NEET Solved Paper 2016 Question 15
Question: A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is $ {U_1}\text{,} $ at wavelength 500 nm is $ {U_2} $ and that at 1000 nm is $ {U_3}. $ Wien’s constant, $ \text{b = 2}\text{.88 }\times\text{ 1}{0^{6}}\text{ nmK}\text{.} $ Which of the following is correct?
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:
Maximum amount of emitted radiation corresponding to $ {\lambda_m}=\frac{b}{T} $ $ {\lambda_m}=\frac{2.88\times 10^{6},nmK}{5760K}=500,nm $ From the graph $ {U_1}{U_2}{U_3} $
BETA
