# Homework Help: Planck's Radiation Law?

1. Sep 22, 2007

### sanitykey

1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

All given in the problem i think.

3. 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.

Last edited: Sep 22, 2007
2. Sep 22, 2007

### Gokul43201

Staff Emeritus
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).

3. Sep 22, 2007

### 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?

4. Sep 22, 2007

### Gokul43201

Staff Emeritus
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.

5. Sep 23, 2007

### sanitykey

Thanks for the help again :)

I was wondering i can see the difference between this:

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

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

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.

6. Sep 23, 2007

### dextercioby

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.