# What is the percentage of ortho- vs. parahelium in the sun?

Parahelium (P) has its 2 electrons with opposite spins and S=0. Orthohelium (O) has parallel spins and S=1. The energy levels of P are a little bit higher than O. Maybe this leads to easier building of P. But can lead also to easier ionization. I tried to find data on net but it doesnt work.

Is there a difference or they are 50-50? If they are not can you tell why?

This is an interesting question! The Sun is a dynamical ball of hot plasma maintaining various types of equilibrium: hydrostatic, thermal, and nuclear. Observations over the last century have improved estimates of chemical abundances in the Sun, but its still an active area of research. The Sun is composed mostly of hydrogen and helium, which has been well established, but the precise relative composition is observationally uncertain and theoretically difficult to compute.

Observationally, the amount of helium in the sun is itself still uncertain, let alone which types of helium are present:

1. Using spectroscopy, one can only derive the solar chemical abundances at the surface of the Sun, not in the interior, since the spectral lines are photons that originated in the photosphere and chromosphere. Certain solar prominances on the Sun's surface, such as solar flares, are more suitable for obtaining observational data and estimating abundances (see more below on this). Orthohelium is the triplet series and parahelium is the singlet series of the neutral Helium atom emission/absorption. Orthohelium shows a fine structure, whereas parahelium does not. One could try to look for the fine structure in the spectroscopy, which would require very high precision spectroscopy that is very difficult for the Sun, and it would not be fair to rely on such measurements for abundance estimation (at least not yet!... maybe in the future). Some work has started to do this for leveraging information about solar flares, coronal rain, etc... but it's far from being useful for global abundance estimation. The amount of orthohelium in the Sun's surface can depend on many things. For example, the plages of a solar flare were studied and shown to contain $$\sim 10$$ times the amount of orthohelium (ie, the amount of helium atoms in the $$2s^3S$$ level) when one assumes an inhomogeneous chromosphere, compared to assuming a homogeneous chromosphere, i.e. $$\sim 10^{11} atoms/cm^2$$ vs $$\sim 10^{10} atoms/cm^2$$ (see pg.s 10-12 of this).

2. There is currently an open debate about what is the correct metallicity of the Sun, due to uncertainties in the amount of oxygen and other heavier elements. The discrepancy, known as the solar-metallicity or solar-abundance problem, is between the predictions of stellar evolution models (which are based on spectroscopy) and (rather recent) results of measurements from helioseismology. Perhaps your question is one that could be answered after this discrepancy is settled with the improved techniques! :)

There are many spectral lines for each of para- and orthohelium, so one would have to possibly consider many of such lines to estimate the abundance of para- and orthohelium. The two brightest orthohelium transitions are at wavelengths (approximately) 1083 nm and 587 nm, which are direct transitions from the 2S and 2P states, respectively, and they're both difficult to measure spectroscopically. There are actually three spectral lines corresponding to orthohelium at approximately 1083 nm (the lowest being the metastable ground state, and the other two are very closer together and are hard to resolve, i.e. see fig. 2 here).

Theoretically, I'm not sure how to do a simple calculation, perhaps someone else could offer one though. I was thinking about using Fermi-Dirac statistics, but I do not know the density of states of helium.

Parahelium (P) has its 2 electrons with opposite spins and S=0. Orthohelium (O) has parallel spins and S=1. The energy levels of P are a little bit higher than O. Maybe this leads to easier building of P.

Not necessarily. Theoretically, computing the abundance of orthohelium and parahelium is difficult, because it depends on temperature, density, and other properties of the plasma that can vary on many timescales, and because helium has a very high ionization energy, and because there are several spectral lines that correspond to orthohelium atomic transitions.

Also, the helium atoms in the solar plasma will exist under high temperatures and densities, making back-of-the-envleope calculations difficult. Plus there are magnetic fields, etc... it's complicated. Numerical simulations of magnetohydrodynamics of the solar plasma might be able to compute theoretical values for ortho- and parahelium relative abundances.

• Thank you very much. I am sorry that it is not easy and there are no data. I am surprised because astronomers measure very distant planet's gases within 1% estimate (which recently were not even close to be observed - the planets!). I hoped on a rough estimate for average conditions. Or at least to a measurement which form is more abundant. Jul 26 at 20:14
• My pleasure! You said, "I am surprised because astronomers measure very distant planet's gases within 1% estimate (which recently were not even close to be observed - the planets!)." Are you referring to orthohelium and parahelium? I'm not aware of any such measurements for them in exoplanet atmospheres. Also, "I hoped on a rough estimate for average conditions." I also did! But it was really hard! ;D If you could specify for what "average conditions" you're interested in, I could try or it could help others in writing an answer. Jul 26 at 23:16