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} $