Solution

The infrared and optical transitions share the same upper level 3. Using the supplied infrared photon energy gives

n_{3} = \frac{1.5 \times 1 0 ^{- 21}}{( 3.3 \times 1 0 ^{- 7} ) h c / ( 314 \times 1 0 ^{- 6} )} = 7.185 \times 1 0^{6} m^{- 3} .

With the same

T_{e}, n_{e}

and three-level fraction

f_{3}

used in the abundance calculation of b,

n (S^{+}) = 5.31 \times 1 0^{7} m^{- 3}, \frac{n ( S ^{+} )}{n _{e}} = 5.31 \times 1 0^{- 5} .

This exceeds the optical abundance by a factor

5.53

. Because the upper level is common, its population cancels in the intrinsic emissivity ratio:

\frac{j _{32}}{j _{31}} = \frac{A _{32} ν _{32}}{A _{31} ν _{31}} = 2.715 \times 1 0^{- 6} .

The observed ratio is

1.5 \times 1 0^{- 5}

. Thus changing only the electron temperature or a common level-population correction cannot reconcile these two lines from the same gas. Dust extinction suppressing the optical line is a natural explanation. If the far-infrared attenuation is negligible, the required optical attenuation is

A_{671.6} = 2.5 lo g_{10} (5.53) ≃ 1.86 mag .

Different apertures, filling factors, optical-depth assumptions or emissivity calibrations could also cause a discrepancy. A total sulfur abundance would still need an ionization correction beyond this ionic comparison.

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