Blackbody Radiation - Peak wavelength

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SUMMARY

The peak wavelength of the Planck blackbody spectrum at temperature T is defined by the equation λ_{max}T=0.29, where T is in Kelvin and λmax is in centimeters. The discussion highlights the use of the bisection method and iterative techniques to solve the transcendental equation derived from the Planck function. The correct approach involves converting the Planck function to a wavelength spectral distribution before finding the maximum. The iterative method can yield a solution close to x=4.95, which is essential for accurately applying Wien's Displacement Law.

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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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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?
 
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## 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.
 
## 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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Additional comment: You already did the hardest part. Using iterative method (see post #6), you can start with ##x=5 ## on the right side and I think you get ## x=4.95 ## on the left. Take this 4.95 and insert it into the right side, and I think you get close to 4 decimal place accuracy for your answer. This is the number you need to write out Wien's law with extreme precision. (## h ## ,## c ##, and ## k_b ## are already known. Plug it in and you should have the right answer.)
 
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