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I have come across many websites which state that proton-proton fusion, which is the dominant type of fusion that powers the Sun, is extremely slow; and that is why the Sun is still burning today. 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 is the D(euterium)-T(ritium) fusion for example compared to proton-proton fusion ?

ProfRob
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Abanob Ebrahim
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2 Answers2

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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.

John Rennie
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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 di-proton. 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. At the solar core, this reaction rate yields just 250 W m$^{-3}$.

For the deuterium-tritium reaction used in controlled fusion experiments on Earth, 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 $$ \frac{r_{DT} \times Q_{DT}}{r_{pp} \times Q_{pp}} \sim \frac{10^{40}\ 10^{-15}}{10^{64}\ 10^{-43}} \times \frac{18\ \mathrm{MeV}}{26\ \mathrm{MeV}} = 10^{4}$$

In other words, controlled nuclear fusion experiments (briefly) yield 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 in an Earth-based fusion reactor compared to the core of the Sun.

ProfRob
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