Intensity and wavelength _ of any emission depends on temperature
Question
Intensity and wavelength _ of any emission depends on temperature
Solution
The intensity and wavelength of any emission indeed depend on the temperature of the emitting body. This is explained by Planck's Law of Black Body Radiation. Here's a step-by-step explanation:
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Planck's Law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T. The law is named after Max Planck, who proposed it in 1900.
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According to Planck's Law, the energy of a photon, the basic unit of light, is directly proportional to its frequency. The higher the frequency, the higher the energy. This relationship is given by the equation E = hv, where E is energy, h is Planck's constant, and v is frequency.
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The frequency of light is inversely proportional to its wavelength. This means that light with a high frequency has a short wavelength, and light with a low frequency has a long wavelength.
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When the temperature of a body increases, it emits photons with higher energy, i.e., higher frequency and shorter wavelength. This is why heated metal glows red at first (long wavelength, low frequency), then turns white as it gets hotter (a mix of all visible wavelengths, high frequency).
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The intensity of the emitted radiation also increases with temperature. This is because an increase in temperature means an increase in the vibration of the particles of the body, which in turn leads to the emission of more photons.
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Therefore, both the intensity and the wavelength of the emission depend on the temperature of the body. The hotter the body, the shorter the wavelength (i.e., the higher the frequency) and the greater the intensity of the emitted light.
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