Q.

A specific dye when dissolved in water was found to absorb maximum light at 4500Ao. Its maximum fluorescence emission is at 5000Ao. On the average, the number of fluorescence quota is 52% of the number of quanta absorbed. Calculate the percentage of absorbed light that is emitted as fluorescence. Use wavelengths of maximum absorption and emission.

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 46.9.

(Unlock A.I Detailed Solution for FREE)

Detailed Solution

(i) Energy (E) absorbed by each photon=hv=hcλ. Hence:E=6.626×10−34Js×3.0×108ms−4500×10−10m=4.4×10−19J(ii) Energy (E) emitted by each photon=hcλ=6.626×10−34Js×3.0×108ms−5000×10−10m=3.97×10−19J∴ Energy fraction emitted=3.97×10−19J4.4×10−19J×52100×100=46.9%
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon