A regular series of lines observed in the visible part of the emission spectrum of a cosmic star has the property that one of its emission lines coincides with the first line of the hydrogen Balmer series, and that every 5th, 10th, 15th, etc. line afterwards coincides with 2nd, 3rd 4th etc. line of the hydrogen Balmer series. Assuming that this particular series in the emission spectrum of the star can be associated with a hydrogen-like ion, what element would be responsible for the observed emission? What is the longest wavelength that can be found in this particular spectral series? Hint: the atomic number should give you a clue about the element in question.​​

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It was ernest rutherford

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

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