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Blackbody Radiation Notes - Please tell me if i'm wrong!! (1 Viewer)

intuiit

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Please read this PDF and tell me if my ideas on blackbody radiation are correct or not!!!

I'm sure i got something wrong in what I said!

Read it here.

:jaw: HHEEEELLLPPP!!!!!!!!!!! :jaw:
 
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Gay Captain

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awesome notes there

way more than you'd need to know for the hsc... stick to the syllabus obviously for that

you might be interested to know about the maths behind it as well, since your notes were all qualitative... to copy an old post of mine:

Consider an ideal reflecting metal box with a small hole. The inner walls contain charges in constant thermal motion. The oscillators (i.e. electrons) emit radiation at all frequencies. The goal is to calculate the spectral energy density. What is that? Well, the graph you've all seen is just spectral energy density as a function of wavelength and temperature.

To work out the spectral energy density in a blackbody for a frequency range between f and f+df, you need two things:

1. The number of standing waves inside the cavity, dN(f) and
2. E(f), the average energy of a wave of frequency f. The E(f) should have a line over it to indicate it's an average but I can't do that on BoS.

The total energy in a range of frequencies f+df is then given by the product
dU(f) = E(f).dN(f)

Classically and in Planck's theory, dN(f) is worked out exactly the same way from a geometrical argument I won't write down here.

The difference came when they worked out the average energy of a given wave inside the blackbody.

In a system of distinct objects (oscillators in the blackbody walls), the probability that a given oscillator has some energy E is
P(E) = e^(-E/kT)
Where k is Boltzmann's constant, and T is temperature. That's nothing special, it just comes from simple thermal physics (which sadly isn't in the hsc). We get the average energy by multiplying one possible value E times the probability of obtaining that value for E, and then sum over all possible values, and divide by the total probability:

E = sum(E*P(E))/sum(P(E)) = sum(E*e^(-E/kT))/sum(e^(-E/kT)

Understanding exactly what that means (it's just a simple statistical argument) is less important than seeing the difference between the classical result and what Planck did:

The classical formulation finishes by taking the integral (i.e. continuous values from zero to infinity)of the above.
Planck substituted in nhf for each "E" in the formula, then he took the sum for each n from zero to infinity (i.e. discrete values) instead of an integral.

After doing the maths for both cases, classically the average energy of a wave E(f) is just kT. In Planck's formulation you get E(f) = hf/(e^(hf/kT)-1)

Classically, frequency is not a factor; any wave might have any energy in a continuum from zero to infinity. Planck's claim was that electromagnetic radiation of frequency f is allowed only energies given by E = nhf.

So the classical result says that the likelihood of a high frequency wave is the same as that of a low frequency wave. Planck says it becomes increasingly unlikely to find waves of higher frequencies.

That's why classical predictions go off to infinity on the left hand side (high frequency) of the blackbody graph, while Planck's formula gets the correct curve, which falls away to zero as frequency gets big.
 

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