# Derivation of E=hf?

1. Oct 4, 2012

### holtto

What is the derivation for E=hf and why did experimental observation of black bodies show that quantization of light was neccessary?

2. Oct 5, 2012

### tom.stoer

E=hf was not 'derived' in the strict sense. In back body radiation one could assume that energy of a certain mode with frequency f is not related to its frequency but to the electromagnetic field strength (squared). That would mean that a 'photon' of frequency f could carry an arbitrary energy E (like a classical electromagnetic wave where the energy has nothing to do with the frequency). Doing state counting that way one finds that the total energy emitted by a black body (of temeperature T) is infinite - which is nonsense. So Planck introduced the idea that the memission process of radiation from the atoms of a black body is quantized as E=hf, i.e. that a photon of frequency f must always carry a fixed amount of energy E=hf (he did not say that these photons must always come in these quanta, he only assumed this for the emission). In that way Planck was able to guess an energy density (per frequency) which interpolated between two known formulas, one for low and one for high frequency, and to obtain a finite result for the total energy emitted by the black body.

http://en.wikipedia.org/wiki/Planck's_law

It was Einstein who assumed (a few years later) for the interpretation of the photoelectric effect that photons of frequency f always carry energy E=hf (and that this is not only due to a specific emission process). He was then able to explain the observed energy spectrum of the photo electrons which only depends on the frequency (not on the intensity of light, what would have been guess based on classical elecromagnetism).

http://en.wikipedia.org/wiki/Photoelectric_effect

3. Oct 5, 2012

### mraptor

I was about to ask exactly this questions..
I found derivatoin of E =mc^2, and de Broigle eq p = h/lambda, but not E = hf

4. Oct 7, 2012

### holtto

the de broglie formula comes from E=pc=hf

5. Oct 11, 2012

### Jay_

I was just reading through the forum.

Can you explain "state counting" in simple words? Lets say a photon does have energy proportional to its EMfield squared. How does that make its energy infinite?

6. Oct 11, 2012

### Staff: Mentor

Do a Google search for "Rayleigh-Jeans law".

7. Oct 12, 2012

### tom.stoer

You have to use two inputs
- density of states
- equipartion theorem

Suppose you black body is an empty cube; you have to calculate the number of electromagnetic waves fitting into the cube i.e. satisfying the boundary conditions. In three dimensions the density of states for a given freqency interval reads

$$N(\nu)\,d\nu \sim v^2\,d\nu$$

The classical treatment goes as follows
- calculate the average energy contained in each mode with a given frequency
- classical statistical physics says that for a system in thermal equilibrium the equipartition theorem applies
- therefore the average kinetic energy per d.o.f. is constant, i.e. ~ kT

So

$$\bar\epsilon_T(\nu) = kT$$

Together the average energy density for a given freqency interval reads

$$u_T(\nu)\,d\nu \sim \bar\epsilon_T(\nu)\, N(\nu)\,d\nu \sim kT\,v^2\,d\nu$$

The total power (= radiation energy per time) is calculated by integrating

$$P(T) = \int_0^\infty u_T(\nu)\,d\nu = \infty$$

Therefore the above mentioned assumpttions are wrong b/c a black body of finite temperature T and finite internal energy cannot produce infinite radiation power.

It was Planck who corrected the assumptions and "guessed" a different energy density. And it was Einstein how derived this energy density using the assumption of equilibrium of absorption + spontaneous emission + induced emission.

8. Oct 12, 2012

### Jay_

I don't understand the equations completely. What is the tilde '~' stand for? But thanks for your explanation :)

9. Oct 12, 2012

### tom.stoer

~ means proportionality i.e. neglecting irrelevant constants b/c in the end the difference between ∞ and const. times ∞ doesn't matter ;-)