Blackbody radiation intensity find maximum

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
Talltunetalk
Messages
8
Reaction score
0

Homework Statement


By direct calculation, derive and expression for the wavelength λ(max) at which blackbody radiation intensity is a maximum

2. Homework Equations

Planck's law for the intensity distribution of blackbody radiation as a function of wavelength λ and the blackbody temperature T is given by

I(λ,T) = (2pi)hc^2/(λ^5)(e^((hc)/λkT) -1)

h is Planck's constant 6.626 x10^(-34)
k is blotzman constant 1.38066 x10^(-23)
c is speed of light

The Attempt at a Solution



(-5(2pi)(hc^2)λ^(-6))/(e^(hc)/(λkT) - 1) + λ^(-5)(2pi)(hc^2)(-1)(e^((hc)/(λkT)) -1)^(-2)((-hc)/(kTλ^2))

I used the product rule to get to the answer. I have not yet simplified I am wondering if I have even started the problem correctly.
 
Physics news on Phys.org
How did you arrive at your expression? And I don't see an equation, just an expression.

(Just show us your starting-off point, we don't need to see the gorty details... )
 
I derived Planck's Law with respect to λ and treated T as a constant. My thoughts were that if I do this and find the derivative I can then set it equal to zero and this will give me an expression for λ(max)
 
That is correct, but I didn't see anything set to zero ... the rest is just 1st-year calculus. You may wind up with an implicit rather than an explicit expression for lambda_max.
 
Yeah the main idea is using the product rule, being careful to use the chain rule for that e. I recently did this problem in my own physics class. A good thing to do is simplify your expression by putting all the constants into one letter

a=hc/kt for example would be a good idea.

If you are frustrated and can't figure out why its not working, go to hyperphysics page: finding the blackbody peak. google it - i'd send you a link but pf won't let me till i have 10 posts