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Based on the first equation at

$$_1^{1}H+_1^{1}H \to _2^{2}He+\gamma$$

Is it correct to say that in the P-P chain two hydrogen atoms fuse to form a helium-2 isotope in an excited state which decays by gamma emission to a helium-2 isotope in a non-excited state, emitting a photon in the process?

$$_1^{1}H+_1^{1}H \to _2^{2}He^* \to _2^{2}He+\gamma$$

Or is it just a one-step process without excited helium-2 intermediate?

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1 Answer 1

up vote 4 down vote accepted

In a very general sense a lot of reaction that are written in one step can also be written in two. I.e. alpha capture on carbon-131 is often written $$\alpha + ^{13}\!\mathrm{C} \to ^{16}\!\mathrm{O} + \text{various photons and leptons} \,,$$ but may be written as one of $$\begin{align} \alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O} \\ \alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O}^* \to ^{16}\!\mathrm{O} + e^+ + e^-\\ \alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O}^{**} \to ^{16}\!\mathrm{O} + \gamma\,(\text{6.05 MeV}) \\ \dots \end{align}$$

However, in order for that to be reasonable, there must be a bound intermediate state to talk about.

In the case of PP fusion there is no bound $^2\mathrm{He}$ state (much less a bound excited state).2 Fusion only results if there is a weak tranformation at a time when both protons are very close to one another.

Nor can we talk about one proton turning into a neutron and then finding the other proton, because that intermediate state is energetically forbidden.

So, long story short, I don't think that you should write the proton-proton fusion process with an intermediate state. It's all or nothing.

1 A reaction selected entirely because I know it well.

2 Contrast this with the oxygen system where the excited states are real and have been studied in detail by neutron knock-out reaction on Oxygen-17.

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Fair enough but my question wasn't "should I write it with an intermediate state" but "is it happening with an intermediate state". So if I understood it correctly you are saying that it does happen with an intermediate state but we don't write it down for obvious reasons (clarity / we can always decompose things further until it takes 50 pages to explain 1 + 1 = 2) ? – Aegis Mar 16 '14 at 15:24
No, there is no intermediate state. That's the point. Is has to happen in one step. – dmckee Mar 16 '14 at 15:26
Oh ok, sorry. Thanks! – Aegis Mar 16 '14 at 15:26
Could you briefly clarify what you mean by "bound intermediate state" and why there is one in the case of gamma emission? (I am a freshman) – Aegis Mar 16 '14 at 15:32
An unbound state is one in which the constituents has enough energy to escape to arbitrarily large distance; bound systems do not. The two protons are always have escape energy because the residual strong attraction between them is less than the Coulomb repulsion. – dmckee Mar 16 '14 at 15:39

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