Question related to Chemistry (Black body rediation)
Asked by | 3rd Apr, 2008, 01:21: PM
Heated bodies radiate by processes just like the absorption described above operating in reverse. Thus, for soot heat causes the lattice to vibrate more vigorously, giving energy to the electrons (imagine them as balls in a pinball machine with strongly vibrating barriers, etc.) and since the electrons are charged they radiate away excess kinetic energy. On the other hand, the electrons in a metal have very long mean free paths, the lattice vibrations affect them much less, so they are less effective in radiating away heat. It is evident from considerations like this that good absorbers of radiation are also good emitters.
At sufficiently high temperatures, all bodies become good radiators. Items heated until they glow in a fire look much more similar than they do at room temperature. For a metal, this can be understood in terms of a shortening of the mean free path by the stronger vibrations of the lattice interfering with the electron's passage.
Planck suggested that the blackbody radiation curves could be understood if the oscillators in the walls of the oven were somehow constrained so that they could not continuously emit radiation, as Maxwell's equations would predict, but could only emit it in chunks - called quanta - of a definite magnitude. For higher frequencies, he suggested, the radiation had to be emitted in bigger chunks. This would explain the lack of radiation in the oven at these higher frequencies - assuming the oscillators in the walls had on average energy kT, it would be very unlikely that one oscillating charge would by random excitation have an energy of, say, 5kT, so if its frequency of oscillation were high enough, it would almost never have enough energy to emit one quantum, so by Planck's theory, it wouldn't emit any radiation at all. On the other hand, if the temperature were increased fivefold, it would often have enough energy to emit a quantum.
Answered by | 12th May, 2008, 09:11: AM
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