Consider the following setup:
A beam of deuterons is accelerated towards a lead target. With sufficient acceleration (in the MeV range, nothing fancy...), I would expect some of the deuterons to fuse with the lead nuclei, producing bismuth, polonium, and astatine.

Looking at the table of nuclides, we see that these fusion reactions quickly reach a wall of very strongly radioactive isotopes. These are mostly $$\alpha$$ radiators, and those which are $$\beta$$ radiators decay into isotopes that will eventually reach the $$\alpha$$ radiators, assuming continued bombardment with deuterons.

So, in effect, this setup would input $$\rm^2H$$ ions, and output $$\rm^4He$$ ions (plus some $$\rm^1H$$ and $$\rm^1n$$ debris from not fully fused deuterons, I guess). This is nuclear fusion, catalyzed by the lead nuclei.

Comparing the lead target to a hydrogen target (beam of ions, gaseous target, paraffin wax, whatever), we see that the projectile ions lose much less energy when they scatter elastically off a lead nucleus than when they scatter off another proton. Simply because the lead nuclei are so much heavier. As such, each deuteron that's fired on a lead target will have several attempts at fusion before it has lost too much energy to allow interaction. As such, the cross section for the eventual $$\rm2\ ^2H\rightarrow{}^4He$$ reaction should be much larger for the lead target than for the hydrogen target.

Unfortunately, I was not able to find any previous work on this idea.

My question is: Has any research been done to measure the involved cross sections?

I'm especially interested in the reactions

\begin{align} ^2_1\mathrm H + {^n_m\mathrm T} &\rightarrow {^{n+1}_m \mathrm T} + {^1_1\mathrm H} \\ ^2_1\mathrm H + {^n_m \mathrm T} &\rightarrow {^{n+1}_{m+1}\mathrm T} + {^1_0 \mathrm n} \\ ^2_1\mathrm H + {^n_m \mathrm T} &\rightarrow{^{n+2}_{m+1}\mathrm T} \end{align}

where $$\rm T$$ is some relatively stable isotope of Pb, Bi, and Po. I guess that the first reaction will be the most important, and that the escaping proton will carry away most of the fusion energy, enabling it to further react with target nuclei. So I'm also interested in the reaction:

$$^1_1\mathrm H + {^n_m \mathrm T} \rightarrow {^{n+1}_{m+1}\mathrm T}$$

• Why deuterium in particular? – Jon Custer Jan 4 '19 at 14:12
• @JonCuster I guess that the cross sections of the Deuteron reactions might be significantly larger than those of the Proton reaction. This is just a hunch which might be proven wrong. If a naked proton is used, it has to tunnel through the potential barrier, and if it does, the resulting nucleus is activated enough to just reemit a proton. The neutron of the deuterium can tunnel from a fully bound state to another fully bound state with lower energy, and leave the energy difference for the proton to carry away. – cmaster - reinstate monica Jan 4 '19 at 14:29
• @JonCuster Also, the energy allowance for the Deuteron is much higher, even though the height of the potential wall surrounding the target is the same. Thus, the classical approach distance is shorter, and the tunneling probability should be higher. And don't forget that the escaping proton can still undergo fusion on its own. – cmaster - reinstate monica Jan 4 '19 at 14:34
• The place to get data and references is nndc.bnl.gov/ensdf/DatasetFetchServlet – Jon Custer Jan 4 '19 at 14:37
• @JonCuster You forget what we are bombarding the target with: $^2_1H$ ions. And what we get out is $^4_2He$ ions. The $Pb$ nucleus is really just the catalyst. And as such, it is expected to get the catalyst back out unchanged. – cmaster - reinstate monica Jan 4 '19 at 18:43

When I need cross sections, I use the Evaluated Nuclear Data File. I find lots of data for protons on lead. However, the data library for incident deuterons seems to only include low-mass targets. The open-access 2018 ENDF reference paper might describe the rationale behind these decisions, but I haven't read it carefully. My guess is that the literature on heavy-beam cross-sections doesn't systematically cover all energies. You have some detective work to do if you want to know more.

So that's a partial answer to your bolded question about cross-sections. However, your lead-catalyzed fusion idea has other problems that will prevent it from ever generating net positive power.

