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Fission divides one Helium atom into two Hydrogen (Deuterium) atoms. And fusion, once again, puts together those two Hydrogen atoms into one Helium atom. In both reactions, overall output energy is enormous.

$ \begin{array}{rcrclclrl} \text{1 Neutron} & + & {}_2^4He & \to & {}_1^1H + {}_1^1H & + & \text{Energy} + \text{3 Neutrons} & \dots & (I) \\ \text{2 Neutrons} & + & {}_1^1H + {}_1^1H & \to & {}_1^2H + {}_1^2H &&(somehow) & \dots & (II) \\ \text{Energy} & + & {}_1^2H + {}_1^2H & \to & {}_2^4He & + & \text{Big Energy} & \dots & (III) \\ \end{array} $

I don't know whether the reaction $(II)$ is possible with today's technology, or not. Neither I know if it is endothermic or exothermic. But, if we could realize it, would it be possible to generate infinite energy by looping these three reactions forever successively in the given order?

I intuitively feel that the answer is "no", but I need an explanation on why it is not possible.

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1 Answer 1

The answer lies in this diagram

enter image description here

It shows the binding energy per nucleon over the number of nucleons n in a nucleus. As you can see, it has a maximum around Iron (n=56), which means that left from iron energy can be released by fusion. For elements heavier than iron you need to put in energy to fuse them together, or equivalantly that fission is energetically favorable. That is why conditions like in a supernova are necessary to produce the heavier elements.

So, fusioning ${}^2H$ to ${}^4He$ releases energy, but the reverse fission process costs energy. There is no net gain, even in a perfect aparatus.

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Generally that is correct, but I believe there is an extra level of details. There are some exothermic fission reactions in which elements lighter than Fe release energy in fission reaction. For example a neutron hitting Lithium-6 generates Tritium, alpha particle and 4.7 MeV. So there has to be more precise rules –  Gotaquestion Oct 3 '13 at 17:01

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