Homework Help: A couple Blackbody Problems?

1. Sep 21, 2004

Nebula

A couple Blackbody Problems???

I'm a little confused about these two problems involving blackbodies, hopefully someone could give me a bit of insight. Thanks in advance.

1. For a blackbody we there is a frequency peak and a wavelennght peak. Lets call em v and w respectively. Now consider the derivations of the maximums dependant on temperature to prove v*w not equal to the speed of light.

Im not sure what to do here. No were in my text do they talk about the derivations for the maximums. So Im not really sure in what direction to head.

next question.

2. The peak value Mw(max) at the wavelength w(max) in the distribution of blackbody radiation increases with T (temperature). Show Mw(max) depends on T as:
Mw(max)=CT^p
where C is some constant and power p
so we have to find the constant and the power.

2. Sep 21, 2004

Tide

The Planck distribution depends on the temperature. To find the maximum you will need to differentiate with respect to the temperature and find the value of T that makes the derivative = 0.

3. Sep 21, 2004

Nebula

Right but what equations do I work with?

4. Sep 21, 2004

Tide

5. Sep 25, 2004

P3X-018

Hey Tide
In that page they give placnk's law as

((8*pi*v^2)/c^3)*...

But look at the following page why do they give planck's law with the term
http://scienceworld.wolfram.com/physics/PlanckLaw.html

(2v^2)/c^2

Why is the term ((8*pi*v^2)/c^3) different than the one giving in scienceworld??

6. Sep 25, 2004

Tide

One might be a flux and the other an intensity - I didn't have time to study them carefully. I recommend the orginal poster refer to his textbook for the correct version!

7. Sep 25, 2004

Tide

P3x,

I think for the problem at hand you should be focusing on the functional dependence which is
$$\frac {\nu ^3}{e^{\frac {h \nu}{kT}}-1}$$
to find the peak.

8. Sep 25, 2004

P3X-018

In the Physics Formulary, it says:
Planck's law for the energy distribution for the radiation of af black body is:

omega(v,T) = ((8pi*h*v^3)/c^3)*(exp(hv/kT)-1)^-1

"Energy distribution" is that the intensity or flux?
They define the flux as P/A
http://scienceworld.wolfram.com/physics/EnergyFlux.html

:S

9. Sep 25, 2004

Tide

Edited:

Yes, I agree with that. It's the same as what I wrote - I just left off the normalization!

Last edited: Sep 25, 2004
10. Sep 25, 2004

P3X-018

So is this the flux:

omega(v,T) = ((8pi*h*v^3)/c^3)*(exp(hv/kT)-1)^-1 ???

11. Sep 25, 2004

Tide

That looks good!