- #1

Zacarias Nason

- 68

- 4

## Homework Statement

After reading the forum stickies I'm not entirely sure where to put this question since it involves using math to solve a question, but is informally stated and isn't a book problem, either-I just started reading Fong's Elementary Quantum Mechanics, and in the first few pages the relationship between the Plank, Wien and Rayleigh-Jean formulas are discussed. I'm assuming that since Rayleigh-Jean and Wien are special cases of Planck's Law that they can be derived from Planck's Law, but I'm having trouble getting from one to the other/proving they are equivalent under some circumstances, assuming only algebra is really necessary for this.

## Homework Equations

[/B]

It displays all of the formulas in the form of energy per unit volume.

[tex] u(v)dv = \frac{8 \pi v^2 k T}{c^3} dv \ \ \ \text{(Rayleigh-Jean's)} [/tex]

[tex] u(v)dv \ \text{~} \ v^3e^{-hv/kT} dv \ \ \ \text{(Wien's)}[/tex]

[tex] u(v)dv = \frac{8 \pi h v^3}{c^3} \frac{1}{e^{hv/kT}-1} \ \ \ \text{(Planck's)} [/tex]

## The Attempt at a Solution

I tried to equate the Wien and Planck formula and ran into some trouble, but may very well be doing this wrong from the outset. Also, I'm not really sure how or if the tilde/asymptote approx. symbol may affect this. Here's my work:

[tex] v^3 e^{-hv/kT} dv = \frac{8 \pi h v^3}{c^3} \frac{1}{e^{hv/kT}-1} dv[/tex]

[tex] e^{-hv/kT} dv = \frac{8 \pi h}{c^3} \frac{1}{e^{hv/kT}-1} dv[/tex]

[tex] e^{-hv/kT} = \frac{8 \pi h}{c^3} \frac{1}{e^{hv/kT}-1} [/tex]

[tex] e^{-hv/kT} = \bigg(\frac{8 \pi h}{c^3}\bigg) (e^{-hv/kT})\bigg( \frac{1}{1-1/e^{hv/kT}}\bigg) [/tex]

[tex] 1 = \bigg(\frac{8 \pi h}{c^3}\bigg) \bigg( \frac{1}{1-1/e^{hv/kT}}\bigg) [/tex]

[tex] 1-1/e^{hv/kT} = \frac{8 \pi h}{c^3} [/tex]

[tex] e^{hv/kT} = 1 - \frac{c^3}{8 \pi h} [/tex]

This...obviously can't be right. Apart from the crazy huge number, I can't take the natural log of a negative number. What am I doing wrong? Does the necessary approach require integral calc or something?