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Vishrant Vasavada Grade: 11
        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.
7 years ago

Answers : (2)

Pratham Ashish
17 Points
										

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

7 years ago
Ramesh V
70 Points
										

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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Regards,

Naga Ramesh

IIT Kgp - 2005 batch

7 years ago
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