A certain dye absorbs `4530A^(@)` and fluoresence at `5080A^(@)` these being wavelength of maximum absorption that under given condition `47%` of the absorbed energy is emitted. Calculate the ratio of the no of quanta emitted to the number absorbed.

Explanation

To solve the problem, we need to find the ratio of the number of quanta emitted to the number absorbed when a dye absorbs light and fluoresces. Here’s a step-by-step solution:

### Step 1: Understand the Energy of Quanta
The energy of a photon (quantum of light) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) = energy of the photon,
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) = speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) = wavelength in meters.

### Step 2: Calculate the Energy of Absorbed Quanta
The dye absorbs light at a wavelength of \( 4530 \, \text{Å} \) (angstroms). First, convert this to meters:
\[ 4530 \, \text{Å} = 4530 \times 10^{-10} \, \text{m} = 4.53 \times 10^{-7} \, \text{m} \]

Now, calculate the energy of the absorbed quanta:
\[ E_{absorbed} = \frac{hc}{\lambda_{absorbed}} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{4.53 \times 10^{-7} \, \text{m}} \]

### Step 3: Calculate the Energy of Emitted Quanta
The dye fluoresces at a wavelength of \( 5080 \, \text{Å} \). Convert this to meters:
\[ 5080 \, \text{Å} = 5080 \times 10^{-10} \, \text{m} = 5.08 \times 10^{-7} \, \text{m} \]

Now, calculate the energy of the emitted quanta:
\[ E_{emitted} = \frac{hc}{\lambda_{emitted}} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{5.08 \times 10^{-7} \, \text{m}} \]

### Step 4: Calculate the Number of Quanta Absorbed and Emitted
Let \( N_{absorbed} \) be the number of quanta absorbed. The total energy absorbed can be expressed as:
\[ E_{total} = N_{absorbed} \cdot E_{absorbed} \]

Given that \( 47\% \) of the absorbed energy is emitted, the energy emitted is:
\[ E_{emitted} = 0.47 \cdot E_{total} = 0.47 \cdot N_{absorbed} \cdot E_{absorbed} \]

Let \( N_{emitted} \) be the number of quanta emitted. The total energy emitted can also be expressed as:
\[ E_{emitted} = N_{emitted} \cdot E_{emitted} \]

### Step 5: Set Up the Ratio
Now, we can set up the ratio of the number of quanta emitted to the number absorbed:
\[ \frac{N_{emitted}}{N_{absorbed}} = \frac{0.47 \cdot N_{absorbed} \cdot E_{absorbed}}{E_{emitted}} \]

### Step 6: Simplify the Ratio
This simplifies to:
\[ \frac{N_{emitted}}{N_{absorbed}} = 0.47 \cdot \frac{E_{absorbed}}{E_{emitted}} \]

### Step 7: Calculate the Ratio
Now, substitute the expressions for \( E_{absorbed} \) and \( E_{emitted} \):
\[ \frac{N_{emitted}}{N_{absorbed}} = 0.47 \cdot \frac{\frac{hc}{\lambda_{absorbed}}}{\frac{hc}{\lambda_{emitted}}} \]
This simplifies to:
\[ \frac{N_{emitted}}{N_{absorbed}} = 0.47 \cdot \frac{\lambda_{emitted}}{\lambda_{absorbed}} \]

Substituting the wavelengths:
\[ \frac{N_{emitted}}{N_{absorbed}} = 0.47 \cdot \frac{5.08 \times 10^{-7}}{4.53 \times 10^{-7}} \]

### Final Calculation
Calculating the ratio:
\[ \frac{N_{emitted}}{N_{absorbed}} = 0.47 \cdot \frac{5.08}{4.53} \approx 0.47 \cdot 1.12 \approx 0.5264 \]

### Conclusion
Thus, the ratio of the number of quanta emitted to the number absorbed is approximately:
\[ \frac{N_{emitted}}{N_{absorbed}} \approx 0.5264 \]