A certain dye absorbs 4530 A and fluoresces at 5080 A these being wavelengths of max. absorption that under conditions 47% of the absorbed energy is emitted. Calculate ratio of quanta emitted to the number absorbed.

let no. of femitted quanta = n1

& of absorbed = n2

frequency of emitted quanta  f1 =  c/ 5080* 10^-10

energy e1  =  h*f1 = h*  c/ 5080* 10^-10

frequency of absorbed quanta f2 = c/ 4530* 10^-10

energy e2 =   h*  c/ 4530* 10^-10

   total energy of emitted quanta =  0.47 0f  absorbed quanta

hence,

 n1* h*  c/ 5080* 10^-10 =   0.47*n2* h*  c/ 4530* 10^-10

n1/5080 = 0.47 * n2 /4530

n1/n2 =  0.527

Quantum efficiency(Q.E) is defined as ratio of quanta emitted to the number absorbed

it is expressed in ratio of energies

quanta absorbed on dye : E = hc/lamda = 1242 eV-nm /453 nm = 2.742 eV

quanta exhibited/undergo fluorescence on dye : E = hc/lamda = 1242 eV-nm /508 nm = 2.445 eV

 

Q.E = 0.47*2.742 / 2.445

       = 0.527

or 52.7 % efficient

I'm not sure if the method was correct, plz verify with someone else

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Naga Ramesh
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Hello Student

Energy of light absorbed in an photon= hc/λ_absorbed​
Letn₁​photons are absorbed
Total energy absorbed= n₁​hc/λ_absorbed​
Now E of light re-emitted out in one photon= hc/λ_emitted​
Letn₂​photons are re-emitted then
Total energy re-emitted out= n₂​hc/λ_emitted​
E_absorbed​×47/100 = E_{re−emittedout​}
hc/λ_absorbed​× n₁ ​× 47/100 = n₂ × hc/λ_emitted​
n₂​/n₁​ = 47/100 × λ_emitted​ /λ_absorbed​
= (47/100) × (5080/4530)
n₂​/n₁​ = 0.527

I hope this solution will help you.