How Is Lambda Max Derived from Planck's Law?

[georgeh]

I have a problem that states
Show that the wavelength Lamba_max=(2892 micro meters*K)/T
hint: set the dS/DLamba=0.
i have no idea how to do this.

 

[Hootenanny]

You need to find the maximum of a function. The maximum of a function is given when f'(x) = 0. So taking Planck's radiation formula for energy density per unit wavelength;

S_{\lambda} = \frac{8\pi hc}{\lambda^{5}} \cdot \frac{1}{e^{\frac{hc}{\lambda kT}} - 1}

The constant term 8\pi hc does not effect the position of the peak and can therefore be ignored. Thus the derivative becomes;

\frac{d}{d\lambda} = \frac{1}{\lambda^5}\cdot \frac{1}{e^{\frac{hc}{\lambda kT}} - 1} \;\; d\lambda = 0

Once you have found the derivative all that remains is to solve the equation. As far as I know there exists no analytical solution to the equation and it must be solved numerically. However, if I have time I may venture into the maths forums and inquire as to whether an analytical solution exists.

HINT: Solve numerically for \lambda_{max}T first.

~H