On a timescale of 1e11 years, all uranium and thorium will have decayed, but planets will still exist, orbiting red dwarf stars.

Would it be possible to use lead as a nuclear fuel? In principle, the curve of binding energy says lead fission is an exothermic reaction. Is there a way to do this while getting out more energy than you put in?

(Getting out more money than you put in, is of course a bigger challenge, but we are talking about a scenario where there is no competition from uranium or thorium let alone fossil fuels. I'm assuming the minimum size for a viable fusion reactor remains large compared to a few gigawatts.)

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    $\begingroup$ By that time I hope, humanity will know how to build a nuclear fusion reactor. $\endgroup$ Commented Feb 9, 2018 at 8:36
  • $\begingroup$ Its not a guarantee that all uranium and thorium will have decayed in the universe. If a star goes supernova it creates new isotopes at that time, and the half life counter starts for them then, not now. These things are a product of supernova events. Now I don't know when stars that go supernova will stop being created, but you would need to wait tell all those stars die. $\endgroup$ Commented Feb 9, 2018 at 8:47
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    $\begingroup$ @Oleg Fusion is the fuel of the future and will always be. $\endgroup$ Commented Feb 9, 2018 at 21:33

4 Answers 4



The issue, as you say, is getting out more energy then you put in. There is no isotope of lead in nature whose decay products will cause further fission, unlike our current nuclear fuels. That means that although an individual lead fission event yields energy that reaction won't propagate, leaving you at the mercy of spontaneous nuclear decay (and lead falls into a magical valley of nuclear stability).

That in turn means that to cause continual fission events you have to break apart the nuclei manually. This requires a lot of energy to do, and has to be done for every single atom of lead. You don't get a helping hand from other decay events (as you do with conventional nuclear fuels).

It's hard to envisage a process by which you can inspire fission in lead without spending more energy than you put in, and if you do have such a process then it would be easier to cause fission in any other element with an atomic weight above iron's (aside from possibly eka-lead, but we've only made that in a lab).

So my answer is no. I'm being simplistic (nuclear physics is a lot more complicated than I just made out), but I'm fairly confident the logic holds together.


In theory you could, see:


enter image description here

Lead is around the 200 mark in mass number on that graph, so energy can be released by fission.

Experimental work on Lead fission has been done. The problem, as others have highlighted, is that that kind of fission is accelerator-driven - basically, you bombard a lead target with neutrons of the appropriate energy, and this makes individual lead atoms undergo fission. You don't get a self-sustaining reaction.

Accelerator driven fission has been looked at as an energy source, and a way of trans-mutating nuclear waste - the lack of a chain reaction means that it can't run away as per Chernobyl. If you could make it work with lead to get a net energy gain is another matter, but if your reaction and neutron sources were highly energy efficient, then in theory you could use lead as a nuclear fuel.

  • $\begingroup$ This is a good answer, the question that I have now is though: Given that you have some efficient neutron source, is lead the way to go or are there better/more efficient/more economical (lead will remain pretty rare) uses? $\endgroup$
    – Raditz_35
    Commented Feb 9, 2018 at 15:01
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    $\begingroup$ Of the heavy elements, Lead is the most abundant (per Harper below, it's the end of many decay chains). Thallium and Bismuth may actually work better due to being less intrinsically stable (odd proton number). Digging up the data on fission cross-sections is a bit tricky.. $\endgroup$ Commented Feb 9, 2018 at 15:19
  • $\begingroup$ The problem with lead is you have this trainwreck of adjacent isotopes (205, 206, 207, 208) within lead. Getting the isotope you need is biblically difficult because the masses are 0.49% apart instead of 1.27% apart with U235/8, (which took the Manhattan Project). This is so difficult to do that reactors which breed both Pu239 and Pu240 are considered propagation safe. $\endgroup$ Commented Feb 9, 2018 at 19:03
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    $\begingroup$ @Harper I think "biblically difficult" and 1e11 years in the future are not contradictory =) $\endgroup$
    – Cort Ammon
    Commented Feb 9, 2018 at 20:34

You couldn't fission lead profitably, but you could transmute it into a fissile material.

Lead is the end (finis) of most heavy-isotope decay chains. That's why there's so much of it. While it may be possible to fission it, it won't be possible to profitably do so, it will be a net energy lose.* Lead is the nuclear equivalent of ash.

Likewise, iron is the end of most fusion. That's why there's so much of that.

If you want fissile materials, you will need to transmute other elements into a fissile material like U233, U235 etc. Decay is always exothermic (or it wouldn't happen). You would be spending energy to reverse the decay: synthetic un-decay.

For instance turning Bi209 into U233 will cost "only six" alpha un-decays instead of the seven needed to turn Pb207 into U235.

Decay is not fission. It's possible for un-decays to cost less energy and resources than fission gains in power, making the exercise profitable generally. It also makes sense to "unprofitably" make U233/5 for a specialty app only nuclear power can do, like a submarine or spaceship.

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    $\begingroup$ Lead is at the end of decay chains, but not fission reactions. When Uranium splits, you don't get lead, you get lighter elements like strontium and iodine and such. So lead could theoretically be split into smaller nuclei, releasing energy. $\endgroup$
    – kingledion
    Commented Feb 9, 2018 at 13:56
  • $\begingroup$ @kingledion thanks for the edit, I wrote it while drowsy. I upgraded the last half. Annoyingly, lead is not only the finis, it's also the only high mass element you'll have in OP's scenario, except for its big brother Bismuth, which saves one un-decay at least. $\endgroup$ Commented Feb 9, 2018 at 18:45

It seems to me that getting your reactor efficiency and neutron production high enough to profitably split lead is a much greater technological challenge than, say, creating (or capturing) and manipulating a miniature black hole.

This means that the best way to get energy from lead might be to drop it into a rotating black hole and collect the gamma rays etc from its acceleration in the accretion disk.


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