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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. 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
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 -- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and 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. Regards, Naga Ramesh IIT Kgp - 2005 batch
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
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and 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. Regards, Naga Ramesh IIT Kgp - 2005 batch
E(absorbed)=1242/453=2.742 E(energy of photon)=1242/508=2.44547% of absorbed energy is re-emitted out:47/100*2.742no. of photons emitted=energy beingemitted/energy of 1 photon=47/100*2.742/2.444=0.527
Hello StudentEnergy of light absorbed in an photon= hc/λabsorbedLetn1photons are absorbedTotal energy absorbed= n1hc/λabsorbedNow E of light re-emitted out in one photon= hc/λemittedLetn2photons are re-emitted thenTotal energy re-emitted out= n2hc/λemittedEabsorbed×47/100 = Ere−emittedouthc/λabsorbed× n1 × 47/100 = n2 × hc/λemittedn2/n1 = 47/100 × λemitted /λabsorbed = (47/100) × (5080/4530)n2/n1 = 0.527I hope this solution will help you.
A certain dye absorbs 4530A∘ and fluoresence at 5080A∘ these being wavelength of maximum absorption that under given condition 47% of the absorbed energy is emitted. Calculate the ratio of the no of quanta emitted to the number absorbed.
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