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My textbook provides intuition of Planck's Quantum theory which is copied right next.

Max Planck proposed that emission or absorption of energy in a blackbody is discontinuous. It is absorbed or emitted in packets $hf$ or integral multiple of these packets $nhf$. Each packet is called Quantum.

Energy lost or gained is given by;

$E=hf$ where $f$ is the frequency of radiations.

I think the equation which is consistent with the definition above is E=nhf. If not, please explain which thing I am missing.

I have searched it on internet but explanation is given in terms of photon however I want to understand how does $E=hf$ is consistent with the brief description given in my book.

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The $E = hf$ is the energy of each packet or photon. So if $n$ photons are emitted, the total energy is $E = nhf$.

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  • $\begingroup$ Does that mean that a blackbody may release several packets of energy at a time? $\endgroup$
    – user104909
    Mar 7, 2016 at 17:44
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    $\begingroup$ @SufyanNaeem Yes. A blackbody emits electromagnetic radiation of a particular wavelength depending on the temperature of the body. The higher temperature a body has, the higher the frequency of these emitted packets of energy(photons) will be which determines the $f$ in Planck's law and $n$ is the number of photons emitted. You can calculate the total lost energy by determining the photon energy density. $\endgroup$
    – Starior
    Mar 7, 2016 at 17:53
  • $\begingroup$ @Starior if an electron emits or absorb radiation of frequency "f" then it would either be demoted or promoted . The energy difference between the orbits, it made transition between, should be given by; $$\delta {E} = nhf$$. But my book states it is given by; $$\delta {E} = hf$$ Explain please. $\endgroup$
    – user104909
    Mar 8, 2016 at 9:08
  • $\begingroup$ @SufyanNaeem Note that every single electron would emit radiation with an energy of $$E = hf$$ but the total lost energy would be $$E = nhf$$. Therefore, since one electron emits radiation with an energy of $$E = hf$$, the energy difference between the initial and final orbit would be $$\delta {E} = hf$$ as your book states. $\endgroup$
    – Starior
    Mar 8, 2016 at 11:19
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    $\begingroup$ @SufyanNaeem Yes, you can make that statement at an elementary level. However, as I stated above to calculate the total energy lost or absorbed by a blackbody, you may need to determine the photon energy density which is governed by Bose-Einstein distribution function. $\endgroup$
    – Starior
    Mar 8, 2016 at 14:27

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