This is just theoretical consideration. I expect that size of the reactor would be impractical to build on Earth, but I'm interested how much. EDIT: perhaps it could be good for some huge spacecraft in far future, this thought was my original motivation

Consider that you do not want neutrons to take away energy from your nuclear reactor because you want to use some form of direct energy conversion with high power density and Carnot efficiency. There are in principle 2 sulutions:

  1. Use aneutronic nuclear reaction ( like $^1$H+$^{11}$B or $^3$He+$^3$He ). However, there the ratio between Bremsstrahlung power loss $P_{Xray}$ and fusion reaction power $P_{Fusion}$ is just $P_{Fusion}/P_{Xray} \approx 0.5 ... 0.8 < 1 $. In such case you would like to confine bremsstrahlung inside your plasma. So the plasma has to be optically thick for X-ray of energy $0.1-10 MeV$. ( let aside non-equlibrium concepts like Inertial electrostatic confinement which are controversial )
  2. Use $^3$H+$^2$H reaction which has the ratio $P_{Fusion}/P_{Xray} \approx 140$. However it produces 80% of energy in form of fast $ 14 MeV $ neutrons. So you can ask how big should be the reactor to be optically thick for neutrons of this energy.
  • $\begingroup$ You already have a very nice, long lived and COMPLETELY FREE reactor of that kind. Go outside during the daytime and look into the sky. Just don't stare at it directly, it's too bright and can hurt your eyes. You do want the neutrons to come out, by the way, without them you won't have the tritium to fuel a practical reactor... unless you want to make fusion reactors an energetic appendage to tritium breeding fission reactors? That's probably what would happen with practical fusion, anyway... IF it was economical, which it is not. $\endgroup$
    – CuriousOne
    May 1, 2015 at 8:27
  • 5
    $\begingroup$ Please no flame-war. In context of spaceship propulsion is high power density (per mass) important, which Sun does not provide. Also it is hard to convert Sun power to propellant exhaust velocity ~1000 km/s, regulate the power or turn it on/off. I guess optically thick reactor can be still possibility for some large spacecraft ( say several kilometres in size ) $\endgroup$ May 1, 2015 at 9:11
  • $\begingroup$ I didn't intend to give you a flame war nor am I going to give you one now. With respect to spacecraft reactors, in case of continuous operation, fission is technologically perfectly adequate and promises higher power density than fusion. It is thermodynamically limited by the radiative cooling of the reactor, which would be an identical problem for both fission and fusion designs. For pulse mode reactors laser or ion-beam driven inertial fusion is a perfectly fine method of operation. $\endgroup$
    – CuriousOne
    May 1, 2015 at 9:22
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    $\begingroup$ I like fission, yet I disagree. The power densities required are HUGE ( like >1MW/kg of whole ship ). With electricity power generation as intermediate it is not feasible. Radiative cooling is effective at high temperatures ( > 5000 K ) where no solid material survives. The propellant exhaust itself is an effective cooling (of thermal rocket). So propellant exhaust directly from reactor core make sense. And fission gas-core or plasma-core nuclear reactors are not good, because low neutron cross-section of gas/plasma. There fusion make sense (more than on Earth) $\endgroup$ May 1, 2015 at 11:57
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    $\begingroup$ Orion project is nice, but it has also many problems ( nuclear weapons are quite complex and expensive, does not have specific impulse as good as pure fusion etc. ). As I said - cooling is not a problem if you (1) cool it by exhaust of propellant itself (that is how chemical rocket engine achieve such high power density) and (2) if plasma cool itself by black-body radiation at ~10000K. If the plasma is kept by magnetic field in free space quite far form ship construction, not much of solid surface is irradiated so not much cooling is needed. 99.9% of waste heat does not touch solid surface. $\endgroup$ May 2, 2015 at 22:30

1 Answer 1


I think that this is an interesting question and I will try to give a very rough answer (warning: it might include some approximations).

Instead of calculating the optical thickness of the D-T plasma, we can try to simply estimate the mean free path $\lambda$ of a neutron in the plasma. It can be estimated as (see for example here for a nice explanation) $$\lambda = (n \sigma)^{-1},$$ where $n$ is the number density of the target nuclei and $\sigma$ the neutron cross section. The plasma density in ITER will be approximately $n=10^{20}\,\mathrm{m}^{-3}$, so we have that value. What about the cross section?

We can look at some old papers [1-3], or use the Ramsauer model to estimate it. Whatever we decide to do, an important thing to note is that the cross section decreases with increasing energy (most pronounced for low energies, for high energies, it can be approximated as constant). As a rough estimation of any of the above mentioned methods, we get something like $\sigma \approx 3\,\mathrm{barn}$ (maybe 1, maybe 5, so let's say 1). $1\,\mathrm{barn}$ corresponds to $1\cdot10^{-28}\,\mathrm{m}^2$.

Using those values, we can approximate the mean free path to $$\lambda \approx (10^{20}\mathrm{m}^{-3} \cdot 10^{-28}\,\mathrm{m}^2)^{-1}=10^8\,\mathrm{m}.$$

Which is indeed rather large. Similar to the sun, as mentioned in the comments by @CuriousOne (note, however, that the number density in the sun is much larger).

Keep in mind that those are rough estimations, but you get the idea.

  1. [1] doi:10.1103/PhysRev.98.666
  2. [2] doi:10.1016/0375-9474(72)90930-X
  3. [3] doi:10.1103/PhysRevC.22.384

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