badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
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.

11 years ago

Answers : (5)

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

11 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

--

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

11 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
one year ago
Yash Chourasiya
askIITians Faculty
246 Points
							Hello Student

Energy of light absorbed in an photon= hc/λabsorbed​
Letn1​photons are absorbed
Total energy absorbed= n1​hc/λabsorbed
Now E of light re-emitted out in one photon= hc/λemitted
Letn2​photons are re-emitted then
Total energy re-emitted out= n2​hc/λemitted
Eabsorbed​×47/100 = Ere−emittedout​
hc/λabsorbed​× n1 ​× 47/100 = n2 × hc/λemitted​
n2​/n1​ = 47/100 × λemitted​ /λabsorbed​
= (47/100) × (5080/4530)
n2​/n1​ = 0.527

I hope this solution will help you.
5 months ago
ankit singh
askIITians Faculty
596 Points
							
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.
4 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details