Fusing Iron Nuclei: Ignition Temperature

[hawkingfan]

I am aware that fusing two iron nuclei together is highly endothermic. But at what temperature does it begin. Does anybody have a clue? I know that fusing two hydrogen nuclei begins at 10 to 14 million Kelvins and that helium fusion begins at 100 million Kelvins. It wouldn't surprise me that iron would be orders of magnitude higher than this (probably in the hundreds of trillions but that's just a random estimate) I don't know enough about chromodynamics or Fermi-Dirac statistics to do the calculation which would allow me to determine this temperature. I tried calculating the fusion temperature of hydrogen using Maxwell-Boltzmann statistics and got a temperature on the order of 10^10. The sun's interior isn't even that hot. I know that it's the quantum tunneling effect that allows fusion at such a low temperature. If anybody reading this is adept enough to calculate the theoretical ignition temperature of iron, can you please write the number.

And no, this is not homework so please don't report it. I'm working on a personal invention/future experiment where the calculations and concepts are too far out of bounds given my current knowledge on relativistic QFT.

 

[bcrowell]

The Coulomb barrier is $kZ^2e^2/r$, where k is the Coulomb constant and r is twice the radius of an iron nucleus (roughly 1 fm multiplied by A^1/3). Calculate this energy and divide by the Boltzmann constant and you have a rough estimate of the minimum temperature. I get about 10^12 or 10^13 K.

[EDIT] Hmm...if I do this for hydrogen fusion, I get about 10^10 K, which is much too high compared to the actual core temperature of the sun, about 10^7 K. I can't believe that tunneling allows fusion at 1000 times less than the Coulomb barrier, so it seems like something is wrong with my calculation.

[EDIT] OK, here's some info: http://burro.cwru.edu/Academics/Astr221/StarPhys/coulomb.html So counterintuitively (for me, at least), the combination of tunneling and a high-energy tail in the Maxwellian distribution does allow fusion to proceed at a temperature corresponding to an energy 1000 times less than the Coulomb barrier.

 

[bcrowell]

Having thought about this some more now, I think I understand what's going on. This estimate requires fairly delicate treatment, the reason being that approximations that work well for fusion of heavy nuclei don't work as well for fusion of light nuclei. For heavy nuclei, I know from experience that dropping the energy even 10-20% below the Coulomb barrier results in a reduction of the tunneling probability by many orders of magnitude. However, this clearly doesn't carry over to hydrogen-hydrogen fusion in the sun.

A more detailed calculation can be done using the Hill-Wheeler approximation to the tunneling probability, which is derived by applying the WKB approximation to a barrier that's shaped like a concave-down parabola. The parabola is parametrized by the height of its top, E_C, and by a parameter $\hbar\omega$ with units of energy, which is defined as the energy quantum of a harmonic oscillator whose potential is the inverted form of the parabola. Basically $\hbar\omega$ measures the width of the barrier. The Hill-Wheeler approximation becomes (for energies significantly below the barrier)
[tex]T \sim \exp\left[-\frac{2\pi}{\hbar}(E_C-E)/\omega\right][/itex].
Taking a nuclear potential plus a Coulomb potential, I get a rough estimate of
[tex]1/\omega=m^{1/2}Z^{1/2}e^{1/2}k^{1/4}a^{-3/4}[/itex],
where m is the reduced mass, k is the Coulomb constant, and a is the slope of the sides of the nuclear potential, which is ~ 1 MeV/fm. The result is that the tunneling probability scales like
$$T\sim\exp\left[-(\ldots)(E_C-E)m^{1/2}Z^{1/2}\right]$$.
If you take m and Z both to be proportional to the mass number A, this becomes
$$T\sim\exp\left[-(\ldots)(E_C-E)A\right]$$.
This shows why the tunneling probability is so much higher for light nuclei than for heavy ones. I think this is basically the correspondence principle at work. The large-A limit is basically the classical limit of quantum mechanics, where tunneling probabilities have to go to zero. So I think our estimates of E_C are about right, and because iron has a fairly big A, you really can't tunnel in at energies significantly below this with any decent probability. I assume this is why elements heavier than iron are only made in supernova explosions, not in the cores of stars.

If I write the tunneling probability in the form $\exp[-(E_C-E)/b]$, then I get b~1 MeV for deuteron-deutron fusion, but b~0.02 MeV for symmetric fusion of medium-mass nuclei with A~100 and Z~50. So in the d-d case, b is comparable to the Coulomb barrier itself, and you can get appreciable fusion probabilities at energies that are very small. But in the iron-iron case, the tunneling probability is going to come out ridiculously small if you're more than a few hundred keV below the barrier.

 

[Nik_2213]

IIRC, Iron/Iron fusing doesn't occur until you have a large, dying star collapsing in the moments prior to a supernova 'bounce'...

 

[hawkingfan]

bcrowell, that is exactly what I did the first time. First I used classical kinetic theory for deriving the Coulomb barrier. The second time, I took some more quantum mechanical factors into consideration. I still don't quite know the exact temperature range but I'll just go with the classical approximation. I'm trying to estimate how much thermal energy must be supplied to a cubic micron of iron (at room temperature) to initiate fusion. Since the classical approximation gives an upper limit, I'll work with that. It's better to overestimate your liabilities than underestimate them. (and yes, I'm assuming a near-steady heat capacity)

 

[bcrowell]

bcrowell, that is exactly what I did the first time. First I used classical kinetic theory for deriving the Coulomb barrier. The second time, I took some more quantum mechanical factors into consideration. I still don't quite know the exact temperature range but I'll just go with the classical approximation. I'm trying to estimate how much thermal energy must be supplied to a cubic micron of iron (at room temperature) to initiate fusion. Since the classical approximation gives an upper limit, I'll work with that. It's better to overestimate your liabilities than underestimate them. (and yes, I'm assuming a near-steady heat capacity)

You're welcome.