Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.