Qs re average and peak wavelength of Planck distribution

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Buzz Bloom
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This thread is prompted by a closed thread which left it’s OP’s original question unanswered.
->https://www.physicsforums.com/threads/average-wavelength-for-blackbody-radiation.423536/
The original question asked:
is the ratio, of (a) the wavelength corresponding to the average energy in the Planck’s Law distribution
PlancksLaw-lambda.png

to (b) the wave length corresponding to the peak (maximum value) of this distribution, independent of temperature?​

I am guessing that the answer to this question is YES, but I am hopeful someone can cite a reference that confirms this. Also, the OP gave 1.84 as the value of this ratio for some unspecified temperature, but did not provide any references, so I would also like see a reference that will validate that this value is correct. If someone who can’t provide a citation knows with confidence that YES is correct or not, and/or that the ratio 1.84 is correct or not, I would much appreciate your letting me know this.

The OP also gave the value 0.29 for the product of peak wavelength and temperature, but gave no units. I found that this product is called the
Wien wavelength displacement law constant,​
and it is represented by “b” and has the value
-b = 2.8977729(17) x 10^-3 m K.
http://physics.nist.gov/cgi-bin/cuu/Value?bwien​
 
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Although I have been hoping someone would save me the effort of calculating my own answers to the questions in the previous post, I was finally, after five days, able to derive the answers for myself.

I have confirmed that the answer to the first question
Buzz Bloom said:
is the ratio, of (a) the wavelength corresponding to the average energy in the Planck’s Law distribution
planckslaw-lambda-png.112253.png

to (b) the wave length corresponding to the peak (maximum value) of this distribution, independent of temperature?
is YES.

The second question was to confirm (or not) that the value of this ratio is 1.84. The value I calculated is 0.73624.

I will post my analysis leading to either of both of these answers if anyone would like to see it.

Regards,
Buzz
 
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