The Sun's energy comes primarily from fusion of light elements in its core.

It is estimated that a very small fraction of mass of the Sun (~$10^{-12}$ times the abundance of hydrogen) is uranium (both 235 and 238 isotopes). But given the huge mass of the sun (~$2*10^{30}$ kg), the mass of uranium in sun will come to around $2*10^{18}$ kg, which is again a significant quantity.

Since uranium is heavier compared to other elements in the Sun, I reckon that most of it will sink to core or at least be concentrated in inner regions of the Sun. This uranium would undergo spontaneous fission, but further, nuclear fusion taking place in the core releases a large amount of neutrons every second (which can cause fission of uranium nearby, similar to what happens in a multistage thermonuclear weapon).

Is there fission of uranium taking place in sun as well? What percent of Sun's energy output can be attributed to fission?

  • $\begingroup$ As you ask about the energy output of the sun by fission (of Uranium & consorts), I guess you are supposing this would be a source of energy. This is not the case. Any Uranium, very short lived in the heat and bombardment with protons and other particles, would have to be built up at first, drawing energy. So the energy via the Uranium route does not really come from Uranium fission, but is only delayed by creating Uranium as a first step. $\endgroup$ Commented Nov 11, 2023 at 15:34
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    $\begingroup$ @GyroGearloose …The sun's not big enough to make uranium, is it? Don't you need a supernova (kilo-, hyper-, uber-utlra-nova, whatever) for that? $\endgroup$
    – Will Chen
    Commented Nov 11, 2023 at 20:01
  • $\begingroup$ @WillChen it looks like I really overestimated the power of our sun, one, in creating heavy nuclei, and, two, in destroying them. Are there any processes in our sun that would accelerate the decay of uranium significantly? $\endgroup$ Commented Nov 12, 2023 at 15:55
  • $\begingroup$ @GyroGearloose Not an expert and haven't really looked into it recently, but IIRC there are not really any parameters that can affect the speed of radioactive decay. Other than radioactive decay, fissioning induced by free neutrons would destroy it too, but as ProfRob's answer points out, those should be exceedingly rare in the proton-heavy Sun. $\endgroup$
    – Will Chen
    Commented Nov 12, 2023 at 16:06
  • $\begingroup$ @WillChen I was thinking of high energy protons, like those that create diprotons and deuterium, that also might smash against any uranium and change or destroy it. But thinking twice, the outer layers of the sun might be too cool for that. $\endgroup$ Commented Nov 12, 2023 at 16:15

1 Answer 1


There must be spontaneous fission of uranium taking place in the Sun. But the long half-life of this process, combined with the very low abundance of uranium means that fission is a totally negligible fraction ($\sim 10^{-26}$) of the average solar output. The fission process is spontaneous rather than induced by neutrons, because there is no significant source of free neutrons in the Sun.

The energy produced by radioactive decay of uranium is far larger, because the lifetimes for radioactive decay are shorter, although the total output is still negligible - a fraction of about $1.4\times 10^{-10}$ of the solar output, averaged over its lifetime.


Spontaneous Fission

There will of course be some fission taking place in the Sun. That fission would have to be spontaneous since there is no plentiful supply of free neutrons inside the Sun. The spontaneous fission process itself is so slow, and the fissile material so sparse that any kind of "chain reaction" is not possible.

The Uranium does not get concentrated significantly towards the core. That is because the Sun is sufficiently well mixed and turbulent that any diffusion processes are rather slow. Asplund et al. (2020) estimate the primordial abundance fraction of uranium, and accounts both for its half-life to radiactive decay and for a modest amount (only 0.06 dex) of concentration towards the centre over its lifetime. They report initial fractional abundances (by number, with respect to hydrogen) of $10^{-12.65}$ and $10^{-12.16}$ for $^{235}U$ and $^{238}U$ respectively.

The primordial mass fraction of hydrogen is estimated to be 71%. Thus if we take 1 kg of the primordial solar material, then it contains about $4.25\times 10^{26}$ hydrogen atoms, $9.5\times 10^{13}$ $^{235}$U atoms and $2.9\times10^{14}$ $^{238}$U atoms.

