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I don't really understand the reason why wave theory of light fails to explain the blackbody radiation. My textbook says the Planck's quantum theory explains blackbody radiation. It says "If we assume, he said, that radiation is emitted in packets of energy instead of continuously as in a wave, then we can explain the black body spectrum."

So, the problem in the wave theory is that energy is absorbed or emitted continuously, not in multiple quanta. But why is continuous emission or absorption a problem when we deal with black body radiation? I was taught in class (or maybe I interpreted it this way) that if energy would be continuously radiated, then the intensity of radiation must increase on heating the black body and wavelength of light would stay same. But from experiment, wavelength changes. Hence it fails to explain it.

But assuming that what I wrote above is correct, why can't the wavelength change? And even in Planck's theory, why can't simply the number of quanta emitted increase and frequency be constant, so it gives radiation of same wavelength?

Please note that I have tried my best to look this up on the Internet, but all I see is explanations for back body radiations in terms of Planck's theory.

If you find this question not framed well/not acceptable in its current form, please leave a comment so that I can edit it, before downvoting it to close it.

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marked as duplicate by Rob Jeffries, Kyle Kanos, ACuriousMind, John Rennie, Martin May 22 '15 at 11:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This is the spectrum of black body radiation, classical, and quantum calculations.

black bodyr

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.

So as you see the wavelengths are in the x axis so all wavelengths are covered.

Black-body radiation provides insight into the thermodynamic equilibrium state of cavity radiation. If each Fourier mode of the equilibrium radiation in an otherwise empty cavity with perfectly reflective walls is considered as a degree of freedom capable of exchanging energy, then, according to the equipartition theorem of classical physics, there would be an equal amount of energy in each mode. Since there are an infinite number of modes this implies infinite heat capacity (infinite energy at any non-zero temperature), as well as an unphysical spectrum of emitted radiation that grows without bound with increasing frequency, a problem known as the ultraviolet catastrophe. Instead, in quantum theory the occupation numbers of the modes are quantized, cutting off the spectrum at high frequency in agreement with experimental observation and resolving the catastrophe. The study of the laws of black bodies and the failure of classical physics to describe them helped establish the foundations of quantum mechanics.

The above explains why the classical assumptions lead to a wrong spectrum.

Here is a link to an answer I gave a while ago, which goes into more detailed links and might be useful.

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If you try to analyze the power radiated by a black body and assume a continuous spectrum of energy, you find that the black body must emit an infinite amount of energy. This is known as the ultraviolet catastrophe and was solved by Max Planck assuming that energy is emitted in discrete units of $hv$ and not in a continuous spectrum.

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  • $\begingroup$ I would very much appreciate if you could explain what you said in detail. But if not, at least tell me how did you get "assume a continuous spectrum of energy, you find that the black body must emit an infinite amount of energy.". And why a non-continuous spectrum would require finite energy. I mean what law or theorem or whatever did you apply. $\endgroup$ – Swapnil Rustagi May 21 '15 at 14:59
  • $\begingroup$ @Swapnil The accepted answer to the duplicate explains exactly what causes the UV catastrophe. $\endgroup$ – Rob Jeffries May 21 '15 at 15:39
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Discreteness of energy reproduces experimental results in a satisfying manner. If you assume discreet spectrum you get to preform summation instead of integration. To be precise, you get to sum geometric series which converges. After that, you get the factor of integration of the form 1/(exp (hν/kT) − 1). This now is OK. If you dont have discreteness assumption you dont get this factor...and it all diverges. It is summing-integrating difference.

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