from Krane: Introductory nuclear physics).


So, to get your protons to fuse, the Coulomb barrier can be approximated by the electrostatic potential between two point charges outside the range of the strong force (${\sim}\mathrm{fm}$). If you put this into the Schrödinger equation, you get a tunneling probability describing how many of the collisions that will have the chance of fusing. Also included is the difference in binding energy per nucleon in the end product, and it is still a crude model to say the least (the diproton is not bound; also see later link on the proton-proton chain).

To accelerate a proton to $1\,\mathrm{MeV}$, you need to accelerate it through a field of electric potential $1\,\mathrm{MV}$, since a proton has a charge of $1\,e$, where $e$ is the elementary charge (an example that shows the motivation for using $\mathrm{eV}$ as an energy unit to begin with). To be certain of tunneling happening, you will need many collision attempts, decided by the calculations above. How much it will cost you in electricity bills depends on how you generate the accelerating field.

The answer to the core of your question is that there is not a "certain" energy where fusing will always take place in QM, but a continuous spectrum of probabilities depending on the potentials and energies involved. That is probably (aha) why the texts you read quickly turn to talking about energy distributions rather than distinct energies. If you set the dial on your machine high enough, the probability might eventually get close to $1$. In practice, other reaction channels might have taken over before that, and there is more to be said on the proton-proton chain and fusing, but that is another question.

For more details on the $\alpha$ decay model, one can start with e.g. Modeling alpha halflife (HyperPhysics).

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  • $\begingroup$ I think you hit the nail on the head with the classical vs quantum. I'm looking for a classical mechanics answer for a quantum mechanics question. But I like the "ball up a hill" analogy, and that gets me what I'm looking for. I also appreciate the sidebar on Alpha Decay. That creates a connection where I didn't have one before. The hardest part of all this is trying to create the "big picture" from all these bits of complicated facts. Thanks. $\endgroup$
    – BLAMM
    Commented Mar 7, 2014 at 15:25
  • $\begingroup$ @BLAMM: Great that you had use of the text! I just want to add that the "ball rolling up the hill" analogy is not something I have thought up personally, but how it was taught to me once, and it is probably a common analogy. Credit where credit is due. But yeah, paraphrasing some old physicist: "nobody understands quantum mechanics". $\endgroup$ Commented Mar 7, 2014 at 16:19

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