Question

As the wavelength of the radiation decreases the intensity of the black body radiation ​

Answer 1

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Answer 2

The wavelength of the radiation decreases the intensity of the black body radiation ​

• Electromagnetic radiation is emitted by all things with a temperature greater than absolute zero (0 K, -273.15 oC).

• A blackbody is a hypothetical or model body that absorbs all incident radiation and does not reflect or emit any. It is a hypothetical item that is a "perfect" radiation absorber and emitter across all wavelengths.

• A blackbody's spectral distribution of thermal energy emitted (i.e., the pattern of the intensity of the radiation over a range of wavelengths or frequencies) is solely determined by its temperature.

• Several rules can be used to describe the properties of blackbody radiation:

---->Planck's Law of Blackbody Radiation is a formula for calculating the spectral energy density of emission at each wavelength (E) for a given absolute temperature (T).

                         Eλ =          8$\pi hc$

                               ---------------------------------

                                      λ$λ^{2}$(ehc/λxT)-1)

---->The frequency of the emission peak (fmax) rises linearly with absolute temperature, according to Wien's Displacement Law (T). In contrast, when the body's temperature rises, the wavelength at the emission peak drops.

                        =   $f_{max}$ $\alpha$ T

--->The Stefan–Boltzmann Law is a relationship between total energy released (E) and absolute temperature (T) (T).

                    =     E $\alpha$ $T^{4}$