NEET Solved Paper 2018 Question 43
Question: The power radiated by a black body is P and it radiates maximum energy at wavelength, $ {\lambda_0} $ . If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $ \frac{3}{4}{\lambda_0} $ , the power radiated by it becomes $ nP $ . The value of n is [NEET - 2018]
Options:
A) $ \frac{256}{81} $
B) $ \frac{4}{3} $
C) $ \frac{3}{4} $
D) $ \frac{81}{256} $
Show Answer
Answer:
Correct Answer: A
Solution:
We know, $ {\lambda _{max}}\text{T=} $ constant (Wien’s law)
So, $ {\lambda _{ma{x_1}}}{T_1}\text{=}{\lambda _{ma{x_2}}}{T_2} $
$ \Rightarrow {\lambda_0}T=\frac{3{\lambda_0}}{4}T’ $
$ \Rightarrow T’=\frac{4}{3}T $
So, $ \frac{P _2}{P _1}={{( \frac{T’}{T} )}^{4}}={{( \frac{4}{3} )}^{4}}=\frac{256}{81} $
BETA
