Need help with blackbody radiation

In summary, the conversation discusses the derivation of expressions for B(f) in the cases of hf << kT and hf >> kT. The first part is easily solved using a Taylor expansion, but the second part is more complicated. It is noted that for hf >> kT, the exponential in the Planck law should be much larger than 1, resulting in the neglection of the 1 in the denominator. This leads to the Wien distribution being derived from the Planck distribution.
  • #1
Nylex
552
2
Can someone help me with this please?

B(f) = [(2hf^3)/c^2]{1/[exp (hf/kT) - 1]} - Planck law in terms of frequency.

"Derive expressions for B(f) for the cases i. hf << kT, ii. hf >> kT".

I've done the first bit, that's just using a Taylor expansion. The second bit is where I'm stuck. In my lecture notes, I have that for hf >> kT, B(f) = [(2hf^3)/c^2]exp(-hf/kT). What I don't understand is why it's like that. If hf >> kT, shouldn't that exponential tend to infinity?

Thanks.
 
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  • #2
Yes,the exponential should much larger than 1 and therefore the 1 in the denominator would be neglected...The exp.would come on top with a "-" in its argument and from the Planck distribution,u'd be getting the Wien distribution...

Daniel.
 
  • #3
OH! Thanks :smile:.
 

1. What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by a perfect blackbody, which is an object that absorbs all incoming radiation and reflects or emits none of it. This type of radiation is a result of the thermal energy of the object and is emitted at different wavelengths depending on its temperature.

2. How is blackbody radiation related to temperature?

The intensity and wavelength of blackbody radiation are directly proportional to the temperature of the object. This means that as the temperature increases, the intensity and peak wavelength of the radiation also increase. This relationship is described by the Planck's law of blackbody radiation.

3. What is the significance of blackbody radiation?

Blackbody radiation is important in many fields of science, including astronomy, thermodynamics, and quantum mechanics. It helps to explain the emission of light from stars and other celestial bodies, and it is also used in the development of technologies such as solar cells and infrared cameras.

4. How is blackbody radiation measured?

Blackbody radiation can be measured using a spectrometer, which is an instrument that separates and measures the different wavelengths of light. The intensity of the radiation at different wavelengths can then be plotted on a graph called a blackbody curve, which provides information about the temperature of the object.

5. Can blackbody radiation be seen?

Yes, blackbody radiation can be seen, but it is only visible at certain temperatures. At lower temperatures, the radiation is in the infrared range and cannot be seen by the human eye. As the temperature increases, the radiation shifts towards the visible spectrum, and at very high temperatures, it can even be seen as visible light.

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