How did Planck conclude $E=h\nu$ from his radiation law?

  • 4
    $\begingroup$ I thought he postulated it and showed that it agrees with experiment. $\endgroup$ – bremsstrahlung Mar 29 '16 at 4:09
  • 1
    $\begingroup$ Its the reverse- the planck's radiation law emerges from his fundamental ,path breaking assumption E=h v . $\endgroup$ – drvrm Mar 29 '16 at 4:37

The spectral radiance of a body, Bν, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body at absolute temperature T is given by

$$B_\nu(\nu, T)= \frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{k_\mathrm B T}}-1}$$

The relationship comes from solving a model of electromagnetic oscillators in a cavity, and h

First recognized in 1900 by Max Planck, it was originally the proportionality constant between the minimal increment of energy, E, of a hypothetical electrically charged oscillator in a cavity that contained black body radiation, and the frequency, f, of its associated electromagnetic wave. In 1905 the value E, the minimal energy increment of a hypothetical oscillator, was theoretically associated by Einstein with a "quantum" or minimal element of the energy of the electromagnetic wave itself. The light quantum behaved in some respects as an electrically neutral particle, as opposed to an electromagnetic wave. It was eventually called the photon.

So the black body model developed to explain the absence of the ultraviolet catastrophe in the data, needs the energy increments leaving the oscillators as electromagnetic radiation to be proportional to the frequency, by construction. This was a brilliant hypothesis that was later confirmed by innumerable experiments.

| cite | improve this answer | |
  • $\begingroup$ surely it was the ultraviolet catastrophe. $\endgroup$ – Peter Diehr Mar 29 '16 at 11:42
  • $\begingroup$ @PeterDiehr thanks . would this be dyslexia? $\endgroup$ – anna v Mar 29 '16 at 11:51
  • $\begingroup$ @annav: Ha! Even I didn't notice it. dyslexia? $\endgroup$ – user36790 Mar 29 '16 at 12:02
  • $\begingroup$ @user36790 dyslexia is transposing numbers and letters. In my case related "concepts" :) $\endgroup$ – anna v Mar 29 '16 at 12:04
  • $\begingroup$ @annav: May be Alzheimer's :P $\endgroup$ – user36790 Mar 29 '16 at 12:11

I don't know if Planck did it this way, but one can separate out the assumption that the energy carried by light is quantized into photons from the assumption that the energy of each photon is $E = h \nu$. Whatever the form of the energy relation is, it can only depend on the frequency because (in a universe with space and time translational and rotational invariance) that is the only local property that characterizes light. If you denote the dependence of the photon energy on its frequency by the function $\epsilon(\nu)$, then you can show that the spectral radiance is given by

$B(\nu, T) = \frac{2 \nu^2}{c^2} \frac{\epsilon(\nu)}{e^{\frac{\epsilon(\nu)}{k_B T}} - 1}$.

(The $\frac{2 \nu^2}{c^2}$ contribution is purely classical and comes from the density of normal modes of a plane wave in a perfectly conducting cavity.)

Finding $\epsilon(\nu)$ is now simply a matter of finding a function that matches this formula to experimental results. Fortunately, the correct answer is quite simple: $\epsilon(\nu) = h \nu$.

In the high-temperature limit $k_B T \gg \epsilon(\nu)$, we can Taylor expand the exponential and the dependence on $\epsilon(\nu)$ drops out (because there are so many photons that the fact that their number is quantized is irrelevant), leading to the classical Rayleigh-Jeans law that holds at frequencies such that $h \nu \ll k_B T$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.