
A certain dye absorbs at 4000 $\dot{A}$ and fluoresces 5000 $\dot{A}$ at these being wavelengths 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.
Answer
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Hint: Quantum yield is the average quanta value extracted and the quanta value absorbed. In simple systems, the amount of quantum yield is usually much higher than the unit and is determined primarily by the amount of energy levels.
Complete Step by Step Solution:
Energy of light absorbed in an photon=$\dfrac{hc}{{{\lambda }_{absorbed}}}$
Let ${{n}_{1}}$ photon are absorbed
Total energy absorbed=$\dfrac{n1hc}{{{\lambda }_{absorbed}}}$
Now $E$ of light re-emitted out in one photon=$\dfrac{hc}{{{\lambda }_{emitted}}}$
Let ${{n}_{2}}$ photon are re-emitted then
Total energy re-emitted out=$\dfrac{n2hc}{{{\lambda }_{emitted}}}$
$\Rightarrow {{E}_{absorbed}}\times 40/100={{E}_{re-emitted}}$
$\Rightarrow \dfrac{hc}{{{\lambda }_{absorbed}}}\times n1\times \dfrac{40}{100}=n2\times hc{{\lambda }_{emitted}}$
$\Rightarrow \dfrac{n2}{n1}=\dfrac{40}{100}\times {{\lambda }_{emitted}}{{\lambda }_{absorbed}}$
$\Rightarrow \dfrac{40}{100}\times \dfrac{5000}{4000}$
$\Rightarrow \dfrac{{{n}_{2}}}{{{n}_{1}}}=0.5$
So, the ratio of the number of quanta emitted to the number absorbed is $0.5$.
Additional Information: Quantum yield is generally understood as the product of the reaction product. It is calculated by the number of reacting moles per unit of time. This type of quantum yield is now very popular, as it can be measured at minimal cost. Its disadvantage is that it depends on the reaction time at which it is determined. For this reason it can be considered a visual quantum yield. The problem becomes apparent when the reaction consists of a few steps.
Note: The concept of quantum has a heuristic meaning only in situations where the output spectrum is independent of the abundance of pleasant light. Situations of this type are found only in changes from low voltage electronic components that are cheerful and independent to the vibrating vibration reserve, which are acquired with excitement.
Complete Step by Step Solution:
Energy of light absorbed in an photon=$\dfrac{hc}{{{\lambda }_{absorbed}}}$
Let ${{n}_{1}}$ photon are absorbed
Total energy absorbed=$\dfrac{n1hc}{{{\lambda }_{absorbed}}}$
Now $E$ of light re-emitted out in one photon=$\dfrac{hc}{{{\lambda }_{emitted}}}$
Let ${{n}_{2}}$ photon are re-emitted then
Total energy re-emitted out=$\dfrac{n2hc}{{{\lambda }_{emitted}}}$
$\Rightarrow {{E}_{absorbed}}\times 40/100={{E}_{re-emitted}}$
$\Rightarrow \dfrac{hc}{{{\lambda }_{absorbed}}}\times n1\times \dfrac{40}{100}=n2\times hc{{\lambda }_{emitted}}$
$\Rightarrow \dfrac{n2}{n1}=\dfrac{40}{100}\times {{\lambda }_{emitted}}{{\lambda }_{absorbed}}$
$\Rightarrow \dfrac{40}{100}\times \dfrac{5000}{4000}$
$\Rightarrow \dfrac{{{n}_{2}}}{{{n}_{1}}}=0.5$
So, the ratio of the number of quanta emitted to the number absorbed is $0.5$.
Additional Information: Quantum yield is generally understood as the product of the reaction product. It is calculated by the number of reacting moles per unit of time. This type of quantum yield is now very popular, as it can be measured at minimal cost. Its disadvantage is that it depends on the reaction time at which it is determined. For this reason it can be considered a visual quantum yield. The problem becomes apparent when the reaction consists of a few steps.
Note: The concept of quantum has a heuristic meaning only in situations where the output spectrum is independent of the abundance of pleasant light. Situations of this type are found only in changes from low voltage electronic components that are cheerful and independent to the vibrating vibration reserve, which are acquired with excitement.
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