- #1
Dorje
- 25
- 0
As a refresher exercise in modern physics, I want to derive Wien's displacement law:
[tex]\lambda_{max}T=2.898x10^{-3}mK[/tex]
from Planck's formula:
[tex]R(\lambda)=(\frac{c}{4})(\frac{8\pi}{\lambda^4})(\frac{hc}{\lambda})(\frac{1}{\exp^(\frac{hc}{\lambda\kT})-1})[/tex]
by differentiating R([tex]\lambda[/tex]) and setting dR/d[tex]\lambda[/tex] = 0. I get to an expression like this:
[tex]\exp^(\frac{hc}{\lambda\kT})(hc - 5kT\lambda)+5kT\lambda=0[/tex]
If it wasn't for the "5kT[tex]\lambda[/tex]" term by itself on the left-hand side of the equation, the solution would simply be:
([tex]\lambda[/tex]) (T) = hc / 5k
which is Wien's law. There must be something wrong though, or maybe there's a trick involved that I'm not seeing?
Thanks
[tex]\lambda_{max}T=2.898x10^{-3}mK[/tex]
from Planck's formula:
[tex]R(\lambda)=(\frac{c}{4})(\frac{8\pi}{\lambda^4})(\frac{hc}{\lambda})(\frac{1}{\exp^(\frac{hc}{\lambda\kT})-1})[/tex]
by differentiating R([tex]\lambda[/tex]) and setting dR/d[tex]\lambda[/tex] = 0. I get to an expression like this:
[tex]\exp^(\frac{hc}{\lambda\kT})(hc - 5kT\lambda)+5kT\lambda=0[/tex]
If it wasn't for the "5kT[tex]\lambda[/tex]" term by itself on the left-hand side of the equation, the solution would simply be:
([tex]\lambda[/tex]) (T) = hc / 5k
which is Wien's law. There must be something wrong though, or maybe there's a trick involved that I'm not seeing?
Thanks