Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !