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物理和生物化学:发光现象的计算

无与伦比652 2018-11-12 17:35:27 369  浏览
  • 1.A chromophore solution is places in a 1 × 1 cm regular fluorescence cuvette. The absorbance of this solution, at 1 cm path length, is 0.80 at 450 nm and 0.30 at 530 nm. When the sample is excited at 450 nm and the fluorescence emission... 1.A chromophore solution is places in a 1 × 1 cm regular fluorescence cuvette. The absorbance of this solution, at 1 cm path length, is 0.80 at 450 nm and 0.30 at 530 nm. When the sample is excited at 450 nm and the fluorescence emission is measured at 530 nm (both were set at a very narrow 5 nm slit for spectral band width), the observed fluorescence emission is 100 arbitrary unit. Calculate the theoretical fluorescence emission intensity (using the same arbitrary unit) after corrections for the inner filter effects on both excitation and emission. Show critical steps of calculation. 2. For a particular Donor:Acceptor pair, the spectral overlap integral “J” between the corrected donor fluorescence emission and the acceptor absorption is 1.58 x10-13 (cm3 M-1) and the efficiency of fluorescence energy transfer is 0.667. Assuming 2/3 for the dipole orientation factor κ2 and 1.4 for therefractive index “n” of the medium, calculate the value of R. Show critical steps of calculation. 展开

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物理和生物化学:发光现象的计算
1.A chromophore solution is places in a 1 × 1 cm regular fluorescence cuvette. The absorbance of this solution, at 1 cm path length, is 0.80 at 450 nm and 0.30 at 530 nm. When the sample is excited at 450 nm and the fluorescence emission... 1.A chromophore solution is places in a 1 × 1 cm regular fluorescence cuvette. The absorbance of this solution, at 1 cm path length, is 0.80 at 450 nm and 0.30 at 530 nm. When the sample is excited at 450 nm and the fluorescence emission is measured at 530 nm (both were set at a very narrow 5 nm slit for spectral band width), the observed fluorescence emission is 100 arbitrary unit. Calculate the theoretical fluorescence emission intensity (using the same arbitrary unit) after corrections for the inner filter effects on both excitation and emission. Show critical steps of calculation. 2. For a particular Donor:Acceptor pair, the spectral overlap integral “J” between the corrected donor fluorescence emission and the acceptor absorption is 1.58 x10-13 (cm3 M-1) and the efficiency of fluorescence energy transfer is 0.667. Assuming 2/3 for the dipole orientation factor κ2 and 1.4 for therefractive index “n” of the medium, calculate the value of R. Show critical steps of calculation. 展开
2018-11-12 17:35:27 369 0
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