Scientific Basis: The concept relies on using fast neutrons from a deuterium-tritium (DT) fusion reactor to transmute mercury-198 into mercury-197, which then beta-decays into stable gold-197.
Feasibility: While the nuclear physics is sound, practical implementation depends on producing a sufficiently intense neutron flux—14.7 MeV DT fusion neutrons qualify.
Marathon Fusion’s Claim: The startup estimates that a 1 GW thermal fusion plant could yield several tonnes of gold per year via this method.
Methodology: Their estimates are derived from a digital twin—a computer simulation of reactor conditions and transmutation pathways. However, no physical reactor currently exists to validate the claim.
Challenges Highlighted:
Scaling neutron flux economically
Managing radioactive gold waste initially produced
Competing technical hurdles in commercial fusion
Potential omission of critical physical effects in the modeling
Conclusion: The idea is theoretically plausible but highly speculative until a working fusion reactor can test it. The economic viability and safety implications remain open questions.
So, they’re starting from a nonexistent feedstock. Mercury-198 is only 10% of natural mercury. It needs isotopic separation—a process that, per unit weight, would cost more than gold itself.
I agree, but maybe they plan to use natural mercury and extract the transmuted gold from whatever else (radioactive crud?) builds up—a much less expensive process.
Regardless, Marathon Fusion’s claim that a 1 GW (thermal) power plant can transmute 5 tonnes of gold per year, without compromising fuel self-sufficiency, is absurd. Let’s do the math.
Given:
Mass = 5,000 kg = 5,000,000 g
Molar mass of gold (Au) = 196.97 g/mol
Avogadro’s number = 6.022 × 10²³ atoms/mol
Number of gold atoms = 1.53×10^{28}
Given:
Energy per DT reaction: 14.7 MeV = 2.35×10^{-12} J
Total energy = (1.53×10^{28})(2.35×10^{−12})=3.6×10^{16} J
Given:
Number of seconds in year = 31557600
Power = 3.6×10^{16} J / 31557600 s≈1.14×10^{9} Watts
So, 114% of the total yearly neutron production from a 1 GW (thermal) power plant would be required to make 5 tonnes of gold. This requires a neutron multiplier (mercury is not). And what about the neutrons needed to breed more tritium?
The plot below shows relevant ENDF‑derived cross‑section data. Consider the 8.5-14 MeV range. It looks like the total neutron-reaction cross section of Hg is ~5 barns, and the (n,2n) cross section looks like ~0.9 barns. Because Hg nuclii are ~200 times more massive than neutrons, it takes about 60 elastic collisions to slow a 14 MeV neutron down to 8.5 MeV. However, because each of these collision has a ~0.9/5 probability of undergoing (n,2n), almost all of the neutrons undergo (n,2n) before they can slow below 8.5 MeV (the cut-off energy for neutron multiplication). That is why Hg is considered to be a neutron multiplier.
Aside from ^{198}Hg(n,2n)\rightarrow^{197}Au, the only other way to make gold from mercury is ^{196-204}Hg(n,p)\rightarrow^{196-204}Au. However, all of the Hg(n,p) gold beta-decays (half-life < 7 day) back to the original mercury isotope.
No problem. These radioactive isotopes of gold have half-lifes shorter than 7 days. Waste older than two months will have already decayed by at least 8 half-lifes, leaving less than 1/2^8 of the original waste.
How much mercury is needed to prevent neutrons from escaping the blanket?
Because of mercury’s large atomic mass (A) = ~200.59, it will take hundreds of elastic collisions to thermalize a DT fusion neutron. The fraction of energy lost per elastic scattering event is only \frac {2A}{(A+1)^2} = 0.0099. If the MFP of each collision is ~4.94 cm, the blanket might need a mercury-thickness of more than a meter.
You forgot the Hg(n,inl) and Hg(n,el) data. I was about to post it myself, but then I thought—didn’t Marathon Fusion simulate this using MCNP on a digital twin? If so, that would be far more accurate. Someone should ask them to post their data.
I used OpenMC to simulate the expected gold & tritium production (per DT fusion neutron originating at the center of a 1 meter radius sphere containing various amalgams of Hg:Li). The simulations were conducted over a range of Hg/(Hg+Li) ratios, using either natural isotopic abundances for both elements or isotopically enriched (100% Li6 & 100% Hg198).
The plot shows that a blanket of natural Hg & natural Li will not work. It fails because the TBR drops below 1.2 for Hg/(Hg+Li) ratios above 0.2, but the Hg/(Hg+Li) ratio needs to be above 0.2 if one wants to prevent more than 30% of the fusion neutrons from escaping the blanket.
Mercury is extremely toxic. It has a high vapor pressure and a low boiling point (357 C), making it a poor choice for a high-temperature coolant material. On top of that, it would need to overcome an extremely high regulatory burden.
Summary of Key Points
Mean Free Path Formula
Given Values
Step-by-Step Calculation
