I was reading Quantum chemistry by McQuarrie and in it he states that according to classic electromagnetic theory (prior to quantum mechanics) intensity of radiation is given by the Rayleigh- Jeans law (which was inaccurate as it showed the ultraviolet catastrophe). In the Rayleigh- Jeans law EM intensity is proportional to frequency squared and the black body temperature.

However in describing the failure of classic EM to deal with the photoelectric effect the book states that the, "the intensity of the radiation is proportional to the square of the amplitude of the electric field." (for classic EM) and the intensity shouldn't depend on frequency (it was Einstein who introduced the frequency dependence of intensity)

So I'm confused - the Rayleigh- Jeans law is a classical law for intensity of EM radiation and depends on freqency, but the book says that according to classical theory the intensity depended on amplitude of the E field and not frequency.

Isn't this a contradiction?


1 Answer 1


Are you confusing black body radiation with the photoelectric effect?

Classically, the black body spectrum is given by $$ f(\omega) \sim \omega^2, $$ which is the Rayleigh-Jeans law.

It is in good agreement with experiment for small $\omega$, but it also obviously can't be true for large $\omega$, because $f(\omega)$ being a probability density has to satisfy $$ \intop_0^{\infty} f(\omega) \, d\omega = 1, $$

but this integral diverges for $f(\omega) \sim \omega^2$. This divergence is known as the ultraviolet catastrophe, and it requires quantum mechanics to be explained. As you know, according to quantum mechanics, $f(\omega)$ is given by the Planck formula which also grows as $\omega^2$ for small $\omega$, but has an exponentially decaying dumping factor for large $\omega$ which makes the integral convergent.

For the photoelectric effect, the intensity that does not depend on the frequency is not the intensity of radiation that comes from a black body, but the intensity of the flow of electrons in a capacitor, aka the photoelectric current.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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