How Does Temperature Affect the Intensity Spectrum in Planck's Radiation Law?

[sanitykey]

Homework Statement

The dependence on wavelength \lambda of the intensity I(\lambda)d\lambda of the radiation emitted by a body which is in thermal equilibrium with its surroundings at temerature T is given by:

I(\lambda)d\lambda = \frac{2 \pi h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}d\lambda

in the interval of wavelength between \lambda and \lambda+d\lambda. In this expression, h is Planck's constant, k is Boltzmann's constant, and c is the velocity of light.

Sketch and clearly label on one figure the dependences of I(\lambda)d\lambda on \lambda for three different temperatures T_{1} < T_{2} < T_{3}.

Simplify the above expression in the limit of (i) short wavelength (\lambda\rightarrow0) and (ii) long wavelength (\lambda\rightarrow\infty).

(The binomial expansion e^{x} = 1+x+x^{2}/2+... may be useful.)

Homework Equations

All given in the problem i think.

The Attempt at a Solution

I found Planck's Radiation Law was almost exactly the same as this i searched for it on wikipedia for more information:

http://en.wikipedia.org/wiki/Planck's_law

On that page is a graph which i thought was showing what the first part of the question is asking but i don't understand what the question means when it says "clearly label on one figure the dependences of I(\lambda)d\lambda on \lambda"?

For the second part i tried to make the formula look simpler first:

\frac{A}{\lambda^{5}(e^{B/\lambda} - 1)}

I think as \lambda\rightarrow0, e^{B/\lambda} - 1 can be simplified to e^{B/\lambda} because the latter expression will be very large giving:

\frac{A}{\lambda^{5}e^{B/\lambda}}

I'm having some trouble posting the rest of my thread but i thought for the last part as lambda goes to infinity the expression would simplify to A/lambda^5 but I'm not sure how to work these out for definite i think this is probably wrong.

 

[Gokul43201]

For large \lambda, you can expand e^{B/ \lambda } to first order, using the Taylor series provided to you. You will get a different answer than A/ \lambda ^5. Your answer for the short wavelength limit is correct.

As for "labeling", I believe that is asking you to make it clear which curve represents which temperature (T1,T2,T3).

 

[sanitykey]

Thanks for the help this is what I've got so far:

\lambda^{5}(e^{B/\lambda} - 1) = \lambda^{5}(1 + \frac{B}{\lambda} + \frac{B^2}{2 \lambda^2} + \frac{B^3}{6 \lambda^3}... -1)

\frac{A}{\lambda^{5}(\frac{B}{\lambda} + \frac{B^2}{2 \lambda^2} + \frac{B^3}{6 \lambda^3}... )}

\frac{A}{B \lambda^4 + 1/2 B^2 \lambda^3 + 1/6 B^3 \lambda^2 + 1/24 B^4 \lambda + 1/120 B^5}

I think all the other terms in the expansion would tend to 0 so they can be ignored, if this is right it's as far as i can get though i can't see how to simplify any further although this makes it look longer and maybe even more complicated?

 

[Gokul43201]

Actually, it's better than that!

B/ \lambda <<1 \implies B^2 \lambda^3 << B \lambda ^4, etc. So feel free to throw away all terms after the first one.

 

[sanitykey]

Thanks for the help again :)

I was wondering i can see the difference between this:

I(\lambda)d\lambda = \frac{2 \pi h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}d\lambda

and Planck's radiation formula:

I(\lambda)= \frac{2 h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}

is d\lambda and \pi. How do you transform one into the other? What does it mean to write it like the question did with d\lambda either side or more specifically what does the d\lambda mean in that context?

If you integrated both sides to get rid of the d\lambda s would the \pi dissapear? I really should try and integrate it myself rather than just asking but it looks complicated :S i'll give it a shot though.

 

[dextercioby]

Thanks for the help again :)

I was wondering i can see the difference between this:

I(\lambda)d\lambda = \frac{2 \pi h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}d\lambda

and Planck's radiation formula:

I(\lambda)= \frac{2 h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}

There' a \pi difference. But only because you forgot to put it in the second formula. The difference between the 2 formulas is important only when you try to change the variable from wavelength to frequency \nu or angular frequency \omega. In that case, the formula involving differentials should be used.