Let's skip the lead part and suppose that you start with a pure bismuth target. Furthermore, let's consider the "fairyland" reaction chain where your deuteron beam transfers a proton to one bismuth nucleus, and the remaining neutron gets moderated by the medium and captured by a different bismuth nucleus. The reaction chain is

\begin{align} && Q\text{-value (MeV)} && \text{delay (days)} \\ \rm^{209}Bi + d &\to\rm {}^{210}Po + n & 2.78 && \text{prompt} \tag1\\ \rm^{209}Bi + n &\to\rm {}^{210}Bi \to{}^{210}Po + \beta & 5.76 && 5.0 \tag2\\ \rm^{210}Po &\to{}\rm^{206}Pb + \alpha & 5.42 && 138 \tag3 \end{align}

I call this the "fairyland" reaction chain because it has a number of problematic issues, not least of which is that it's exceedingly unlikely that you would productively use both of the nucleons in your incident deuteron.

The total energy released in this chain is $$Q_\text{fairyland} = Q_1 + Q_2 + 2Q_3 \approx\rm20\,MeV$$. (I computed $$Q$$-values using the tabulated mass excesses.) Suppose for the sake of argument that you're able to drive this chain using a $$\rm1\,MeV$$ beam (which seems unlikely, since nucleon separation energies are typically 5-10 MeV). Then we come out ahead only if the fraction of the beam that undergoes this fairyland process is more than $$\rm(1\,MeV/20\,MeV) = 5\%$$. That seems ... improbable. If the fairyland process is more efficient with a 10 MeV beam, then you need 50% efficiency to come out ahead energetically, which is just silly. And those minimum estimates assume no other inefficiencies in producing your beam or extracting your heat, and neglect e.g. the delay of a few hundred days of continuous beamtime before your polonium concentration reaches secular equilibrium with the rest of the target.

You suggest starting with a lead target, and thinking of this process as catalyzed deuteron-deuteron fusion. The stable lead isotopes are 204, 206, 207, 208. (Lead-205 is stable from a practical point of view, with lifetime 20 My, but not from a geological point of view.) Single-nucleon transfer reactions on the lightweight lead isotopes make either a heavier stable lead isotope or a bismuth which decays to lead by electron capture. So with a lead target, you wouldn't be able to access the fairyland reaction chain above until you had converted a chemically significant fraction of the target to bismuth. Figuring out how much time this would take using a 1 MeV beam with a thermal power that wouldn't vaporize the lead is a cute little homework problem; it's longer than a person's career, even with optimistic assumptions.

In a comment you compare this to the CNO cycle in stars. That process has similar issues. It doesn't occur in our Sun, because the average temperature in the core isn't high enough. In fact, solar fusion is a pretty inefficient process overall. The power density in the Sun's core is only about $$\rm100\,W/m^3$$, which you could replicate in your house using a grid of incandescent light bulbs. Stars get away with being inefficient by also being enormous.

There's some overlap between your catalyzed-fusion idea and this other question.

After thinking about it some more, here's an estimate of the relative cross sections for these processes that's based on more than a gut feeling. It doesn't 100% match up with the assumptions above, but it'll give you a feeling.

(Unfortunately one of the references that I was looking for is temporarily offline due the current US federal government shutdown, so this is a little more ad-hoc than I'd like.)

Here's a Rutherford scattering calculator, for computing the scattering due to the long-distance electromagnetic interaction between the projectile and the nucleus. The Rutherford cross section diverges at forward angles, because "no deflection" is the mostly likely outcome of a charged particle passing through a thin foil. Picking a 10 MeV beam with charge $$Z=1$$, the cross section for scattering more than $$20^\circ$$ from a bismuth nucleus is about 40 barns. If your target were a one-micron foil (so that we don't have to worry about multiple scattering) then about $$10^{-4}$$ of your beam would scatter at this angle or more, so an order-of-magnitude estimate for scattering of the entire beam would be about a centimeter.

Here's a plot from the ENDF of the cross section for (p, anything) on a bismuth target. (I think. Sometimes I discover after the fact that the ENDF hasn't done what I want.)

