A main sequence star will fuse some of its hydrogen, but not all. In massive stars ($>1.5M_\odot$) the core is convective but the rest of the atmosphere radiative and hence does not mix much: as it undergoes shell fusion it will produce an onion-like structure with unused hydrogen on top. Solar mass stars only do this up to helium, but again leave a mantle of unused hydrogen. Stars less than $0.35M_\odot$ are fully convective and can in principle use up all hydrogen. However, I suspect this is not complete except for very low-mass M dwarves that have trillions of years to mature.

Looking at planetary nebulae, I have seen statements that they are about 90 percent hydrogen, 10 percent helium. This seems to fit with papers I have found (example, example) although in some cases helium may reach 29% (example). Given that for a G star about half of the mass is ejected, that would suggest a fraction $0.9\times 0.5 = 0.45$ of unused hydrogen. But surely the fractions will be different for other masses.

So, to sum up, what is known about the fraction of hydrogen that is never fused over a stellar lifetime?


1 Answer 1


The answer is as complex as stellar evolution. H-burning takes place in the core during the main sequence; in a shell after the star has left the main sequence until it reaches the tip of the red giant branch and then ignites He in the core; then it also burns intermittently in shells during the asymptotic giant branch.

The later phases are much shorter than the main sequence phase, but the luminosities and hence rate of H-burning are orders of magnitude larger. Thus one would have to integrate the rate of H-burning over the lifetime of each phase.

A further complication is that much of the mass of a star is lost through stellar winds in the later phases of evolution. That material will be mostly hydrogen and is unavailable for burning.

All this will be very sensitive to the initial mass of the star.

We can put some sort of limits on this for a solar-type star of initial mass $1M_\odot$. Such stars will end their lives (or at least end their nuclear burning lives) as white dwarfs with a mass of about $0.5M_\odot$.

These white dwarfs have a composition that is almost entirely carbon and oxygen. Almost none of this will have been present when the star was born, so assuming an initial composition of about 74% H and 25% He by mass, then the $0.5M_\odot$ of C/O must have been synthesised by burning about $0.74\times 0.5M_\odot$ of H.

This will be a lower limit to the amount of H that has been burned, since some of the nucleosynthesis products will be lost as part of the $0.5M_\odot$ that escapes into space via the stellar wind. But assuming this is a small component because mixing is quite ineffective on the main sequence, we have that at least $0.74\times 0.5\times 100/0.74 = 50$% of the initial H would be burned.

If you calculate this fraction for shorter lived, higher mass stars then it is lower. This is because the relationship between initial mass and final white dwarf mass is very non-linear. For example, a $8M_\odot$ main sequence star might produce a $1.2M_\odot$ white dwarf, meaning the fraction of H burned is at least $(1.2/8\times 0.74)\times 0.74 \times 100 = 15$%. We would have to factor in though that more nucleosynthesis products will be lost into space in the vigorous winds from high-mass stars.

Even higher mass stars will likely burn more hydrogen as a fraction than $8M_\odot$ stars since they will ignite their C/O cores and there will be additional opportunities for mixing and H-burning as the star evolves towards a supernova and the fully-burned core mass grows substantially with mass.

The burnt fraction will increase towards lower masses because the white dwarfs they form will not be much less massive than $0.4M_\odot$. The fraction will become approximately 100% when stars are fully convective at $\leq 0.35 M_\odot$. In practice, the H is simply locked up in these stars at the moment because their low rate of H-burning means they have lifetimes of $10^{11}-10^{12}$ years.


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