The spontaneous fission decay constants, $\lambda$, of $^{235}$U and $^{238}$U can be calculated as $2.0\times10^{-19}$ year$^{-1}$ and $8.3\times 10^{-17}$ yr$^{-1}$ respectively.

The number of spontaneous fissions that have taken place over the lifetime of the Sun will be $$N_0[1 - \exp(-\lambda\ \tau)] \simeq N_0\lambda\ \tau\ ,$$ (assuming $\lambda\ \tau \ll 1$), where $N_0$ is the initial number of nuclei and $\tau$ is the age of the Sun ($\tau \simeq 4.57\times 10^{9}$ years). This is an upper limit because the uranium nuclei will be decaying other than by fission on shorter lifetimes. Each spontaneous fission ultimately releases about 200 MeV of energy.

Taking the mass of the Sun as $2\times 10^{30}$ kg, then the total number of decays is $3.8\times 10^{25}$ and $4.8\times 10^{28}$ for $^{235}$U and $^{238}$U respectively (i.e. the contribution from the light isotope is negligible because of its lower abundance and slower spontaneous fission rate). This number should probably be reduced by a small factor because the $^{238}$U has a radioactive decay half-life roughly equal to the age of the Sun.

The total energy released by U fission is therefore about $10^{31}$ MeV or $1.5\times 10^{18}$ J over the lifetime of the Sun at an average power of $\sim 10$ Watts.

This is roughly the amount of energy that the Sun releases in 4 billionths of a second!

Radioactive decay

Far more energy is generated from the radioactive decay of uranium into lead isotopes. The rate-determining step here is the half-life of the initial decays of the uranium isotopes ($\lambda = 9.8\times10^{-10}$ year$^{-1}$, and $1.55\times 10^{-10}$ year$^{-1}$ for $^{235}$U and $^{238}U$ respectively.)

The number of decays is given by the equation above (the approximation cannot be used here), thus 99% of the $^{235}$U and about 50% of the $^{238}$U has decayed.

Each decay chain (to lead) releases about 100 MeV of energy. Thus the total energy released by both isotopes is $(0.99\times 9.5\times 10^{13} + 0.5\times 2.9\times 10^{14})\times 2\times 10^{30} \times 100$ MeV $= 7.6\times 10^{33}$ J - an average of $5\times 10^{16}$ W over the solar lifetime, or about $1.4\times 10^{-10}$ of its total output, with the fraction declining from an initially higher value to a value lower than this today (as all of the $^{235}$U has decayed).

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    $\begingroup$ @Mithoron can you point me to exactly where that figure is given for a fission weapon, it is hard to find. I find that mass of U in the Sun to be about $3\times 10^{20}$ kg. $\endgroup$
    – ProfRob
    Commented Nov 11, 2023 at 14:09
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    $\begingroup$ @Mithoron the 7th row of the table says that the Ivy King bomb produced 2100 TJ (not 2 TJ) from 60 kg of hightly enriched uranium. Thus $1.8 \times 10^{26}$ nuclei, producing an average of just 73 MeV each, if they we all to fission. I used 200 MeV. Perhaps you've forgotten the factor that hardly any of the uranium undergoes spontaneous fission in the lifetime of the Sun? $\endgroup$
    – ProfRob
    Commented Nov 11, 2023 at 14:31
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    $\begingroup$ Oh, it's this comma vs dot thing, sorry, but that's giving 3 orders more! Most of U undergoes spontaneous fission in the lifetime of the Sun. Longest half-life of U is "only" 4.5 billion years. $\endgroup$
    – Mithoron
    Commented Nov 11, 2023 at 14:38
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    $\begingroup$ @Mithoron, that is the half life for radioactive decay. Got it. The question asks specifically about fission - so that it what I gave in my answer. I will add an answer based on radioactive decay which is more significant. $\endgroup$
    – ProfRob
    Commented Nov 11, 2023 at 14:46
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    $\begingroup$ @AndrewSteane the calculation referred to didn't account for the capture of free neutrons by protons, which I think reduces their numbers by 13 orders of magnitude. $\endgroup$
    – ProfRob
    Commented Nov 11, 2023 at 15:20

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