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.

  • $\begingroup$ The second law of thermodynamics says no. Now to read the question. $\endgroup$
    – rob
    Aug 15, 2016 at 18:56

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.

  • $\begingroup$ 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 $\endgroup$ Oct 3, 2013 at 17:01
  • $\begingroup$ @Gotaquestion just saw this . no need for more precision. if you see the position of lithium in the binding energy curve the energy balances work. $\endgroup$
    – anna v
    Sep 24, 2015 at 3:07
  • 1
    $\begingroup$ Sorry to be a bore, but your sentence about supernovae is a bit loose. Heavier elements can be produced by neutron capture (obviously endothermically) without a supernova. And indeed it is thought about half of them are. See physics.stackexchange.com/questions/7131/… $\endgroup$
    – ProfRob
    Sep 24, 2015 at 6:05

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