A certain dye absorbs $4000 ˚ A$ and fluoresces at $5000 ˚ A$ these being wavelength of maximum absorption that under given conditions $40$ of the absorbed energy is emitted. Calculate the ratio of the number of quanta emitted to the number absorbed.

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Solution

Energy of light absorbed in an photon $= h c / λ_{a b s o r b e d}$
Let $n_{1}$ photons are absorbed
Total energy absorbed $= n_{1} h c / λ_{a b s o r b e d}$
Now E of light re-emitted out in one photon $= h c / λ_{e m i t t e d}$
Let $n_{2}$ photons are re-emitted then
Total energy re-emitted out $= n_{2} h c / λ_{e m i t t e d}$
$E_{a b s o r b e d} \times 40 / 100 = E_{r e - e m i t t e d o u t}$
$h c / λ_{a b s o r b e d} \times n_{1} \times 40 / 100 = n_{2} \times h c λ_{e m i t t e d}$
$n_{2} / n_{1} = 40 / 100 \times λ_{e m i t t e d} λ_{a b s o r b e d}$
$= 40 / 100 \times 5000 / 4000$
$n_{2} / n_{1} = 0.5$

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A certain dye absorbs 4000˚Aand fluoresces at 5000˚A these being wavelength of maximum absorption that under given conditions 40 of the absorbed energy is emitted. Calculate the ratio of the number of quanta emitted to the number absorbed.

A certain dye absorbs $4000 ˚ A$ and fluoresces at $5000 ˚ A$ these being wavelength of maximum absorption that under given conditions $40$ of the absorbed energy is emitted. Calculate the ratio of the number of quanta emitted to the number absorbed.