A black body is at a temperature of 2880 K . The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U 1 , between 999 nm and 1000 nm , is U 2 and between 1499 nm and 1500 nm nm is U 3 . Wien's constant, b

Question

A black body is at a temperature of 2880 K . The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U 1 , between 999 nm and 1000 nm , is U 2 and between 1499 nm and 1500 nm nm is U 3 . Wien's constant, b = 2.88 × 10 6 nm-K, then,

Quantrex Academy · Accepted Answer

Correct answer: D. U 2 &#x003E; 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 U 2 > U 1 (in fact U 2 maximum)

A black body is at a temperature of 2880 K . The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U 1 , between 999 nm and 1000 nm , is U 2 and between 1499 nm and 1500 nm nm is U 3 . Wien's constant, b = 2.88 × 10 6 nm-K, then,

Options

Correct answer

Step-by-step solution

More from Thermal Properties of Matter