A black body is at a temperature of 2800 K. The energy of radiation emitted by this object 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 is `U_(3)`.Wien's constant `b = 2.88 xx 10^6 nm - K`, then

Options

A. `U_(1) = 0`
B. `U_(3) = 0`
C. `U_(1) gt U_(2)`
D. `U_(2) gt U_(1)`

Answer

D

Explanation

At `2880 K`: `lambda_(m) = (b)/(T) = (2.88 xx 10^(6) nm -K)/(2880K) = 1000nm`

Therefore `U_(2) gt U_(1)` & `U_(2) gt U_(3)`