Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The photosensitive surface is receiving the light of wavelength 5000 A at the rate of 10-8  s-1The number of photons received per second is (h=6.62×10-34 J s, c=3×108 m s-1)
PhysicsDual Nature of MatterJEE Main
Options:
  • A 2.5×1010
  • B 2.5×1011
  • C 2.5×1012
  • D 2.5×109
Solution:
1263 Upvotes Verified Answer
The correct answer is: 2.5×1010
Energy of photon

    E=hcλ

Given,   λ=5000  = 5×10-m

     E=6.6×10-34×3×1085×10-7

= 3.96×10-19 J

Energy received per second = 10-8 s-1

Number of photon's received per second

= Energy received per second Energy of one photon 

= 10-83.96×10-19=2.5×1010

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.