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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:

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

rob
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cmaster - reinstate monica
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  • Why deuterium in particular? – Jon Custer Jan 04 '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 04 '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 04 '19 at 14:34
  • The place to get data and references is https://www.nndc.bnl.gov/ensdf/DatasetFetchServlet – Jon Custer Jan 04 '19 at 14:37
  • The key thing about fusion is that it generates energy. Here one is getting a radioactive isotope that decays to a stable form with emission of alpha particles. So I guess most people would not want to call this fusion. – Dr S T Lakshmikumar Jan 04 '19 at 14:42
  • And, for reference, you will be hard pressed to find a larger fusion cross section than D-T (primarily because of a strong resonance in the He-5 nuclear structure). Yet that cross section is still a maximum of 5 barns. (And, no, in beam experiments a proton will not undergo a fusion reactions). – Jon Custer Jan 04 '19 at 14:42
  • Note that you are only getting a radioactive isotope since you added the energy in the first place to a non-radioactive isotope. No such thing as a free lunch... – Jon Custer Jan 04 '19 at 14:43
  • Semantics apart i guess use of slow neutrons would be even better for creating these isotopes since there is no electrostatic repulsion at all. – Dr S T Lakshmikumar Jan 04 '19 at 14:43
  • @DrSTLakshmikumar True enough, I had that thought myself. However, you don't find free neutrons in nature, while you do find heavy hydrogen. – cmaster - reinstate monica Jan 04 '19 at 14:57
  • @JonCuster As to the "free lunch": You do get the fusion energy, just retarded. You expend some energy to create the radioactive elements, but you get more energy back when those nuclei decay. It's somewhat similar to the CNO cycle (en.wikipedia.org/wiki/CNO_cycle ) that happens in stellar fusion, just with a much heavier catalyst. – cmaster - reinstate monica Jan 04 '19 at 17:09
  • If you start with Pb, where is the 'more energy back' coming from since Pb isotopes are the terminus of the major decay chains? – Jon Custer Jan 04 '19 at 17:14
  • @JonCuster Exactly. You start with Pb, you get Pb. In between you've added hydrogen, and you've got $\alpha$, $\beta$, and $\gamma$ radiation out. $\alpha$ radiation is helium ions, $\beta$ will annihilate to produce more $\gamma$, heating the target. Because the $2{^2_1H}$ input is heavier than the $^4_2He$ output, the output energy must be more than the energy that you expended in the acceleration. As a matter of fact, most of the reactions of the forms that I ask about have lighter products than inputs, the difference must end up as kinetic or radiation energy. – cmaster - reinstate monica Jan 04 '19 at 17:57
  • But all that energy came from your energetic particles in the first place... – Jon Custer Jan 04 '19 at 18:09
  • @JonCuster Yes, of course. Partly from their kinetic energy, and partly from their rest mass. That's the whole point of fusion: Munch some particles together to get a particle that's lighter than the particles that were munched together. The mass difference is eventually released as heat. Whether there is a $Pb$ nucleus involved in the process does not change the energy balance, as long as you get the $Pb$ back out unchanged. – cmaster - reinstate monica Jan 04 '19 at 18:32
  • But if you get the Pb back out unchanged, where is any energy coming from except from what you are putting in? Fusion/fission isn't some kind of magic. You can't start with Pb, end with Pb, and get excess energy out. – Jon Custer Jan 04 '19 at 18:35
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    @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 04 '19 at 18:43
  • So, yes, all your output energy comes from your (inefficient) input energy. I think I'm done commenting here... – Jon Custer Jan 04 '19 at 18:52
  • I agree with cmaster that it is in fact in theory possible to take out energy (as you do indirectly fuse deuterium to helium) but it is still quite probable that one could not won back all the energy expended by the inefficient acceleration. And this idea lacks one of the greatest joys of other fusion schemes: the lack of permanents radioactivity (other than neutron activation) – b.Lorenz Jan 04 '19 at 21:51
  • @b.Lorenz I'm happy that at least some people seem to understand my question :-) And, yes, the strongly radioactive target is a drawback. This is somewhat alleviated by the fact that there are no radiators with several thousand years half-life time in the relevant region of isotopes. Also, there would only be waste if we decided to throw away a target, which would be pointless to do. A target can be used indefinitely. On the plus side, this setup can be build in a relatively small laboratory. The effort is not even comparable to the huge experiments we need for ITER & co. – cmaster - reinstate monica Jan 05 '19 at 07:51
  • @cmaster I like this idea. Couldn't you reformulate your question so that it only asks for the cross sections, while not raising suspicion by a "fishy looking" (for some) fusion scheme? – b.Lorenz Jan 05 '19 at 09:51

2 Answers2

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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.)

proton on bismuth cross section

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.

rob
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  • 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 06 '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 06 '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 06 '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 07 '19 at 00:34
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binding energy curve

Fusion and fission regions.

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

u235fission

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.

If you start with lead and end with lead there is no binding energy per nucleon released.

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.

Please note that energy balances for nuclear reactions are not statistical, indvidual nuclei have to interact, in the example with uranium above, an individual neutron has to hit an individual uranium for the reaction to happen. Energy has to be balanced for each individual interaction.

anna v
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    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 04 '19 at 19:41
  • @cmaster but two deuterons do not hit one lead atom at the same time! Interactions happen one by one in releasing energy, and the individual lead that becomes a T which decays back to alpha and a lead has to take energy, the incoming deuteron energy, which it releasesback in the decay of your T. only in counting you have two deuterons into an alpha. In your scheme there is a one to one correspondence of a deuteron with a produced alpha. – anna v Jan 05 '19 at 03:52
  • 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 05 '19 at 07:32
  • @cmaster How on Earth can a deuteron deposit a neutron in one lead nucleus and a proton in another? Fissioning a deuteron isn't cheap. I guess the 1st Pb could absorb the deuteron & emit a proton, but I don't expect that proton to have the energy to penetrate another Pb. – PM 2Ring Jan 05 '19 at 09:30
  • @Chappo it is a copied plot, there is a link below – anna v Jan 05 '19 at 09:40
  • @cmaster sure alphas will come out statistically , the energy is not the energy given by fusion but the energy that the original deuteron carried, no gain in energy, interaction by interaction, no fusion. – anna v Jan 05 '19 at 09:42
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    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 05 '19 at 10:28
  • @cmaster it is energy from different deuterons distributed , it is not one reaction. or two consecutive ones. Your T nucleon gives off one alpha but it has been created from one (deuteron +lead) interaction. It is "fission" energy of the T. fission gives energy going from heavier to lighter. Whether the energy of this T decay is large or small has nothing to do with deuteron-deuteron fusion. – anna v Jan 05 '19 at 11:51
  • Yet, whenever I take any such sequence of reactions that yields back the same $Pb$ target nucleus, I find that the sum of the input particles (some count of deuterons) is heavier than the sum of the output particles (some protons, seldomly a neutron, but most importantly $\alpha$ particles). I repeat: This difference must go somewhere, and it goes into free energy. – cmaster - reinstate monica Jan 05 '19 at 13:36
  • yes, but it is not an energy of fusion, it is energy of the splitting T, which might be anything but very small, looking at the binding energy curve, with respect to deuterium fusion. – anna v Jan 05 '19 at 13:43
  • Ok, just try for a second to ignore the $Pb$. Just look at the black box, which is the target of the deuterion beam. Into this box go deuterions, out come some $^4_2He$ nuclei, some protons, a neutron every now and then, and a lot of $\gamma$ radiation and heat. Do you honestly want to tell me that it matters to the energy balance what's inside of this black box? As long as the black box is in a steady state, we can calculate $E_{out} = E_{in} + \frac{m_{in} - m_{out}}{c^2}$. – cmaster - reinstate monica Jan 05 '19 at 15:28
  • @cmaster in principle i agree. IF two deuterions do enter a Pb nucleus and emit an alpha particle, the excess mass has to be converted to energy either gamma rays or heat. Whether this isv fission or fusion is irrelevant. Given the track record of fusion which is always said to deliver twenty years from.now, it may not be a bad idea to investigate the posdiBility. But frankly getting deuterium from.nature is likely to be far more energy consuming than neutrons – Dr S T Lakshmikumar Jan 05 '19 at 15:45
  • @DrSTLakshmikumar Ok. Now consider the worst case, that four deuterons just deposit a single proton each. Input four deuterons: $m_{in} = 4\cdot 1875.612MeV = 7502.448MeV$, output four neutrons plus one $\alpha$: $m_{out} = 4\cdot 939.565MeV + 1\cdot 3727.379MeV = 7485.639MeV$. The mass difference is $m_{in} - m_{out} = 7502.448MeV - 7485.639MeV = 16.809MeV$. Ample amounts of excess heat that is generated by the reaction inside the black box. – cmaster - reinstate monica Jan 05 '19 at 16:03