Question

a certain dye absorbs light of wavelength 4500 Ao and the fluroscence light of 5000 Ao. assuming that under given conditions 50 % of the absorbed energy is re emitted out as flurosence. cal the ratio of quanta emitted to no of quanta absorbed

Answer 1

let no. of emitted quanta = n1 and of absorbed = n2 Frequency of emitted quanta F1 = c/ 5080 × 10⁻¹⁰ Energy e1 = h × F1= h × c/ 5080 × 10⁻¹⁰ Frequency of absorbed quanta F2 = c/ 4530 × 10⁻¹⁰ Energy e2 = H × c/ 4530 × 10⁻¹⁰ Total energy of emitted quanta = 0.47 0f absorbed quanta hence: n1 × h × c/ 5080 × 10⁻¹⁰ = 0.47 × n2 × h × c/ 4530 × 10⁻¹⁰ n1/5080 = 0.47 × n2 /4530 n1/n2 = 0.527

Answer 2

Answer:

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