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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.`
8 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
```
8 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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we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

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Naga Ramesh
IIT Kgp - 2005 batch
```
8 years ago
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