# Black Body radiation

1. Oct 16, 2011

### slft

In black body radiation, there are two regions of low intensity. One is at the high frequencies and one is at the low frequencies. I understand that there is lower probability to emit radiation at high frequencies because it requires higher energy than the average thermal energy provided. However, why is there also a low intensity in lower frequencies? Is it also the because of low probability of receiving the low energy? If so, does it mean that thermal energy is also discrete?

Last edited: Oct 16, 2011
2. Oct 16, 2011

### Pete99

I'd say that the low probability of emitting in the low frequency region is because the smaller number of modes in which you can emit. You can find a good explanation in http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html" [Broken]

Last edited by a moderator: May 5, 2017
3. Oct 16, 2011

### slft

so is thermal energy discrete?

4. Oct 17, 2011

### Pete99

My understanding is that thermal energy is the mean energy of the random motion of the particles in a system. So I would say that the energy of the individual particles is quantized, but that thermal energy, being the mean value, does not necessarily have to be quantized.

But I am actually not sure about this, so maybe someone have a better answer.

5. Oct 17, 2011

### chrisbaird

If you mean by thermal energy the total random kinetic molecular energy, than this number will be discrete because it is a sum of discrete numbers. But a typical object has so many billions upon billions of molecules that the total thermal energy of an object is going to look so close to continuous that for practical purposes you might as well treat it that way. If you mean by thermal energy the average random kinetic molecular energy, than averages of discrete number sets are not discrete.

To get the correct model of blackbody radiation (one that matches experiment), you have to assume the molecular oscillators are quantized.

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