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Most fusion reactions create energy in form of hard gamma radiations. A single gamma photon carries energy orders of magnitude higher than any ionization level. Even if we develop some hardcore version of "solar" panel based on photoelectric effect, most energy will still be scattered or pass through. In effect the reactor with glow in hard-X. Even if we don't care about losing significant portion of energy it's still not good idea to have a X-ray lighthouse anywhere near humans.

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  • $\begingroup$ The proton-proton chain in stars does produce gammas, but the reactions preferred in fusion reactors (mostly) don't, the energy is released as the KE of the products. I could elaborate on that in an answer, if you like. But I think your question about getting useful energy from gammas is interesting in its own right. $\endgroup$ – PM 2Ring Jun 19 at 9:52
  • $\begingroup$ oh, that makes sense. This would also mean that any reactor designed around stellar-like reactions including p-p and CNO is going to suffer from this. $\endgroup$ – Milo Bem Jun 19 at 10:26
  • $\begingroup$ Yes. But they also suffer from insanely slow reaction times. Even at solar core pressure & temperature, the p+p step almost never produces a deuteron. Just overcoming the Coulomb repulsion is hard enough, but the odds of the diproton to turn into a deuteron are like 1 / Avogadro's number. Usually, it just splits up into 2 protons. $\endgroup$ – PM 2Ring Jun 19 at 10:35
  • $\begingroup$ For details on the low rate of solar deuterium formation, see Ben's excellent answer here: physics.stackexchange.com/q/460310/123208 $\endgroup$ – PM 2Ring Jun 19 at 10:43
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    $\begingroup$ It is worth remembering that the centre of the Sun produces as much heat, per cubic metre, as a well-constructed compost heap. This is why the fusion reactions we hope to get power out of have to be very different from the solar ones. $\endgroup$ – Martin Kochanski Jun 19 at 14:23
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From the released energy of 17.6 MeV per fusion reaction (Deuterium+Tritium) are 14.1 MeV in the form of kinetic energy of the neutron and 3.5 MeV in kinetic energy of the helium. The neutrons are unaffected by the magnetic field and reach the blanket, where they release their energy as heat by collisions. The heat can be used to turn steam turbines to generate electricity, as with any other power plant.

The gammas can also be converted into heat. The photons generated have energies in the range of 1 MeV. At that photon energy the absorption coefficient for water is $\mu \approx 0.07 \,\text{cm}^{-1}$, which means that one meter of water shielding will absorb $1-e^{-\mu x}\approx 99.9$% of all gammas and turn them into heat.

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  • $\begingroup$ oh, that's interesting. When they say absorption what does actually happen to the photons? There are no atoms that can take the whole 1MeV photon in one collision. Do they lose some energy and shift to lower energy visible spectrum? $\endgroup$ – Milo Bem Jun 19 at 12:15
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    $\begingroup$ @MiloBem Yes. In this energy range the process is mostly compton scattering.The incident gamma photon ionizes an atom and the the remainder of the original photons energy is emitted as a new, lower energy photon. $\endgroup$ – Azzinoth Jun 19 at 12:26
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Fusion reactors (when they are perfected, which must be at least 20 years away),won't operate in exactly the same way as the sun, mainly because we can't reproduce the pressure at the sun's core. The most likely reactions fuse deuterium and tritium. The main problem seems to be the loss of neutrons, which also damage the lining of the torus. The problem of gamma radiation also occurs in fission reactors and can be safely contained. In both types of reactor some energy is lost by neutrino emission, but there is no remedy for that. A solution to the energy loss due to gamma radiation may be found, but I don't know of one.

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    $\begingroup$ Incidentally, “fusion reactors are 20 years away” was true in the 1950s as well. It may be one of those statements that are true forever. $\endgroup$ – Martin Kochanski Jun 19 at 14:21
  • $\begingroup$ I think 1950s estimates were more like 50 years away,& we've made a lot of progress since then,but there's some truth in what you say. Its like chasing the end of the rainbow,it's always just a bit further away than you thought. I could be 10 years out either way. $\endgroup$ – Michael Walsby Jun 19 at 17:49
  • $\begingroup$ "some energy is lost by neutrino emission, but there is no remedy for that" - what, you don't have a half light year of lead handy? $\endgroup$ – Maury Markowitz Jun 19 at 19:22
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    $\begingroup$ @MichaelWalsby- having studied the history extensively, there was widespread belief in the mid-1950s that we would hit breakeven in a few years. Then ZETA happened. By the mid-60s we thought it would be never. Then T-3 happened. In the mid-1970s we thought it would be in the early 1980s. Then TFTR happened. In the 90s we thought it might be never again. Then ITER "happened" - for some definitions of "happen". Now we say 30 years. So the estimates have varied widely and continually. $\endgroup$ – Maury Markowitz Jun 19 at 19:24
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    $\begingroup$ I think we will have an experimental reactor which works after a fashion within 10 years or less, but how long it will be before we can build a reactor which is commercially viable, ie can produce large amounts of energy at a competitive price, is anyone's guess. It may never be possible, but I think it will. $\endgroup$ – Michael Walsby Jun 19 at 19:38

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