Question

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 $ is $ {U_3} $ . Wien’s constant, $ b = 2.88 \times {10^6}nm - k $ . Then
(A) $ {U_1} = 0 $
(B) $ {U_3} = 0 $
(C) $ {U_1} = {U_2} $
(D) $ {U_1} > {U_2} $

Answer 1

Hint: to solve this question, we have to know what Wien’s Law is. We know that, Wien's law, additionally called Wien's displacement law, is the connection between the temperature of a blackbody (an ideal substance that radiates and assimilates all frequencies of light) and the frequency at which it emanates the most light. We can say,
Wien's law is applicable to some ordinary encounters:

Complete answer:
A piece of metal warmed by a blow light initially becomes "intensely hot" as the longest noticeable frequencies seem red, at that point turns out to be more orange-red as the temperature is expanded, and at high temperatures would be depicted as "white hot" as more limited and more limited frequencies come to prevail the dark body outflow range. Before it had even arrived at the scorching temperature, the warm discharge was for the most part at longer infrared frequencies, which are not noticeable; in any case, that radiation could be felt as it warms one's close by skin
Step by step solution: we know that, Wien’s displacement law is,
 $ {\lambda _m}T = b $ , here b is Wien’s constant.
Or, $ {\lambda _m} = b/T = \dfrac{{2.88 \times {{10}^6}nm - K}}{{2880K}} $
 $ \lambda = 1000nm $
If we plot the energy distribution with wavelength graph we will see that $ {U_1} > {U_2} $
So, option D. is correct.

Note:
We have to keep that in mind, inferring the Wien's Displacement Law from Planck's Law. Wien's relocation law expresses that the dark body radiation bends for various temperatures at a frequency conversely corresponding to the temperature.