Fusion in the sun for 4 hydrogen to Helium-4. How is the energy produced

So please correct me if I am wrong but: First 2 proton’s (each with an electron) fuse together, The mass stays the same so no energy is produced? Then 1 of the protons turns into a neutron with a Neutrino and a positron. The positron and the electron collide and produce energy equivalent to the mass of 2 electrons (E=MC^2). The neutrino flys away and doesn’t interact with anything (at least in the fusion process) again. So now we have 1 Proton, 1 Neutron and 1 electron aka Deuterium. Then another proton and electron fuse with the Deuterium to form Helium-3. How does energy work here because it isn’t the mass of 1 proton and electron cause that is over 100MeV which is wrong as the entire process (according to google) is only meant to be around 24Mev, and there isn’t an apparent gain/loss of mass. Then the 2 Helium-3 fuse and have the same problem as before (the mass stays the same but energy is still produced). The 2 Helium-3’s then lose 2 protons and 2 electrons to form Helium-4.

My question is how to work out how the energy is produced in a hydrogen -> helium fusion (the type of fusion that happens in the sun)

• Are you familiar with MathJax? Reading reactions as sentences feels like something common core would do, so for instance: $p + p +e^-\rightarrow {}^2_1D + \nu_e + 1.442\,{\rm MeV}$ is preferred.
– JEB
Commented Feb 29 at 14:24
• You seem to be neglecting binding energy. physics.stackexchange.com/questions/tagged/binding-energy Commented Feb 29 at 15:43

(0) This process is called "the proton-proton chain", and should be referred to as such in question titles and what-not.

Two protons do not fuse into $${}^2_2{\rm He}$$, that has a half-life in the 1e-22 second region...compare that with the mean collision time for protons in the core of the Sun to see that it cannot contribute.

Rather, two protons undergo a weak interaction:

$$p + p \rightarrow {}^2_1{\rm D} + e^+ + \nu_e + 0.42\,{\rm MeV}$$

(that it is weak means it is unlikely, hence a 10 billion year lived Sun).

The initial (final) mass-sum on the RHS (LHS) is:

$$M_0 = 2m_p = 1876.544163\,{\rm MeV}$$

$$M_1 = m_d + m_e +m_{\nu_e}= 1876.1239416\,{\rm MeV}$$

(where I've ignored the neutrino mass). Note that

$$M_1 - M_0 = -0.420221\,{\rm MeV} < 0$$

So the total mass is reduced. This is generally referred to "binding energy", which is negative. The deuteron binding energy (the only binding energy every nuclear physicist has memorized) is 2.2 MeV, which gives the deuteron a lower mass than that of a free proton plus a free neutron. (Here, some of the 2.2 MeV is required to turn a proton into a neutron and create the final state leptons).

That mass difference is generally liberated as kinetic energy according to:

$$E = mc^2$$

The positron goes onto annihilate an unrelated plasma electron, releasing:

$$2m_e = 1.022\,{\rm MeV}$$

as 2, 3, 4, ... gamma rays. Note that this is not fusion, and does not require temperature of pressure to occur, though number density is obviously a factor.

The deuteron is stable, and finds a proton:

$$d + p \rightarrow {}^3_2{\rm He} + \gamma + 5.493\,{\rm MeV}$$

which is a fairly hard gamma.

https://en.wikipedia.org/wiki/Proton–proton_chain says the mean residence time of the helium-3 is 400 years.

From here, there are 4 branches to helium 4. The main one is:

$${}^3_2{\rm He} + {}^3_2{\rm He} \rightarrow {}^4_2{\rm He} + 2p + 12.859\,{\rm MeV}$$

(So I retract my hard gamma quip, and now apply it here).

You can see the above reference for details on other branches. They are:

Lithium Burning (https://en.wikipedia.org/wiki/Lithium_burning)

The pp-III branch, involving berylliums-7,8 and boron-8, which is dominant above 25 MK.

p-IV (Hep), which is a theoretical weak interaction branch: $${}^3_2{\rm He} + p \rightarrow \alpha + {\rm appropriate\ leptons}$$.

It's important to appreciate the difference between strong and weak interaction fusion processes, as the time-scale differ by orders of magnitude in the exponent of "orders of magnitude".

Finally, to address your question, "How to work out the energy released". Find the difference between initial mass and final mass:

$$E_{pp} = \Big[4(m_p + m_e)\Big] - \Big[m_{\alpha} + 2(m_e + m_{\nu_e})\Big]$$

Of course the neutrino mass isn't known (and the electron neutrino isn't even a mass eigenstate), but it is tiny ($$m_{\nu_e} \approx 0.07\,{\rm eV}$$)...200 times smaller than the binding energy of hydrogen.

Since the neutrinos carry away their mass and kinetic energy, a full analysis of available energy for heating would require detailed analysis of the neutrino spectrum. See https://jila.colorado.edu/~pja/astr3730/lecture21.pdf . IIRC, the neutrino luminosity is around 1% of the solar luminosity.

Note: IMHO, an interesting, but often overlooked role the neutrinos play is radiating lepton number. In the Standard Model, lepton number is conserved.

The core contains $$0.34 \times M_{\rm solar symbol}/{\rm grams} \times N_A \approx 4\times 10^{56}$$ protons and electron, each. Over the lifetime of the sun, that becomes $$2\times 10^{56}$$ protons, neutrons and electron, each (assuming 100% burning, idk if that is correct), so baryon number is conserved, but $$L=2\times 10^{56}$$ lepton number has "gone missing" and is radiated via neutrinos.

• Thank you for your reply, However I am still confused where the 0.42MeV (in the proton-proton chain), 5.493MeV (in the deterium-proton part) and 12.859MeV (in the helium3-helium3 part) comes from and how to calculate it. Commented Feb 29 at 15:37
• @BobbieSpace - binding energy of the resulting nucleus. Commented Feb 29 at 15:55
• @BobbieSpace I'll edit parts, I included a sum already.
– JEB
Commented Feb 29 at 17:47