# Where is nuclear fusion occuring in the Sun?

My understanding is that the sun is basically a sphere of hydrogen with a helium core, and that the hydrogen is undergoing nuclear fusion to produce helium.

There are many images and cross-sectional schematics on Google but I can't find any actual numbers for the radii.

Are the nuclear reactions occurring where the helium meets the hydrogen?

What radius are the nuclear reaction occurring at?

How thick is this spherical shell of material undergoing nuclear fusion?

• "the sun is basically a sphere of hydrogen with a helium core" - no, that's not quite right. See e.g. the Wikipedia article on the solar core - the inner core drops to 33% hydrogen by mass, but it's still a far cry from pure helium. Mar 15 '19 at 12:46

There is much misconception in the question, but I'll hazard an answer. I am not going to give any links (no point linking to Wikipedia, it's just there), but I'll highlight the important terms in bold. If you want to research more, search for these. Any good general astronomy course textbook will cover these topics, too, if you want a bit more systematic approach.

My understanding is that the sun is basically a sphere of hydrogen with a helium core, and that the hydrogen is undergoing nuclear fusion to produce helium.

Basically, the Sun is a ball of hydrogen and helium, but this is not all there is. Being a Population I star, the Sun contains heavier elements (called metals in stellar astrophysics; anything lithium and heavier is considered metal in this sense). These elements already came with the gas cloud the Sun has formed from, and were produced by previously burst older stars. Despite low abundance, the metallicity plays an important role in the Sun's core power stability.

At some depth the gas ball compresses its inner area enough to heat it up so much that hydrogen fusion into helium begins. This area is called the core. This is where practically all fusion happens, and what is responsible for the star's energy production. For a Sun-mass star and below, the proton-proton chain dominates. The pp-chain energy output is approximately proportional to $$T^4$$. The good news is, if reaction rate drops, then the outer layer of the star will compress the core, so it heats up, and the renewed energy output compensates for the compression. So this highly-sensitive dependency on the temperature is what gives the star its long term stability.

It is also notable that the center of the core is hotter and therefore more energetic than its periphery, and turns hydrogen into helium faster. Absent any mixing, the core would develop an inert helium ball in the middle (helium cannot be fused by a Sun-mass star, its core is too cold for that): A pp-chain core is entirely non-convective. However, there is another multistage reaction that fuses protons into helium nuclei, the CNO cycle. This cycle requires metals ($$C$$, $$N$$ and $$O$$, naturally) be present in the core. They are not consumed, but participate in stages of the reaction and are ultimately recycled. The rate of this reaction depends on the temperature as $$T^{20}$$. It's a huge dependency! The CNO-dominant core has so much temperature gradient that it's fully convective, so it mixes the material very thoroughly.

It happens so that for a star with the mass $$M=1.5 M_\odot$$ the core is fully convective, but this is not an on/off phenomenon. Even in the Sun, the CNO cycle produces roughly 10% of core's output power, and is responsible for intermittent mixing of the core material. The dependency on temperature for this reaction is so large that the reaction is practically irrelevant at $$M=0.9 M_\odot$$ and $$T=14.5\times 10^6 K$$, and becomes dominant at $$M=1.5 M_\odot$$ and $$T=17.5\times 10^6 K$$. The Sun is at the very lower end of this range.

There is not a huge difference in the lifetime of the star even absent the CNO mechanism; it only changes the hydrodynamics of the core and its reactivity to temperature variation. But for the short-term stability it's very important; it amplifies the negative feedback loop that stabilizes the core reaction rate. It is probable (so models tell us) that the Sun's energy output would be much more variable on the scales of $$\sim 10^3$$ years. So we are lucky to have gotten enough "metal" in our home star, in the end--our ice ages have been bad enough already!

There are many images and cross-sectional schematics on Google but I can't find any actual numbers for the radii.

About $$0.2\,R_\odot$$.

Are the nuclear reactions occurring where the helium meets the hydrogen?

A Sun-mass star does not fuse helium. Helium fusion is a much more energetic process, and happens only in more massive and shorter-living stars. Helium is the embers of the combustion in the Sun, not its fuel.

As a side note, the Sun is a very calm reactor by Earthling's standards. The core's energy output is about $$300\, W/m^3$$, far too low for any practical fusion reactor on Earth. You need a chunk of the Sun's core $$10^7\,m^3$$ in size to match the power of a large coal-fueled electrical plant, and that's the volume of a ball about $$300\,m$$ in size. No way we could contain such a fireball at 15 million K; terrestrial fusion projects aim at much higher temperatures and thus reaction rates.

What radius are the nuclear reaction occurring at?

In the Sun while on the main sequence, all throughout the core. The core is essentially isolated against hydrogen supply from outer layers by the radiative zone, where the high thermal gradient stabilizes the gas against convection. As the Sun exhausts its fuel in the core, it transitions into the red giant phase (and the Sun will do it twice!). This happens when the core mostly turns into unburnable helium and cools down. What was previously the hydrogen-rich material in the radiative zone collapses on the surface of the helium core, and restarts the hydrogen reaction when its temperature reaches the ignition point. This reaction occurs only on the surface of the inner helium ball, in a spherical shell. There won't be any mixing mechanism this time that could disturb the inner inert ball.

All numbers come entirely off the top of my head, the best I could recall. Please double-check after me!