8
$\begingroup$

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 ?

$\endgroup$

2 Answers 2

12
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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 ? $\endgroup$ Commented 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. $\endgroup$ Commented Sep 12, 2013 at 13:03
  • $\begingroup$ Actually the stars would be bigger because the interior temperature would be lower. $\endgroup$
    – ProfRob
    Commented Jan 23, 2020 at 15:12
5
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ And by the way, the number density of the deuterium-tritium doesn't look correct. Please take a look at it. $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented Dec 23, 2015 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.