What element is the main product of H fusion and what particles are formed?

Answer 1

Short answer: helium, positrons, neutrinos, gamma rays.

nb, due to the limitations of raw text, I am decribing these atoms in the form "(Atomic Mass) / (atomic number) (element)". "e+" denotes a positron, "e-" an electron, "v" a neutrino and, "y" a gamma ray photon. If this is for your homework, you probably want to use the correct symbols with the appropriate subscript and superscript and without all the parentheses. You may also be using "p" instead of "1/1 H" and possible "2/1 D" or simply "2 D" for deuterium instead of 2/1 H. Onward.

Hydrogen fuses into helium, by the following processes:

ppI chain:

i: (1/1 H) + (1/1 H) -> (2/1 H) + (e+) + (v)

ii: (2/1 H) + (1/1 H) -> (3/2 He) + (y)

iii: (3/2 He) + (3/2 He) -> (4/2 He) + 2(1/1 H)

In words; two protons (hydrogen nuclei) fuse to form deuterium, in doing so, one of the protons becomes a neutron via beta decay with the emission of a positron and neutrino (if you need to go deeper than this, it's decay via the weak interaction with the emission of a W- boson which spontaneously decays into a positron and neutrino).

This deuterium nucleus fuses with another proton to form 3/2 helium. Here no protons are decaying into neutrons so there is no e+/v emission, however because the rest energy per nucleon in 3/2 He is lower than in the hydrogen nuclides 1/1 H and 2/1 H, the leftover energy escapes as a gamma ray. This is the stage at which the vast majority of energy is liberated from the material.

Then, in this particular chain, two 3/2 He nuclei fuse to form a 4/2 He nucleus (an alpha particle) with the return of two protons to the soup.

ppII chain

(i) and (ii) as per ppI

(iii) (3/2 He) + (4/2 He) -> (7/4 Be) + (y)

(iv) (7/4 Be) + (e-) -> (7/3 Li) + (v)

(v) (7/3 Li) + (1/1 H) -> 2(4/2 He)

Here, our 3/2 He from the ppI chain fuses with a 4/2 He to create 7/4 Be, and again the difference in rest energies is carried away by another gamma ray. An electron is captured (this process is rather unimaginatively called "electron capture) transmuting the 7/4 Be into 7/3 Li (one proton becomes a neutron) with the ejection of a neutrino. This 7/3 Li nucleus fuses with a proton to create 8/4 Be which is unstable and almost immediately breaks down into two 4/2 He alpha particles.

ppIII chain

(i), (ii) and (iii) as per ppII

(iv) (7/4 Be) + (1/1 H) -> (8/5 B) + (y)

(v) (8/5 B) -> (8/4 Be) + (e+) + (v)

(vi) (8/4 Be) -> 2(4/2 He)

In this instance the 7/4 Be we made in the ppII chain fuses with another proton to create 8/5 B, with the emission of yet another gamma ray. The boron nucleus then beta decays into 8/4 beryllium, which again breaks down into two alpha particles.

In summary, and to answer your question directly:

ppI: 4(1/1 H) -> (4/2 He) + 2(e+) + 2(v) + 2(y)

ppII: 4(1/1 H) + (e-) -> (4/2 He) + (e+) + 2(v) + 2(y)

ppIII: 4(1/1 H) -> 2(e+) + 2(v) + 3(y)

There is also an hypothesised ppIV chain in which:

(3/2 He) + (1/1 H) -> (4/2 He) + (e+) + (v) + (y)

And also a rare means of creating 2/1 H:

(1/1 H) + (e-) + (1/1 H) -> (2/1 H) + (v)

In our Sun, which I presume is why you are interested in this, the ppI chain dominates, with roughly 85% of hydrogen burning following this route. The ppII chain accounts for about 15%, and the ppIII chain a pathetic 0.02%.

But wait, there's more. In larger, hotter stars there is another important mechanism called the CNO cycle. It looks like this:

(i) (12/6 C) + (1/1 H) -> (13/7 N) + (y)

(ii) (13/7 N) -> (13/6 C) + (e+) + (v)

(iii) (13/6 C) + (1/1 H) -> (14/7 N) + (y)

(iv) (14/7 N) + (1/1 H) -> (15/8 O) + (y)

(v) (15/8 O) -> (15/7 N) + (e+) + (v)

(vi) (15/7 N) + (1/1 H) -> (12/6 C) + (4/2 He)

I'm sure I don't need to explain what's happening this time, so we are left with the summary which is:

CNO: 4(1/1 H) -> (4/2 He) + 2(e+) + 2(v) + 3(y)

You are now a fully qualified nuclear physicist.