Black Body Radiation Arghhhhhhhh (1 Viewer)

[speedie]

ok this part in the syllabus is really gay in my opinion.

firstly, how did explaining that light in is quanta fit the experimental results in the graphs? because if ur using E=hf, the smaller the frequency the larger the energy, which does not explain the left hand side "dip" in black body radiation curves.

secondly, what did einstein contribute to the quantum theory, and how does it relate to black body radiation, and assess his contribution.. --->for this part, all the summaries i've read relate to photoelectric effect, but it doesnt say that, it says in relation to balck body radition and quantum theory!

ok thanks, this has been bugging me for a long time

 

[passion89]

In response to your 2nd question,
Planck hypothesised that EMR must be quantised in order to obey the law of conservation of energy (a completely new idea to the classical theory).
Einstein took this idea and realised that electron energy is directly proportional to its frequency. He also saw that when a light is shone on a metallic surface, a photon (light packet or quanta) collides with and is absorbed by an electron in the metal plate, leaving it more positive. It's energy, hf, is added to the energy of the electron and the electron is ejected with a certain kinetic energy.

Ek = h(f - fo)

Where fo is the threshold frequency and hfo is the work function.

So Einstein took the theory of the photoelectric effect and used it to explain the quantisation of light.

 

[Ennaybur]

if im thinkin what you're thinking, i think the 'dip' is to do with the fact that electrons can only absorb 'quanta' of energy- any less than that and no matter how much is shon on them it wont liberate them etc..

 

[speedie]

if im thinkin what you're thinking, i think the 'dip' is to do with the fact that electrons can only absorb 'quanta' of energy- any less than that and no matter how much is shon on them it wont liberate them etc..

yeh but the thing is, the dip is the left hand side bit, which is the region with lower wavelength, therefore higher frequency, giving them more energy...

 

[zeek]

Think of the graphs as a kind of "census" for energy from some body.
The graph tells you that there is varying frequencies of energy being emitted from the body. This is possible because we know that energy occurs in "packets" or quanta and the energy is being emitted in this manner because at high temperatures, not all of the energy being emitted is in one wavelength, so what happens is that some packets recieve more energy than others and some recieve less. It just so happens that there is a greater intensity of packets with a smaller amount of energy than those with a higher amount of energy at some temperature T. This reflects the left hand "dip" of the curve.

 

[speedie]

The graph tells you that there is varying frequencies of energy being emitted from the body. This is possible because we know that energy occurs in "packets" or quanta

How does energy occuring in quanta make energy being emitted in varying frequencies possible? even if light wasnt in quanta wouldnt it be emitted in a lot of frequencies?

 

[zeek]

No that wouldn't be possible without quanta. Light is transmitted in several frequencies because of the different energy "allocated" to each photon. If the quantum theory wasn't around, then we would be thinking of light as one steady stream of energy, where an energy change somewhere along its path would be "felt" across the whole beam of light. With the quantum theory, we view light as individual particles called photons, where each photon can have its own energy (i.e. frequency) because each photon can be subjected to varying conditions which could either increase or decrease its total energy e.g. half a beam of light passes through a perspex window results in half of the photons losing energy as they become refracted while the other half remain at the same energy level, so when they're measure by a spectrometer we get varying frequencies.
Now if you apply the same principle to the black body radiation, not all quanta are subjected to the same conditions, thus, some would experience more heating or less and will either gain or lose energy. This means that they would either have a higher frequency or lower frequency. It just so happens, that for certain temperatures, there is a large number of quanta which are emitted at a certain frequency and a minority that fall around this frequency, either higher or lower. This is what the black body radiation curve shows.

 

[rama_v]

http://www.phys.unsw.edu.au/~jw/FAQ.html#black

 

[ianc]

No that wouldn't be possible without quanta. Light is transmitted in several frequencies because of the different energy "allocated" to each photon. If the quantum theory wasn't around, then we would be thinking of light as one steady stream of energy, where an energy change somewhere along its path would be "felt" across the whole beam of light. With the quantum theory, we view light as individual particles called photons, where each photon can have its own energy (i.e. frequency) because each photon can be subjected to varying conditions which could either increase or decrease its total energy e.g. half a beam of light passes through a perspex window results in half of the photons losing energy as they become refracted while the other half remain at the same energy level, so when they're measure by a spectrometer we get varying frequencies.
Now if you apply the same principle to the black body radiation, not all quanta are subjected to the same conditions, thus, some would experience more heating or less and will either gain or lose energy. This means that they would either have a higher frequency or lower frequency. It just so happens, that for certain temperatures, there is a large number of quanta which are emitted at a certain frequency and a minority that fall around this frequency, either higher or lower. This is what the black body radiation curve shows.

dude u should write a textbook! Best explanation ever

http://www.phys.unsw.edu.au/~jw/FAQ.html#black

^^ great site!!! thanks.