That's probably a good order-of-magnitude proxy for (d,p) or (d,n), in that it levels off around a couple of barns. Unfortunately that's nowhere near the 50% of the Rutherford cross section that we decided above we would need to come out ahead with a 10 MeV beam and our deeply problematic "fairyland" process. Furthermore we're on the wrong side of the "corner" in the cross section at around 15 MeV, below which it falls precipitously. I read about 0.2 barn at from the plot for (p, anything) on bismuth at 10 MeV, but the curve is steep and the reactions we actually care about are different, so I wouldn't take that too seriously.

There are a bunch of hand-waving problems with my analysis here, and several places where I look at the things I've written and think "but what about..." However, fixing those issues at this point starts to get into the territory of including the cross-section data in some kind of simulation of the beam and target, which is more work that I'm willing to expend. I wish you good luck with it.

• I agree 99% with what you say, many of the thoughts I've had myself already. The 1% nitpick is just that you never lose the energy that you expend to accelerate the deuterons: This energy is converted 100% into heat, which you can capture again. Turning the heat into electricity again, however, is lossy, so we need a large surplus of energy that's created by fusion over the energy that we put in. Calculating with a heat to electricity efficiency of 20% and 22MeV being freed by creating one $\alpha$, we get an allowance of about 5.5MeV input per successful creation of one $\alpha$. – cmaster - reinstate monica Jan 6 '19 at 21:23
• I know that the required reaction rates seem improbably high. However, I'm still asking this question because: 1) Since a deuteron that scatters off a Pb nucleus without reacting retains most of its energy, it has quite a few attempts at merging before its energy drops too low for the neutron to tunnel into the nucleus. It's really a race between energy loss to scattering and successful reaction. 2) Irrespective of whether the final answer is "Yes, this works" or "No, this won't work, ever", I think it's worth considering this approach, and checking what the answer is. – cmaster - reinstate monica Jan 6 '19 at 21:41
• One final note: It would be nice if you just replaced "fairyland" with "proposed". I find the term "fairyland" a bit derisive for physics.stackexchange ... – cmaster - reinstate monica Jan 6 '19 at 21:48
• Hey, I apologize: I in no way meant to deride you. If I didn't think your question was interesting, I wouldn't have answered it. I used "fairyland" to indicate some unreasonable assumptions on my part --- mostly the capture of both nucleons from a single fast deuteron, but several subtler things as well. I don't think that recovering 20% of the waste heat really changes anything; I'm about to elaborate in the answer. – rob Jan 7 '19 at 0:34

Fusion and fission regions.

The normal way of talking , for example slow neutrons on Uranium, it is not called fusion, but fission.

The energy released is in the difference of the binding energy per nucleon,of the original nucleus with the product neuclei, with a small energy input of a neutron.

So, in effect, this setup would input $$^2H$$ ions, and output $$^4He$$ ions (plus some $$^1H$$ and $$^1N$$ debris from not fully fused Deuterons, I guess). This is nuclear fusion, catalyzed by the lead nuclei.

Count the energy: A helium atom has 4 nucleons, where does the extra mass come from if not from the energy of the incoming $$^2H$$ ?. So there is not a gain in energy, to be called either fusion or fission.
• A Deuteron weights 2.0141u, two of them are 4.0282u, an $\alpha$ particle weights 4.0015u. This is 0.0126u less than the two Deuterons that we are putting in. This is the source of the extra energy that is freed. Of course, the count of nucleons stays the same. And of course, the $Pb$ remains unchanged, it is a catalyst. It only takes part in the reaction, but it is not changed by it. – cmaster - reinstate monica Jan 4 '19 at 19:41
• One Deuteron may deposit one neutron in one $Pb$ nucleus, and one proton in another, yielding a heavy $Pb$ and a $Bi$. These are still stable. Another Deuteron may merge fully, yielding $Bi$ or $Po$. And a third Deuteron may deposit the 4th nucleon, yielding $Po$. If a surplus of neutrons get deposited, the nucleus will become a $\beta -$ radiator, if a surplus of protons get deposited, it's $\beta +$, both removing the surplus. It does not matter that these reactions happen one by one, once we get back to the original $Pb$ isotope, we have created at least one $He$ via $\alpha$ radiation. – cmaster - reinstate monica Jan 5 '19 at 7:32
• And where does the mass difference go, if not into free energy? The principles of energy conservation and energy-mass equivalence hold. So, this mass difference has to go somewhere, and kinetic energy + $\gamma$ rays are the only place it can go! – cmaster - reinstate monica Jan 5 '19 at 10:28