I have come across many websites that states that the proton-proton fusion which is the dominant type of fusion that powers the suns, is extremely slow and that is why the sun is still burning to that day. But also I have read that the sun fuses 620 million tons of Hydrogen every second, so that is considered fast for us on the earth scale.

So my question now is : comparing the time it takes to release the same amount of energy, how much faster the D-T fusion for example is compared to proton-proton fusion ?


2 Answers 2


The problem with proton-proton fusion is that there is no bound state of two protons. For the fusion to occur one of the protons has to turn into a neutron by beta plus decay. This is mediated by the weak force so it's a slow process and the probability of it happening while the protons are close enough to form a deuteron is very low. By contrast a deuteron and tritium nucleus readily form $^5$He.

The proton proton fusion cross section isn't known from experiment so only calculated values are known and these are somewhat uncertain. I found this paper that summarises the reaction rates. The p-p fusion is about $10^{26}$ times slower than the D-T fusion.

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    $\begingroup$ So if the reaction rate was as fast as the D-T, does that necessarily mean that the sun will fuse all its available fuel almost instantly ? $\endgroup$ Sep 12, 2013 at 9:21
  • $\begingroup$ If the p-p reaction was as fast as the D-T reaction stars would be much much smaller than they are. If you magically tweaked the reaction rate the Sun, and indeed every star, would immediately explode in something resembling a super nova. $\endgroup$ Sep 12, 2013 at 13:03
  • $\begingroup$ Actually the stars would be bigger because the interior temperature would be lower. $\endgroup$
    – ProfRob
    Jan 23, 2020 at 15:12

To small numerical factors, the fusion reaction rate is $r_{AB} \propto n_A n_B <\sigma_{AB}\, v>$, where $<\sigma_{AB}\, v>$ is the (temperature-dependent) "reactivity" for the reaction, formed from the averaging the cross-section over an appropriate Maxwellian velocity distribution, and $n$ are the number densities of the reactants.

The proton density at the centre of the Sun is about $10^{32}$ m$^{-3}$, the temperature is $1.5\times 10^{7}$ K. The initial (rate-determining) step in the p-p chain is the formation and subsequent beta decay of a diproton. The cross-section for this reaction is $\sim 10^{-23}$ barns and $<\sigma_{pp}\, v> \sim 10^{-43}$ cm$^{3}$ s$^{-1}$ at the solar core temperature.

For the deuterium-tritium reaction used in controlled fusion experiments, the temperatures are $\sim 10^{8}$ K and the number densities of the reactants $\sim 10^{20}$ m$^{-3}$ (roughly correct for the JET and ITER reactors). The cross-section of the reaction at this temperature is a few barns, much higher than the p-p reaction, and the reactivity is $<\sigma_{DT}\, v> \sim 10^{-15}$ cm$^3$ s$^{-1}$.

Putting these order of magnitude estimates together, the ratio of reaction rates multiplied by the energy released per reaction is $$ \frac{r_{DT} \times Q_{DT}}{r_{pp} \times Q_{pp}} \sim \frac{10^{40} 10^{-15}}{10^{64} 10^{-43}} \times \frac{18\ MeV}{26\ MeV} = 10^{4}$$

In other words, controlled nuclear fusion experiments (briefly) yielding roughly $2.5 \times 10^{6}$ W m$^{-3}$ Thus the energy released per unit volume and the reaction rate per unit volume are about 4 orders of magnitude larger/faster than in the core of the Sun.

  • $\begingroup$ And by the way, the number density of the deuterium-tritium doesn't look correct. Please take a look at it. $\endgroup$ Dec 23, 2015 at 14:06
  • $\begingroup$ @aAbanobEbrahim edited. It was $2.5\times 10^{10}. However, you were right about the number density. More like 1e20, so the rate comes down to 1e4 times solar. $\endgroup$
    – ProfRob
    Dec 23, 2015 at 14:50
  • $\begingroup$ How did you calculate this number density ?! I said the number density was incorrect because I think its too small rather than too large. My calculations for the number density of deuterium-tritium gives $6 \times 10^{28}$ $m^{-3}$, more than 6 orders of magnitude above yours. Your $10^{22}$ number would be correct if it was per cubic centimeter not per cubic meter. $\endgroup$ Dec 23, 2015 at 15:11
  • $\begingroup$ @AbanobEbrahim ITER and JET use $\sim 1$ mg of fuel per cubic meter. So $n_{D,T} \sim 0.5 \times 10^{-6}/(2.5\times 1.67\times 10^{-27}) \sim 10^{20}\ m^{-3}$. $\endgroup$
    – ProfRob
    Dec 23, 2015 at 15:20
  • $\begingroup$ That makes sense now. I calculated the number density for deuterium and tritium at their STP mass density values. So wouldn't this be more representative of a thermonuclear explosion rather than a fusion reactor ? Because that is what I had in mind when I asked this question. $\endgroup$ Dec 23, 2015 at 15:26

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