How can one derive the amount of light as well as the shortest wavelength of the emitted light from the temperature of an object due to black body radiation? Shouldn't the amount of light emitted (different wavelengths) be proportional to the temperature, and therefore the shortest wavelength too?
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$\begingroup$ There is no shortest wavelength, although short wavelengths (with photon energy large compared to the temperature), are exponentially suppressed. In other words, your oven creates minute (!) amounts of Röntgen radiation. $\endgroup$– Sebastian RieseCommented May 13, 2015 at 15:59
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The spectrum of the black body radiation is given by Planck's law. The total amount of radiation is given by the Stefan-Boltzmann law.
In principle there is no shortest wavelength, because the radiated intensity remains non-zero at arbitrarily small wavelengths. However, at the low wavelength end of the spectrum the radiated intensity falls exponentially with decreasing wavelength so the intensity becomes negligable at low enough wavelengths. You'd have to define the shortest wavelength using some condition like the wavelength at which the intensity falls to x% of the maximum.
If you define a shortest wavelength like this then you're quite correct that the shortest wavelength will be temperature dependant.
answered May 13, 2015 at 15:59