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Why we can't make chain reaction in fusion or we can't use fusion products .

Deuterium $-$ Helium$-3 $ fusion : $$2D + 3He → 4He + 1p + 18.3 \text {MeV}$$ (I found in Wikipedia)

If we can do the above reaction can we use this proton energy to make another reaction ?

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2 Answers 2

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The usable energy of fusion is in the form of the kinetic energy of the charged products of the reaction, primarily. The chain reaction can occur, and this is the goal (in all systems, it is a goal of each target or pulse; in some systems, it is a goal for the overall operation as well - the sun is an example of the latter), but only if we can contain the products for

  1. enough time
  2. at high enough density, and
  3. high enough temperature

to overcome the barriers (Coulomb) of subsequent reactions. This is very difficult since, when we add so much thermal energy to the components of the reaction, they all very quickly are driven away from one another.

This

  1. decreases the time of interaction,
  2. decreases the density, and
  3. decreases the temperature,

all contrary to the goals stated above. This fact motivates all fusion research.

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If we can do the above reaction can we use this proton energy to make another reaction?

Yes. All common fusion reactions release at least some of their energy in a form that can be cycled back into the fuel to heat other ions to the required temperatures and thereby keep the reaction going. The point where the amount of energy being recycled is equal to all the various losses is known as "ignition". It is the key goal of all fusion devices, and it's why they called it the National Ignition Facility.

The most commonly studied reaction, as it is the easiest, is the D-T reaction. This releases about 25% of its reaction energy as a high-energy alpha. In a magnetic system like a tokamak, the alpha is captured by the confinement magnets and stays in the fuel mass where it undergoes collisions with the D's and T's and quickly thermalizes, heating up the plasma. In ICF like NIF, the super-dense fuel causes so many classical collisions that it thermalizes almost instantly.

Working against this process is the fact that the majority of the energy in this reaction goes into a neutron, which does not get captured and flows out of the fuel. So at a minimum, you need to lose energy about four times slower than the reaction rate or the plasma will cool off.

It gets more complex when you consider the other losses, mostly X-ray production from the hot particles. This has a temperature dependence, but also an atomic mass one. So this means you want to get that alpha out of the fuel as quickly as you can after it's thermalized, otherwise it will help radiate away energy. And other atoms, like those spalled off the walls of the vacuum chamber, really kill your energy balance. Many years of effort went into figuring out how to pull those heavier atoms out of the mix without also removing the hot fuel you want to keep.

In the reaction you noted, both the 4 He's (alphas) and the proton will all be captured and help heat the fuel. This makes the balancing simpler, but has the disadvantage of also having a much lower reaction rate, so even with more of the output being captured it's still much harder to ignite the fuel because of the constant radiative losses.

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