This is science not chemistry

Answer is:     molar solubility of  yttrium fluorid is 4.23·10⁻⁶ m. balanced chemical reaction: yf₃(s)  → y³⁺(aq) + 3f⁻(aq). ksp = 8.62·10⁻²¹. ksp = [y³⁺]  · [f⁻]³.[y³⁺] = [yf₃] = s.[f⁻] = 3[yf₃] = 3s.ksp(yf₃)  = s  · (3s)³.8.62·10⁻²¹ = 27s⁴.s(yf₃)  =  ⁴√(8.62·10⁻²¹  ÷ 27).s(yf₃)  = 0.00000423 m.s(yf₃) = 4.23·10⁻⁶ m.

When the electrons fall back down and leave the excited state, energy is re-emitted in the form of a photon. the wavelength (or equivalently, frequency) of the photon is determined by the difference in energy between the two states. these emitted photons form the element's spectrum.