# Deriving the rayleigh-jeans limit of planck law of radiation

1. Sep 21, 2014

### Jonsson

Hello there,

$$B(\nu) = \frac{2\,h\,\nu^3}{c^2(e^{h\nu/kT}-1)}$$

I want to show that for small frequencies, Reyleigh-Jeans law:
$$B(\nu) = \frac{2\nu^2kT}{c^2}$$
is correct.

I take the limit of Planck law as $\nu \to 0$ using l'hopital rule:
$$\lim_{\nu \to 0} \frac{2\,h}{c^2} \frac{\nu^3}{e^{h\nu/kT}-1} \stackrel{\text{l'H}}{=} \lim_{\nu \to 0} \frac{2\,h}{c^2} \frac{3\nu^2kT}{e^{h\nu/kT}h} = 0$$

I am off by a factor of 3. What is wrong with my maths?

Oh, I got it worked out. I just write, $e^{h\nu/kT} \approx 1 + h\nu/kT$, and substitute this into Planck law of radiation.