Differentiating Planck's radiation law

[Piano man]

Planck's radiation law:
[tex]I(\lambda)=\frac{2\pi hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}[/tex]

I'm trying to calculate the peak of a graph, so setting the derivative equal to 0, I've gotten it down to
[tex]\frac{\lambda(e^{\frac{\alpha}{\lambda}}-1)}{e^{\frac{\alpha}{\lambda}}}=\frac{\alpha}{5}[/tex]

where [tex]\alpha=hc/kT[/tex]

Is it possible to solve for [tex]\lambda[/tex]?

 

[Mute]

Not really, at least in terms of functions you've heard of before. You can solve for it in terms of a special function, but otherwise it's best to just solve it numerically.

If you're interested, the special function is called Lambert's W function, W(z), defined by the relation [itex]z = W(z)\exp(W(z))[/itex]. Taking [itex]\alpha/\lambda = w[/itex], you can hammer your expression into looking like [itex]we^w = \mbox{constant}[/itex], so that [itex]w = \alpha/\lambda = W(\mbox{constant})[/itex]. The result is

[tex]\frac{\alpha}{\lambda} = 5 + W_0(-5e^{-5})[/tex]
and so

[tex]\lambda T = \frac{hc/k_B}{5 + W_0(-5e^{-5})}.[/tex]

Plugging values in for h, c and kb and using wolfram alpha to compute gives ~0.0029 m K, which agrees with the experimental value to the precision I used for the other constants.

Some notes: W(z) is a multivalued function. I used the 0th branch to compute the solution. Also, in mathematica, the function is referred to as the "ProductLog[k,x]", where k is the branch and x is the variable.

 

[Piano man]

That's great thanks a million. I'll have to look further into that.