Transmission Of Heat Question 111
Question: A black body is at a temperature of $ 2880\ 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} $ . The Wein’s constant$ b=2.88\times 10^{6}\ nm,K $ . Then
[IIT 1998]
Options:
A) $ U _{1}=0 $
B) $ U _{3}=0 $
C) $ U _{1}>U _{2} $
D) $ U _{2}>U _{1} $
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Answer:
Correct Answer: D
Solution:
Wein’s displacement law is $ {\lambda _{m}}T=b $
Therefore $ {\lambda _{m}}=\frac{b}{T}=\frac{2.88\times 10^{6}}{2880}=1000,nm. $ Energy distribution with wavelength will be as follows From the graph it is clear that U2 > U1.
BETA
