Beryllium Abundances in Halo Stars from Keck/HIRES ObservationsThe Astronomical Journal
PublisherThe American Astronomical Society
AbstractWe have determined the abundance of Be in stars with an array of metal abundances in order to enhance our understanding of the chemical evolution of the Galaxy, cosmic-ray theory, and cosmology. Observations of the Be II resonance lines at lambda3130 and lambda3131 were made at the Keck telescope with the HIRES spectrometer at a resolution of 46,000 and signal-to-noise ratios of 60-110 (per pixel) typically. Our sample includes 22 halo dwarfs and five disk stars (including the Sun). We have taken special care in determining the stellar parameters for these stars in a consistent manner. The Be abundances were found (1) from the measured equivalent width of the relatively unblended Be II line at 3131.065 Å with an analysis that included 11 weak atomic and molecular lines near that wavelength and (2) from spectrum synthesis that included newly derived enhanced O (relative to Fe) in the synthesis calculations. The two methods are in excellent agreement. We find straight-line fits between Be and Fe: logN(Be/H)=0.96(+/-0.04)[Fe/H]-10.59(+/-0.03); and between Be and O: logN(Be/H)=1.45(+/-0.04)[O/H]-10.69(+/-0.04). It seems that Be and Fe increase at the same rate during the course of the evolution of the Galaxy. But as O increases by a factor of 100, Be increases more rapidly, by a factor of 800. Traditional models in which energetic cosmic rays interact with ambient CNO nuclei in the interstellar medium to produce Be are consistent with this finding, as long as certain chemical evolution effects (such as mass outflow from the halo) are taken into account. However, models predicting a linear relationship between Be and O, such as those producing Be in the vicinity of Type II supernovae, are less consistent with our result. There is some evidence for an intrinsic spread in Be at a given [Fe/H] or [O/H]. There is currently no evidence of a primordial plateau level of Be down to log N(Be/H)=-13.5.
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