# Blackbody Radiation - Peak wavelength

## Homework Statement

The Planck blackbody spectrum is given by

$$u(ω,t)=\frac{ħω^3}{π^2c^3(e^{βħω}-1)}$$

Show that the peak of the Planck spectrum for a blackbody at a temperature T occurs at the wavelength

$$λ_{max}T=0.29$$

where T is in Kelvin and λmax is in cm.

## Homework Equations

$$\frac{d(ω,T)}{dω}=0$$

## The Attempt at a Solution

So using the equation above I get down to:

$$(3-βħω)(e^{βħω})-3=0$$

This is a transcendental equation. I said:

$$x=βħω$$

I've tried using the bisection method to solve but my answer doesn't match up with what it should as from the Wien Displacement Law it should be 0.29 for λ in cm. If anyone could help explain the solving for x, I would greatly appreciate it.

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Homework Helper
Gold Member
This one is somewhat complicated by the fact that you first need to convert the Planck function to a wavelength spectral distribution before you find the maximum in the wavelength spectral distribution (density) function. You get a different answer if you try to work the problem (finding the maximum) with the distribution (density) function in frequency space.

If its converted to the wavelength I got.

$$x=\frac{hc}{λkT}$$

$$5(e^x-1)=xe^x$$

Which is a similar equation to solve. Where would I go from here?

• Homework Helper
Gold Member
## e^{-x}=1-x/5 ##. Numeric solution with x not equal to zero. x=4.95 approximately.

I don't understand how you solved for x. I know normally with an exponential, a natural log would be involved.

Homework Helper
Gold Member
## e^{-x} ## is going to be nearly zero. On a first iteration, ## x=5 ##. Taylor series can get you about 2 or 3 decimal places in one or two iterations. Otherwise iterate ##x=5(1-e^{-x} ) ## without any Taylor series. That's probably much quicker.

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