Answer
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Hint: A blackbody is a theoretical body that absorbs all electromagnetic radiation falling on it. It is a hypothetical object which is a perfect absorber and a perfect emitter of wavelengths radiation. The term originates because for a blackbody all visible light will be absorbed rather than reflected, and therefore the surface will appear black.
Complete solution:
First, let us understand what are the two laws to explain the black body.
Wien’s law: Blackbodies emit radiation across multiple wavelengths. The cooler the object, the longer the wavelength at which most of the radiation will be emitted. Conversely, the hotter the object, the shorter the wavelength at which the peak radiation is emitted. This relationship is represented by Wien’s Displacement Law.
${\lambda _m} = \dfrac{A}{T}$ , where ${\lambda _m}$ is the wavelength of maximum radiation, $A = 2898\mu m.K$ is a constant, and $T$ is temperature.
The wavelength at which maximum energy is emitted is inversely related to the blackbody’s temperature. The spectrum of a blackbody is continuous, meaning it gives off some energy across all wavelengths, but it has a peak at a specific wavelength. The peak of the blackbody curve in a spectrum moves to shorter wavelengths for hotter objects.
Stefan-Boltzmann law: How much energy an object radiates (per a given surface area) is a function of the surface temperature of the object. This property is expressed by the Stefan–Boltzmann law.
$M = \sigma {T^4}$ , where $M$ is the total energy emitted from the surface of the material, $\sigma $ is the Stefan-Boltzmann constant, and $T$ temperature of the emitting material.
The Stefan-Boltzmann Law gives the total energy being emitted at all wavelengths by the blackbody. It is important to note that the total energy increases rapidly with temperature since it varies to the fourth power of the temperature. The primary parameter that determines how much light the blackbody gives off, and at what wavelengths, is its temperature. No object is an ideal blackbody, but many objects behave approximately like blackbodies.
Note: In practical life there exist no such body that can absorb all the incident radiation but some surfaces are very close to the ideal body. Similarly, there does not exist a body that can reflect or transmit all the incident radiation. Every surface absorbs less or more radiation. It is obvious that if a black body absorbs all the incident radiation then it will not reflect or transmit any amount of radiation.
Complete solution:
First, let us understand what are the two laws to explain the black body.
Wien’s law: Blackbodies emit radiation across multiple wavelengths. The cooler the object, the longer the wavelength at which most of the radiation will be emitted. Conversely, the hotter the object, the shorter the wavelength at which the peak radiation is emitted. This relationship is represented by Wien’s Displacement Law.
${\lambda _m} = \dfrac{A}{T}$ , where ${\lambda _m}$ is the wavelength of maximum radiation, $A = 2898\mu m.K$ is a constant, and $T$ is temperature.
The wavelength at which maximum energy is emitted is inversely related to the blackbody’s temperature. The spectrum of a blackbody is continuous, meaning it gives off some energy across all wavelengths, but it has a peak at a specific wavelength. The peak of the blackbody curve in a spectrum moves to shorter wavelengths for hotter objects.
Stefan-Boltzmann law: How much energy an object radiates (per a given surface area) is a function of the surface temperature of the object. This property is expressed by the Stefan–Boltzmann law.
$M = \sigma {T^4}$ , where $M$ is the total energy emitted from the surface of the material, $\sigma $ is the Stefan-Boltzmann constant, and $T$ temperature of the emitting material.
The Stefan-Boltzmann Law gives the total energy being emitted at all wavelengths by the blackbody. It is important to note that the total energy increases rapidly with temperature since it varies to the fourth power of the temperature. The primary parameter that determines how much light the blackbody gives off, and at what wavelengths, is its temperature. No object is an ideal blackbody, but many objects behave approximately like blackbodies.
Note: In practical life there exist no such body that can absorb all the incident radiation but some surfaces are very close to the ideal body. Similarly, there does not exist a body that can reflect or transmit all the incident radiation. Every surface absorbs less or more radiation. It is obvious that if a black body absorbs all the incident radiation then it will not reflect or transmit any amount of radiation.
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