Dash
ReSpEcTeD
First of all... What do these curves represent?
And why is there more than one???
And why is there more than one???
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Black Body Radiation Curve (1 Viewer)
- Thread starter Dash
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The graph is of the amount of energy radiated vs the wavelength. There is more than one function because they are at different temperatures. For example, from the graph it can be seen that at 2400K, there is a greater amount of energy radiated across the frequency (or wavelength) spectrum than at 6000K. Also, for all temps the max energy radiated is at ~200-300nm.
Rahul
Dead Member
basically, the peak intensity moves to a lower wavelength and higher frequency radiation with increasing temperature. in other words, the dominant wavelength in a blackbody when the temperature is 24 000 K will be shorter than the dominant wavelength when the temperature is 6 000 K.
24 000K
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Rahul
Dead Member
the graph isnt really related to the idea of a quanta of energy, but blackbody radiation is.
when plank investigated blackbodies he postulated that EMR was not emitted continuously, but in packets of energies, a quanta.
the amount of energy packaged in each of these packets was proportional to the frequency, E # f(#- proportional to). so E = hf, h- constant.
E- quantum of energy in joules
f- frequency of radiation
h- planck's constant
since c=f.(lambda),
E= hc/(lambda)
planck suggested that energy only occured in packets of quantity hf. different levels of energy came in the fom of multiples of hf, ie- 2hf or 4hf. thus we say its quantised.
and lastly, looking at the dot-point related to this:
identify planck's hypothesis that radiation emitted and absorbed by the walls of a black body cavity if quantised
so you dont need to know this in much detail. at most you will have to re-guritate the hypothesis itself.
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Dash
ReSpEcTeD
Ok, Im still not getting it. Why on earth is there more than one
line? I undertand that the graph measures the energy radiated
from a blackbody with repect to its frequency. BUT WHY! WHY are
there 3 curves when we're talking about a single black body?
And what do you mean that they are at different temps? Explain
This is a question from the independent paper which relates to
my query. Someone wanna give it a shot?
Describe the differences that a graph of intensity distrobution of
different wavelenths for a hot object at 6000K would have...
There are three curves to give you an idea of the wavelength range that give max intensity...you can consider each curve as a separate black body: one is at a temperature of 24000K, another has 6000K and the other one has 1000K...
Rahul
Dead Member
because at different temperatures, blackbodies will radiate different dominant wavelengths.
each of the lines represents a different scenario. the lines are independent of each other. i think this is where you may be confused....
they could have quite easily put each line on its own graph, putting them on all on one graph allows them to be analysed.
Dash
ReSpEcTeD
Yeh.... That is actually what I wanted to know.
Thanx ppl
Thanx ppl
Dash
ReSpEcTeD
Sorry to bring this thread back, but I feel this is still a little bit hazy to me...
Can someone answer this question for me?
Describe the differences that a graph of intensity distrobution of
different wavelengths for a hot object at 6000K would have.
How would you answer this question??
Note: Pretend that you can only see the intensity distrobution of the
1000K object on the graph above
Can someone answer this question for me?
Describe the differences that a graph of intensity distrobution of
different wavelengths for a hot object at 6000K would have.
How would you answer this question??
Note: Pretend that you can only see the intensity distrobution of the
1000K object on the graph above
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Dash
ReSpEcTeD
Ok, since I've done some reading on this matter, I'll answer my own question
Tell me if this is correct...
As the black body becomes hotter (in this case, from 1000K to 6000K), its emission of radiation becomes more intense, resulting in a shorter wavelength radiation. This then suggests that the 6000K intensity distrobution graph will have its peak at a higher intensity than than the 1000K graph, and has a slight displacement to the left since the the increased emission intensity results in a shorter wavelength.
Ok, now this is pretty god dam long for a one mark question
Can anyone make this shorter or whatever?
Thnx
Tell me if this is correct...
As the black body becomes hotter (in this case, from 1000K to 6000K), its emission of radiation becomes more intense, resulting in a shorter wavelength radiation. This then suggests that the 6000K intensity distrobution graph will have its peak at a higher intensity than than the 1000K graph, and has a slight displacement to the left since the the increased emission intensity results in a shorter wavelength.
Ok, now this is pretty god dam long for a one mark question
Can anyone make this shorter or whatever?
Thnx
Rahul
Dead Member
its distribution
otherwise looks alright.
otherwise looks alright.
Dash
ReSpEcTeD
Argh.... picky barstard
Neways, I'm happy, I think I deserved some sleep now...
2morrow is a brand new day... well today... errr... forget it
Neways, I'm happy, I think I deserved some sleep now...
2morrow is a brand new day... well today... errr... forget it
Also here's a couple of extra facts about black body radiation, if you're interested:
1) l varies inversely with T (Wien's Law)
where l is the wavelength with maximum intensity, and T is the temperature in Kelvins. e.g. if I double the temperature (in Kelvins NOT Celcius) of a black body, then I'm halving the wavelength where the peak intensity occurs.
2) I varies with T^4, where I is the peak intensity, and T is the temperature in Kelvins.
(ever wondered why the black body graph shoots up so fast at the peak area, even if you increase the temperature a little bit?)
3) The total radiation power emitted by a black body is:
P = ekA*T^4 (Stefan Boltzmann law)
P = power (W)
e = emissivity (0 < e < 1, e=1 for a perfect black body, and e=0 for a perfect white body reflector)
k = Stefan Boltzmann constant, which I can't remember
A = surface area of the black body (m^2)
T = temperature of black body (K)
It's probably useful to know points 1 and 2 if you want to answer question such as your one quantitively rather than qualitively, but there's no need to know point 3.
1) l varies inversely with T (Wien's Law)
where l is the wavelength with maximum intensity, and T is the temperature in Kelvins. e.g. if I double the temperature (in Kelvins NOT Celcius) of a black body, then I'm halving the wavelength where the peak intensity occurs.
2) I varies with T^4, where I is the peak intensity, and T is the temperature in Kelvins.
(ever wondered why the black body graph shoots up so fast at the peak area, even if you increase the temperature a little bit?)
3) The total radiation power emitted by a black body is:
P = ekA*T^4 (Stefan Boltzmann law)
P = power (W)
e = emissivity (0 < e < 1, e=1 for a perfect black body, and e=0 for a perfect white body reflector)
k = Stefan Boltzmann constant, which I can't remember
A = surface area of the black body (m^2)
T = temperature of black body (K)
It's probably useful to know points 1 and 2 if you want to answer question such as your one quantitively rather than qualitively, but there's no need to know point 3.
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Dash
ReSpEcTeD
Na, they were in last years syllabus. They were removed so you don't have to talk about them.