A blackbody is at a temperature of 2880K. The energy of radiation emitted by this object with wavelength between 499nm and 500nm is `U_1`, between 999nm and 1000nm is `U_2` and between 1499 nm and 1500 nm is `U_3`. The Wien constant `b=2.88xx10^6nmK`. Then

Options

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

Answer

D

Explanation

From Wien's law, `lambda_(m)T` = constant where `T` is the temperature of black body and `lambda_(m)` is the wevelength corresponding to maximum energy of emission. Energy distribution of black body radian is given below:
(i) `U_(1)` and `U_(2)` are not zero because a black emits nearly radiations of all wavelengths.
(ii) Since `U_(1)` corresponding to lower wavelength `U_(3)` corresponding to higher wavelength and `U_(2)` corresponds to medium wave length, hence `U_(2) gt U_(1)`.