The dye acriflavine when dissolved in water has its maximum light absorption at 4530 Å andhas maximum florescence emission at 5080 Å. The number of florescence quanta is about 53% ofthe number of quanta absorbed. What percentage of absorbed light energy is emitted asflorescence?

The energy of a photon is given by the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the light.

1. First, we need to calculate the energy of the absorbed light. The wavelength of the absorbed light is given as 4530 Å. Convert this to meters: 4530 Å = 4530 x 10^-10 m.

2. Plug these values into the equation to find the energy of the absorbed light: E(absorbed) = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s) / (4530 x 10^-10 m) = 4.39 x 10^-19 J.

3. Next, calculate the energy of the emitted light in the same way. The wavelength of the emitted light is given as 5080 Å = 5080 x 10^-10 m.

4. Plug these values into the equation to find the energy of the emitted light: E(emitted) = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s) / (5080 x 10^-10 m) = 3.91 x 10^-19 J.

5. The problem states that the number of fluorescence quanta is about 53% of the number of quanta absorbed. This means that the total energy emitted is 53% of the energy calculated in step 4: E(total emitted) = 0.53 * 3.91 x 10^-19 J = 2.07 x 10^-19 J.

6. Finally, calculate the percentage of absorbed light energy that is emitted as fluorescence. This is given by the ratio of the total energy emitted to the energy absorbed, multiplied by 100%: (2.07 x 10^-19 J / 4.39 x 10^-19 J) * 100% = 47.15%.

So, about 47.15% of the absorbed light energy is emitted as fluorescence.