A black body is at a temperature of `2800K` The energy of radiation emitted by this object with wavelength between `499nm` and 500 nm is `U_(1)` between 999nm and 1000nm is `U_(2)` and between 1499 nm and 1500 nm is `U_(3)` The Wien's constant `b =2.80 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)`