# Planck's Law. It's $E=hf$ or $E=nhf$?

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.

The $E = hf$ is the energy of each packet or photon. So if $n$ photons are emitted, the total energy is $E = nhf$.
• @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. Mar 7, 2016 at 17:53
• @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.
• @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. Mar 8, 2016 at 11:19