
The radiations emitted by a perfectly black body is proportional to
A) Temperature on Ideal gas scale
B) Fourth root of temperature on Ideal gas scale
C) Fourth power of temperature on ideal gas scale
D) Square of temperature on ideal gas scale
Answer
577.2k+ views
Hint: Recall Stefan’s law for the black body radiation, it gives the relation of temperature with energy emitted. A black body is a physical body that can absorb all incident radiation, of all the frequencies and any angle of incidence. The term name black body was coined because it can absorb different frequency radiation. Here we have to use the formula for power radiated by a black body.
Complete step by step solution:
The Stefan–Boltzmann law provides the relation temperature of a black body with the power radiated by it. Specifically, the Stefan–Boltzmann law gives us the total energy radiated of a black body per unit surface area for different wavelengths per unit time (also known as the black-body radiant emittance).
According to Stefan’s law, the amount of energy emitted by a black body is proportional to the 4th power of temperature.
$E = \dfrac{P}{A} = e\sigma {T^4}$.
Where, $E$ = Energy emitted,
$P$ =power
$\sigma $=Stefan’s constant and $\sigma = 5.670373 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$
$T$ =Absolute temperature,
$A$ =Area of the surface
$ \Rightarrow E \propto {T^4}$
Since radiations are emitted in the form of energy waves hence option (C) is the correct option and all other options are incorrect.
Additional Information:
The blackbody has three properties:
(a) It is the surface that emits most for a given temperature and wavelength.
(b) Blackbody radiation does not depend on the direction, that is, blackbody radiation is diffuse
(c) Total blackbody radiation in a vacuum depends only on temperature. Since the blackbody is the perfect absorber and emitter, it will be used as a reference to compare the radioactive properties of real surfaces.
Note:
The most common mistake done in this type of problem is the candidates most of the time are confused between the perfect black body and black body. As in the case of this problem we are given a perfect black body, where $\varepsilon $=1. But if the question only had a black body given then we cannot write $\varepsilon $=1.
Complete step by step solution:
The Stefan–Boltzmann law provides the relation temperature of a black body with the power radiated by it. Specifically, the Stefan–Boltzmann law gives us the total energy radiated of a black body per unit surface area for different wavelengths per unit time (also known as the black-body radiant emittance).
According to Stefan’s law, the amount of energy emitted by a black body is proportional to the 4th power of temperature.
$E = \dfrac{P}{A} = e\sigma {T^4}$.
Where, $E$ = Energy emitted,
$P$ =power
$\sigma $=Stefan’s constant and $\sigma = 5.670373 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$
$T$ =Absolute temperature,
$A$ =Area of the surface
$ \Rightarrow E \propto {T^4}$
Since radiations are emitted in the form of energy waves hence option (C) is the correct option and all other options are incorrect.
Additional Information:
The blackbody has three properties:
(a) It is the surface that emits most for a given temperature and wavelength.
(b) Blackbody radiation does not depend on the direction, that is, blackbody radiation is diffuse
(c) Total blackbody radiation in a vacuum depends only on temperature. Since the blackbody is the perfect absorber and emitter, it will be used as a reference to compare the radioactive properties of real surfaces.
Note:
The most common mistake done in this type of problem is the candidates most of the time are confused between the perfect black body and black body. As in the case of this problem we are given a perfect black body, where $\varepsilon $=1. But if the question only had a black body given then we cannot write $\varepsilon $=1.
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