Black body radiation

[[Tycho?]]

Is there a way to find the frequency of emmited radiation of a black body when only the temperature is known? Does it depend on the material, another quality, or is it possible at all? Perhaps this doesn't reffer to black body radiation at all, but from what I've read it seems to be what I'm looking for.

 

I just think of something like the tungsten in light bulbs. At very high temperatures, it emites light in the visible range. Is there a way to (preferably simply) explain this relationship?

[cyeokpeng]

NB: We consider an ideal blackbody so that the range of frequencies emitted

is only due to its own temperature and NOT through reflection off the

body's surface.

An ideal blackbody has this property that it will absorb all the frequency

of EM radiation falling on it and none gets reflected.

 

'']Is there a way to find the frequency of emmited radiation of a black body when only the temperature is known?

 

The frequency (or wavelength for that matter. wavelength=c/frequency) of emitted radiation due to an ideal blackbody according to quantum theory consists of a range/multitude of frequencies in the whole EM spectrum. However, the range of frequencies plotted against the intensity of the emitted radiation has this bell shape curve (NB: Correct me if I m wrong this bell-shape curve follows the Rayleigh distribution) which has the intensity of radiation peaked at a certain frequency or wavelength following the Wien's Displacement law:

 

lamda(max)*T = 0.2898*10^-2 m.K

 

where lamda(max) = wavelength of the emmited radiation having the highest

intensity

T = temperature in Kelvins

 

Note that the above only calculates the frequency of radiation emmited with the highest intensity. There are other range of frequencies emitted too, but the intensity tapers off quickly as you move away from the lamda(max) value following the bell-shape.

This is the reason why in total darkness, you will not see your buddy emitting visible light from his body, shining like a star. Most of the intensity of EM radiation emmited by a person's body due to its own temperature 300K is at the frequency range of infrared region, and negligible intensity of EM radiation emmited is in the visible light region. (You can calculate it yourself using the Wien's dispacement law shown above) This is also the reason why many high-end operations to discover the enemies position uses infrared imaging detectors.

 

The actual relationship of the intensity-wavelength plot (Your bell-shape curve) is given by the Planck's blackbody radiation law:

 

I(lamda,T) = (2*pi*h*c^2)/(lamda^5*(exp((hc/lamda)*kT-1)))

 

where I(lambda,T) = intensity of emmited radiation at a particular wavelength

of radiation and temperature of the blackbody.

lamda = wavelength of radiation in m

T = temperature of blackbody in K

h = Planck's constant 6.626*10^-34 J.s

c = speed of light/EM waves 3*10^8 m/s

k = Boltzmann constant 1.38*10^-23 J/K

 

'']Does it depend on the material, another quality, or is it possible at all? Perhaps this doesn't reffer to black body radiation at all, but from what I've read it seems to be what I'm looking for.?

 

No, it does not depend on the characteristic of the material. The Planck's law of blackbody radiation is independent of the property of the blackbody material unless its atomic structure of material is degenerate, as in the case of neutron stars. This is because normal object containing matter in our universe are not degenerate in their atomic structure, and therefore can use the same Boltzmann distribution statistics to model its atomic structure, together with the application of Quantum theory since it involves small quantum particles like the atoms, and get the famous Planck's law of blackbody radiation. If the microstructure of the blackbody is degenerate (like in neutron stars where they are neutron-degenerate), then Planck's law does not hold and the blackbody radiation intensity-wavelength curve is obtained by other theories.

 

'']I just think of something like the tungsten in light bulbs. At very high temperatures, it emites light in the visible range. Is there a way to (preferably simply) explain this relationship?

 

If you have understood the above, this should be easy to answer.

The tungsten in light bulbs is normal matter and not degenerate matter, so the Planck's law is applicable here.

To get an idea in which region of frequencies most of the intensity of emmited EM radiation is in, we can apply the Wien's displacement law first to get the wavelength of max. intensity, which is found to be in the high-frequency end of infrared region if you use the max. steady temperature of the filament to be roughly 1000 K. Since the peak is nearer to the visible spectrum, you can deduce that there is observable intensity of emmited EM radation that is in the visible light range. This explains why it lights up as tungsten filament heats up. But careful here, the intensity of radiation peaks at infrared, which means that a large portion of power is waste heat (infrared) and this explains why light bulb using tungsten is not efficient to be used as lighting. More efficient ones uses metallic gas lamps known as flourescent lamps.

[[Tycho?]]

This totally answers my question, thank you very much, great explanation.

[5614]

agreed, great explanation.

 

cheers

[cyeokpeng]

I actually dig out my previous Physics notes and textbook to refresh my concepts. No much of a problem since I also learn a lot in the process. We help each other, that's the spirit.

[1veedo]

I thought background radiation (which is blackbody isnt it) was measured using raidio recievers. How do you measure the background radiation? Is it degenerate?

[swansont]

I thought background radiation (which is blackbody isnt it) was measured using raidio recievers. How do you measure the background radiation? Is it degenerate?

 

If you mean the cosmic background, measured by Penzias and Wilson, they noticed it was still there when they pointed their antenna into deep space, and had a siderial day period.

[cyeokpeng]

I wonder how they calculate the temperature of cosmic background, after measuring it using radio receivers?

[Radical Edward]

I wonder how they calculate the temperature of cosmic background, after measuring it using radio receivers?

 

well we can see from the spectral distribution that the CMB is a black Body spectrum. and since Black Body spectra are dependant only on the temperature of the object, and not on anything else, such as colour or material properties of the materials emitting the spectrum, then we adjust the temperature in the calculations in order to find the temperature that produces the same spectral profile. alternatively you can differentiate it and find the point where the differential of the BBS hits zero in the same place as the peak of the observed spectrum. these are basically equivalent, but it is easier finding a zero on a graph.

 

basically just use Weins displacement law as described above.