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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.
10 years ago

Answers : (3)

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

10 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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10 years ago
Shounak yadav
16 Points
							
E(absorbed)=1242/453=2.742 
E(energy of photon)=1242/508=2.445
47% of absorbed energy is re-emitted out:
47/100*2.742
no. of photons emitted=energy beingemitted/energy of 1 photon
=47/100*2.742/2.444=0.527
11 months ago
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