Question

The power radiated by a black body is P and it radiates maximum energy at wavelength. $\lambda \circ $. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\dfrac{8}{4}\lambda \circ $, the power radiated by it becomes Np. The value of n is?
A. $\dfrac{{256}}{{81}}$
B. $\dfrac{3}{4}$
C. $\dfrac{{81}}{{256}}$
D. $\dfrac{4}{3}$

Answer 1

Hint the emitting radiation problems can be solved by the Stefan’s Boltzmann law or a Wien’s displacement law. We can apply the Wien’s displacement law. Apply this law at the maximum intensity and substitute the given value of the wavelength and the temperature to the value of the unknown temperature.
Formula used: $\mathop \lambda \nolimits_{\max } = \dfrac{C}{T}$

Complete step by step answer:
In Wien’s displacement law, it is the ratio of the temperature of a black body and the wavelength at which it emits the light. The temperature of a black body is an ideal substance which can emit and absorb all frequencies of light.
$\mathop \lambda \nolimits_{\max } = \dfrac{C}{T}$
C= Wein’s displacement law
T= temperature
$\mathop \lambda \nolimits_{\max } $= it is the maximum wavelength
So, $\mathop \lambda \nolimits_{\max } T = $constant
To find the ratio we can use Wein’s displacement law.
$i.e.,\mathop \lambda \nolimits_{\max 1} \mathop T\nolimits_1 = \mathop \lambda \nolimits_{\max 2} \mathop T\nolimits_2 $
\[ \Rightarrow \lambda \circ T = \dfrac{{3\lambda \circ }}{4}T'\]
Therefore, $T' = \dfrac{4}{3}T$
Now the power radiated by it is $P \propto \mathop T\nolimits^4 $
So, $ \Rightarrow \dfrac{{\mathop P\nolimits_1 }}{{\mathop P\nolimits_2 }} = n$
$ \Rightarrow n = \mathop {\left( {\dfrac{{T'}}{T}} \right)}\nolimits^4 $
$ \Rightarrow n = \mathop {\left( {\dfrac{4}{3}} \right)}\nolimits^4 = \dfrac{{256}}{{81}}$
$ \Rightarrow n = \dfrac{{256}}{{81}}$
Then the value of n is= $\dfrac{{256}}{{81}}$.
Hence, the option (A) is correct.

Note:- We have used the Wein’s displacement law in the case for the given wavelength is at the maximum intensity of the emission and we can not use the same equation if the intensity is not maximum and we take care of this condition while we solve the problem. Candidates are advised to cross-check the answer after the calculations since the option contains similar answers.