# Thermodynamically speaking, in the absence of proton decay will matter decay into Iron-56 or Nickel-62?

I understand there are already questions with similar premises but neither of them actually have the answer to the question I have. Thus I'd like to split my question into two more specific sub-questions,

A. Is the chain of reasoning I use sound, in that there are no mathematical errors, the logic used is valid, and the initial assumptions accurate?

B. Does that chain of reasoning imply that Ni-62 is favored over Fe-56?

The general consensus seems to be that in the absence of proton decay matter will transmute into Iron-56, the isotope with the lowest mass per nucleon, via quantum tunneling over extremely long timescales. In the far future it is believed that this will turn all matter into "iron stars". This seems to have been popularized in the 1978 paper "Eternity is Unstable" by John D. Barrow and Frank J. Tipler as well as the 1979 paper "Time without end: Physics and biology in an open universe" by Freeman J. Dyson of Dyson sphere fame. However when I attempted to do the math myself I couldn't make the numbers add up in a way where decay to Fe-56 was the most energetically favorable option, instead indicating that Nickel-62 would be more favorable. Wikipedia justifies the transmutation of Ni-62 into Fe-56 by claiming that the conversion of 28 Ni-62 nuclei into 31 Fe-56 nuclei would release 0.011 Da of mass-energy. This seems to be approximately true, using Wikipedia's values for nuclei mass

$$(61.9283449\ Da*28)-(55.9349375\ Da*31)=0.01059\ Da=9.8645\ MeV/c^2$$

(Wolfram link) However the reaction as given is impossible, as 28 atoms of Ni-62 has a total of 784 protons (28*28) while 31 atoms of Fe-56 has a total of 806 protons(31*26). Thus the end product has an excess of 22 protons and thus an excess positive charge of 22 $$e$$. In order for this reaction to take place either conservation of charge must be violated, 22 nearby positrons must be absorbed in a process analogous to electron capture, or 22 electrons must be produced to balance the excess charge. The first is almost certainly impossible given our current understanding of physics, the second requires the contrived and improbable presence of 22 positrons that somehow get close enough to be absorbed despite the fact that they should be strongly repelled from positively charged nuclei, leaving the third option the only viable possibility. The electron has a mass of $$510.9989461 \ keV/c^2=5.48579905×10^{-4}\ Da$$. (Wolfram link)

$$5.48579905×10^{-4}\ Da*22=0.0120687579\ Da$$ which is more than 0.01059 Da, meaning that the end result of 31 iron atoms and 22 electrons actually outweighs the initial 28 nickel atoms by 0.00147 Da or 1.369 MeV/c^2, making the reaction endothermic and implying that the reverse reaction, of 31 iron atoms capturing 22 electrons and turning into 28 atoms of nickel, is the energetically favorable route.

In addition the calculation only considers the bare mass of the nuclei without accounting for the electron clouds that surround them. The negative gravitational binding energy of a spherical mass is

$${3 \over5}{G M^2 \over R}$$

while the positive potential energy of a uniformly charged sphere is

$${3 \over5}{k_e Q^2 \over R}$$

thus we can expect electrostatic force begin to overwhelm gravity when

$${3 \over5}{k_e Q^2 \over R}={3 \over5}{G M^2 \over R}$$

some rearrangement gives us

$${k_e Q^2}={G M^2}$$

$${Q^2\over M^2}={G \over k_e}$$

$${Q\over M}=\sqrt{G \over k_e}$$

$${Q}=M*\sqrt{G \over k_e}$$

setting M to 1 kg gives us

$$Q = 86.18 pC$$

This means a charge excess or deficit of 86.18 pC or more per kilogram should be enough to disassociate any gravitationally bound body. For a star composed solely of hydrogen this corresponds to an excess or deficit of one electron per 1.11*10^18 hydrogen atoms, or less than one part in a quintillion. Thus we should expect any realistic system to have a nearly equal number of protons and electrons to balance out the charge with only negligible charge imbalances. As such we can expect each nucleus to have a corresponding electron cloud surrounding it to equalize its charge, a cloud which would have a non-negligible impact on the mass of the system. Modifying our measurement of mass/nucleon from nucleus mass/number of nucleons to nucleus mass + electron cloud mass/number of nucleons seems to be a reasonable change.

The mass per nucleon of Fe-56 is thus $$(55.9349375\ Da+26*5.48579905×10^{-4}\ Da)/56=0.99909287 \ Da$$ (Wolfram link)

While the mass per nucleon of Ni-62 is then $$(61.9283449\ Da+28*5.48579905×10^{-4}\ Da)/62=0.99909202 \ Da$$ (Wolfram link)

The difference being $$0.99909287\ Da-0.99909202\ Da = 8.5 × 10^{-7}\ Da = 791.8\ eV/c^2$$ in favor of Ni-62.

The difference is tiny and influences on electron ionization energies, low temperature effects, pressure, gravitational effects or some combination thereof might tip the scales in favor of iron. However without further analysis it is just as probable that those effects will magnify the energy difference. I have been unable to find any independent derivation of what isotope matter would tunnel into, only references to previous material which also assume Fe-56 is the natural end state of matter.

To restate my original sub-questions,

A: Is there anything wrong with my calculations and logic? And,

B: Does that reasoning point towards a future dominated by nickel stars instead of iron ones?