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
A black body is at a temperature of 2880K . The energy of radiation emitted by this body with wavelength between 499nm and 500nm is U 1, between 999nm and 1000nm , is U 2 and between 1499nm and 1500nm nm is U 3 . Wien's constant, b=2.88× 10 6 nm-K, then,
PhysicsThermal Properties of MatterJEE Main
Options:
  • A U 1 =0
  • B U 3 =0
  • C U 1 > U 2
  • D U 2 > U 1
Solution:
1974 Upvotes Verified Answer
The correct answer is: U 2 > U 1
Wien's displacement law is

             λ m T = b       (b = Wien's constant)

                 λ m = b T = 2.88 × 1 0 6 nm - K 2 8 8 0 K

                   λ = 1 0 0 0 nm

Energy distribution with wavelength will be as follows :


            


From the graph it is clear that

 U2 > U1              (in fact U2 maximum)

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.