
A black body, which is at a high temperature $T$K, emits thermal radiation at a rate of E $W/{m^2}$. If the temperature falls to T/4 K, the thermal radiation emitted in $W/{m^2}$ will be
$
{\text{A}}{\text{. }}E \\
{\text{B}}{\text{. }}\dfrac{E}{4} \\
{\text{C}}{\text{. }}\dfrac{E}{{64}} \\
{\text{D}}{\text{. }}\dfrac{E}{{256}} \\
$
Answer
561.3k+ views
Hint: According to Stefan-Boltzmann law, the energy of the radiation emitted by a blackbody is directly proportional to the fourth power of the temperature of that blackbody. We can use this law to relate the given energies and temperatures and find out the energy of radiation emitted by the blackbody.
Formula used:
The Stefan-Boltzmann law for a blackbody is given as follows:
$E = \sigma {T^4}$
Complete answer:
We are given a blackbody which is at a high temperature equal to $T$K. The energy of the radiation emitted by the blackbody is given as E $W/{m^2}$. These two quantities can be related to each other through the Stefan Boltzmann law which can be written for this blackbody in the following way.
$E = \sigma {T^4}$
Here $\sigma $ is known as the Stefan’s constant and we can just write that
$E \propto {T^4}$ …(i)
Now the temperature of the given blackbody falls down to T/4 K. Let E’ be the thermal radiation emitted by the blackbody at this new temperature. Using the Stefan-Boltzmann law, we can write that
$E' \propto {\left( {\dfrac{T}{4}} \right)^4}$ …(ii)
Dividing equation (ii) by equation (i), we get
$
\dfrac{{E'}}{E} = \dfrac{{{{\left( {\dfrac{T}{4}} \right)}^4}}}{{{T^4}}} = \dfrac{1}{{{4^4}}} = \dfrac{1}{{256}} \\
\Rightarrow E' = \dfrac{E}{{256}}W/{m^2} \\
$
This is the required value of the energy of the blackbody at the temperature of T/4 K.
Hence, the correct answer is option D.
Note:
It should be noted that a blackbody can be defined as a body which can absorb or emit radiations of all wavelengths at a certain temperature. A blackbody is always characterized by its temperature. The Stefan-Boltzmann law gives us the relation between the amount of energy released or absorbed by a blackbody and the temperature of the blackbody. We study the temperature of stars in the universe based on the energy of the radiations emitted by them using the Stefan-Boltzmann law.
Formula used:
The Stefan-Boltzmann law for a blackbody is given as follows:
$E = \sigma {T^4}$
Complete answer:
We are given a blackbody which is at a high temperature equal to $T$K. The energy of the radiation emitted by the blackbody is given as E $W/{m^2}$. These two quantities can be related to each other through the Stefan Boltzmann law which can be written for this blackbody in the following way.
$E = \sigma {T^4}$
Here $\sigma $ is known as the Stefan’s constant and we can just write that
$E \propto {T^4}$ …(i)
Now the temperature of the given blackbody falls down to T/4 K. Let E’ be the thermal radiation emitted by the blackbody at this new temperature. Using the Stefan-Boltzmann law, we can write that
$E' \propto {\left( {\dfrac{T}{4}} \right)^4}$ …(ii)
Dividing equation (ii) by equation (i), we get
$
\dfrac{{E'}}{E} = \dfrac{{{{\left( {\dfrac{T}{4}} \right)}^4}}}{{{T^4}}} = \dfrac{1}{{{4^4}}} = \dfrac{1}{{256}} \\
\Rightarrow E' = \dfrac{E}{{256}}W/{m^2} \\
$
This is the required value of the energy of the blackbody at the temperature of T/4 K.
Hence, the correct answer is option D.
Note:
It should be noted that a blackbody can be defined as a body which can absorb or emit radiations of all wavelengths at a certain temperature. A blackbody is always characterized by its temperature. The Stefan-Boltzmann law gives us the relation between the amount of energy released or absorbed by a blackbody and the temperature of the blackbody. We study the temperature of stars in the universe based on the energy of the radiations emitted by them using the Stefan-Boltzmann law.
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