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What are the requirements for hydrogen atoms to go through fusion? Is it a ratio of heat to pressure or are there specific heat and pressure values that must be met (per atom or per mole?) Are there other requirements?

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2 Answers 2

The Wikipedia article answers most of your questions.

What are the requirements for hydrogen atoms to go through fusion?

Two atoms must overcome the coulomb barrier, which can be done by forcing two atoms very close together, or by leaving them moderately close for long periods of time, which allows them to tunnel through the barrier.

Is it a ratio of heat to pressure or are there specific heat and pressure values that must be met (per atom or per mole?)

It's fairly complicated but it does require very high temperatures and very high pressures.

For reference fusion happens at the centre of the sun, but does not produce very much energy per unit volume (about the same as a living mammal) so for a viable reactor we need much higher energies than the centre of the sun.

Wikipedia quote:

The core produces almost all of the Sun's heat via fusion ... The energy production per unit time (power) of fusion in the core varies with distance from the solar center. At the center of the Sun, fusion power is estimated by models to be about 276.5 watts/m3. Source

Are there other requirements?

There are many different ways of producing nuclear fusion and each has its own set of requirements see Muon catalysed fusion as a way of causing nuclear fusion without the requirement of heat or pressure.

Really nuclear fusion is not difficult, there are many reactors that fuse elements, the difficulty is the fact that they require so much energy as an input that they are uneconomical as a power source.

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I'm confused. You say "... (the sun) does not produce very much energy per unit volume (about the same as a living mammal) ..." If it has the same production as homeotherms, then it shouldn't produce the millions of joules that it does. Do you have a reference for your statement? –  LDC3 Apr 18 '14 at 5:24
If you take the amount of heat it generates and divide it by its mass, that number may make more sense. The sun generates enough heat to warm the entire solar system, but it is also by far the heaviest thing in the solar system. –  Hoytman Apr 18 '14 at 12:45
Speaking of other fusion consents, what about the palladium electrolysis concept? –  Hoytman Apr 18 '14 at 12:47

The simplest reaction deuterium and tritium. Tritium is common in big labs (like NIF, JET, Omega) [1]. Tritium sucks - practically speaking. It is expensive, radioactive, and hard to stockpile. Omega spent millions and years on a tritium facility. It may even never be used in fusion power [2]. The next easiest reaction is deuterium with itself. This is the fuel for amateur fusion (e.g. Taylor Wilison, Jamie Edwards, ect etc…). It costs about 150 grams for 100 dollars [3]. Here are these reactions:

D + D → T + H

D + D → He + n

D + T -> He + n

The nuclei have to collide. When they hit they must overcome the coulomb barrier. Quantum tunneling helps a little. There is also a likelihood of fusion. Call it a “fusibility” rate. It is known the cross section. It is specific to that reaction at that energy. Cross sections are normally measured, using particle beams. A is fired at B and fusion occurs. Neutrons are made and measured – and the cross section is calculated from this. We have mapped out fusion in huge elements (like cesium), in this manner, since the 70’s. These numbers go into a volumetric fusion equation [4]:

Volumetric Fusion = Density A * Density B * Collision Velocity * Energy released * Cross Section(Velocity)

Surprisingly, this equation has very different meanings, to different fusion communities. The Tokomak people will say it is all about temperature. The laser people will say it is all about density. Both are macroscopic methods – both try raise the number of high velocity collisions. The fusor crowd will focus on collisions instead of macroscopic effects. Finally, the Muon people will say its all about cross section. Who is right? We do not know yet. We will not know till we have fusion power.

I expect deuterium fusion to start at roughly 10,000 electron-volts; but amateurs have measured neutrons at lower values. The National Ignition Facilities original goal was to get plasma to 10 KeV under high density [5]. The electron-volt a unit of energy. It is the kinetic energy that one electron has after falling down one volt. A nucleus gaining an electron-volt of energy, has gone up 11,604 degrees Kelvin. So 10 KeV is millions of degrees. Fusors do fusion by exploiting this. They drop an ion down a 10,000 volt well. The electric field does work on the ion, heating it to fusion conditions. Most of the ions are not hot enough to fuse. (In fact most hit the wire cage). But, if a handful of nuclei collide at 10 KeV: fusion can happen in a middle school in England.

Cheers! http://thepolywellblog.blogspot.com/


  1. http://www.efda.org/newsletter/jet-returns-to-operation-with-tritium/

  2. https://www.linkedin.com/groups/Problem-Tritium-as-Fusion-Fuel-4810092.S.5857561488729874432?view=&gid=4810092&type=member&item=5857561488729874432

  3. http://unitednuclear.com/index.php?main_page=product_info&cPath=16_17_69&products_id=135&zenid=7c0e4884f11add50bbe2a59310061a1b

  4. “Some Criteria for a Power Producing Thermonuclear Reactor” By John Lawson, 1956

  5. "Development of the Indirect‐drive Approach to Inertial Confinement Fusion and the Target Physics Basis for Ignition and Gain." John Lindl. Page: 3937. AIP Physics of Plasma. American Institute of Physics, 14 June 1995.

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"It costs about 150 grams for 100 dollars" Most people say it in reverse; it makes more sense (i.e. $100 for 150 g). –  LDC3 Apr 20 '14 at 20:50

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