Planck quantised the energy of light in order to solve the black body radiation problem. However, I am confused as to exactly what he quantised. On one hand, I have seen that he quantised the energy of oscillators in the black body (source below). enter image description here

Source of photo

However, the second version of the story I've heard is that it is simply that light energy is given off in discrete chunks (where every photon itself can have any value of energy it likes depending on frequency and is not quantised).

To me, the first idea doesn't make sense because $E=hf$ and as $f$ can be any real number that means the energy of each photon can be any real number. I would believe the second version, however I don't understand how this version of quantisation leads to reducing the number of high frequency waves emitted and thus solving the UV catastrophe. Could you please explain exactly how releasing energy in chunks (of arbitrary size themselves) means less will be released in the higher frequencies?

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    $\begingroup$ Light isn't quantised in the sense buried in your wording. The electromagnetic spectrum is a continuum, indeed. Is the energy of each chunk that is quantised, chunk that is emitted by an oscillator in the case of BB, or by whatever system having quantised levels like in optical electronic transition. f can be "any" number, but must be that number. Volume can be any number, but once you have standard bottles, then it comes as n times the volume of the bottle, as this analogy might helps many readers. $\endgroup$
    – Alchimista
    Apr 13, 2019 at 9:30
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    $\begingroup$ Your question is unclear. Your two accounts boil down to the same. Can you clarify? $\endgroup$
    – my2cts
    Apr 13, 2019 at 9:34
  • $\begingroup$ I could copy this for an answer to your last question hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c2 (second page) $\endgroup$
    – anna v
    Apr 13, 2019 at 9:50
  • $\begingroup$ Very similar question here: Where is quantization used in deriving Planck's law? $\endgroup$ Apr 13, 2019 at 10:27
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    $\begingroup$ $E=hf$ is the energy of one photon with frequency $f$. The energy of one photon can indeed be continuously varied by varying its frequency. The thing that's quantized is the number of photons of each frequency. For any given frequency $f$, the energy of $N$ of that collection of photons is $E=Nhf$. The $N$ is quantized, not the $f$. $\endgroup$ Apr 13, 2019 at 13:17

1 Answer 1



First of all, $E=hf$ is the energy of a photon. Now, if you take two photons, the enegy is twice that. You cannot have $1.5E$, since it does not make sense to talk about $1.5$ photons. Also, you misunderstood "discrete chunks" which is actually photons themselves. The energy is absorbed or emitted as photons.

Obviously $hf$ could be any real number , but fractional values of $hf$ cannot exist. Only integral multiples are allowed. That is all.

enter image description here

If a body is heated (say a blackbody), it would radiate in all frequencies. This is common. What you would get if you used a detector to say apart the types of EM waves emitted, there would be a peak intensity at a certain frequency. There would be smaller intensities at all other frequencies.

Here, $E=hf$ is not used. Instead, the curve for the blackbody radiation law uses the following one for a plot $B(\lambda,T)$ against $ \lambda$.

$$B(\lambda, T)= \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}$$

This peak is not a constant for a specific body, but only at that specific temperature. If you vary the temperature around, you would get peaks at different frequencies.


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