# Homework Help: Planck's Law vs Rayleigh-Jeans Law (Blackbody radiation)

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1. Apr 20, 2015

### WyzZero

First time here, and looking for help on this. The 2nd part of this problem, I have seen some posts on and am still reviewing, but haven't found much on the 1st part.

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

1) Use l'Hopital's Rule to show that
$${\lim_{\lambda\rightarrow 0^{+}}=0}\text{ and }\lim_{\lambda\rightarrow \infty}=0$$
for Planck's Law. So this law models blackbody radiation better than the Rayleigh-Jeans Law for short wavelengths.

2) Use a Taylor polynomial to show that, for large wavelengths, Planck's Law give approximately the same values as the Rayleigh-Jeans Law.

2. Relevant equations

Rayleigh-Jeans Law expresses the energy density of blackbody radiation of wavelength $\lambda$ as
$$f(\lambda) =\frac{8{\pi}kT}{{\lambda}^{4}}$$
In 1900 Max Planck found a better model (known now as Planck's Law) for blackbody radiation:
$$f(\lambda) =\frac{8{\pi}hc{\lambda}^{-5}}{e^{\frac{hc}{{\lambda}kT}}-1}$$
where $\lambda$ is measured in meters, T is the temperature (in kelvins), and

$h$ = Planck's constant $= 6.6262 * 10^{-34} J*s$
$c =$ speed of light $= 2.997925 * 10^{8} m/s$
$k =$ Boltzmann's constant $= 1.3807 * 10^{-23} J/K$

3. The attempt at a solution
For part one, I used L'Hopital's Rule on
$$f(\lambda) =\frac{8{\pi}hc{\lambda}^{-5}}{e^{\frac{hc}{{\lambda}kT}}-1}$$
and got
$$\frac{40{\pi}kt{\lambda}^{-4}}{e^{\frac{hc}{{\lambda}kT}}}$$

Finding this, I decided to do this a few times, slowly removing the $\lambda^{-x}$ until I got:
$$\frac{960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{{\lambda}kT}}}$$

Using $${\lim_{\lambda\rightarrow 0^{+}}} \frac{960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{{\lambda}kT}}}$$ I compare this to $\lim_{\lambda \rightarrow 0^{+}} \frac{1}{e^{\frac{1}{\lambda}}}$ and $e^{\frac{1}{\lambda}}$ decreases to 0 so this goes to 0, but how do I show the same, as the limit of $\lambda$ goes to $\infty$?

AM I doing something wrong?

Also, I am looking at one of the old posts for part 2 (I found on this board), but if you have anything to add to it here, it would be much appreciated.

Last edited: Apr 20, 2015
2. Apr 20, 2015

### WyzZero

Noticed an error in my post, I fixed in the original post. I got it as it approached 0, not $\infty$, to make sense.

Last edited: Apr 20, 2015
3. Apr 20, 2015

### WyzZero

Ok, next update.
Looking into part b) and reading the other post and other info I've found.

I see, to bring it about equal to R-J's Law, I need to find the Taylor polynomials of
$$e^{\frac{hc}{kT\lambda}}$$

Using the Maclaurin series which gives $$e^{x} = \sum_ {n=0}^\infty \frac {x^{n}}{n!}$$
so
$$e^{\frac{hc}{kt\lambda}} = \sum_ {n=0}^\infty \frac {(\frac {hc}{kt\lambda})^{2}}{n!}$$
which gives:
$$1+\frac{hc}{kt\lambda}+\frac{(\frac{hc}{kt\lambda})^{2}}{2!}+\frac{(\frac{hc}{kt\lambda})^{3}}{3!}+...$$
using only the first 2 polynomials:
$$1+\frac{hc}{kt\lambda}$$
as a substitute, I get:

$$\frac{8{\pi}hc\lambda^{-5}}{1+\frac{hc}{kt\lambda}-1} = \frac{8{\pi}kT}{\lambda^{4}}$$

4. Apr 20, 2015

### Staff: Mentor

No, your bottom derivative is wrong.
$\frac{d}{d\lambda}e^{\frac{hc}{{\lambda}kT}}$ requires the use of the chain rule. You have oversimplified things and arrived at an incorrect answer.

5. Apr 20, 2015

### WyzZero

Thanks, As I was walking out of the office, I was running through it in my head and thought the same thing, perhaps i simplified too far. I think i have the answer, will know once i get home to put it to paper.

6. Apr 20, 2015

### WyzZero

k, going from the fact that I over simplified, I will step back a few steps.

So, with l'Hopital's Rule applied to:
$$\frac{8{\pi}hc\lambda^{-5}}{e^{\frac{hc}{kT\lambda}}-1} = \frac{-40{\pi}hc\lambda^{-6}}{-\frac{hc}{kt\lambda^2}*e^{\frac{hc}{kt\lambda}}}$$
I use this to get
$$\frac{-40{\pi}hc}{\lambda^{6}}*\frac{-kt\lambda^{2}}{hce^{\frac{hc}{kt\lambda}}}$$
so
$$\frac{40{\pi}kt}{\lambda^{4}e^{\frac{hc}{kt\lambda}}}$$
Now, I attempted to use this for both $\lambda \rightarrow 0^{+} \text{and} \lambda \rightarrow \infty$

For $\lambda \rightarrow \infty$ I get that $e^{\frac{hc}{kt\lambda}} \rightarrow 1 \text{ and } \lambda^{4} \rightarrow \infty$ so the function $\rightarrow$ 0
For $\lambda \rightarrow 0$ I get that $e^{\frac{hc}{kt\lambda}} \rightarrow \infty \text{ and } \lambda^{4} \rightarrow 0$ so wouldn't this function be undefined or something? Seems broken by 0*$\infty$

7. Apr 20, 2015

### Staff: Mentor

That's what I got as well, although in a slightly different form:
$$\frac{40{\pi}kt \lambda^{-4}}{e^{\frac{hc}{kt\lambda}}}$$
This is still indeterminate, as both the numerator and denominator are unbounded, so I applied L.H. once more and got this:
$$\frac{160\pi K^2 t^2 \lambda^{-3}}{hce^{\frac{hc}{\lambda Kt}}}$$
If my work is correct (you should verify), this is still indeterminate, so one could apply L.H. yet again. Hopefully each time we get closer to something with $\lambda^0$.

8. Apr 20, 2015

### WyzZero

I see, I was on the right track by using l.h. Multiple times but I made an error in my math.

So if I understand this correctly, I use L.H. once for $\lambda \rightarrow \infty$ and multiple times til $\lambda^{x} = \lambda^{0}$ for $\lambda \rightarrow 0$

9. Apr 20, 2015

### Staff: Mentor

I missed something in your first post, that you want to take the limit as lambda --> ∞. I misread it, thinking you wanted lambda going to zero from either side. If you get an expression that you can evaluate, you're done.

For the first one above, that looks good. For the second one, though, [0 * ∞] is indeterminate. If you move the lambda factor to the top, you'll get an indeterminate form that you can use LH on.

10. Apr 20, 2015

### WyzZero

Thank you very much.
This is making a lot more sense now.

I will red the calculation and post my results tomorrow, perhaps if you have a second you could check it over real quick before I submit the assignment.

11. Apr 21, 2015

### WyzZero

ok, so maybe my math wasn't off, I did read you said you misunderstood that part of the problem so I'm hoping the following is correct.

In regards to Planck's Law, when using L.H.:
$$\lim_{\lambda \rightarrow \infty} \frac{40{\pi}kT\lambda^{-4}}{e^{\frac{hc}{kt\lambda}}} = \lim_{\lambda \rightarrow \infty} \frac{40{\pi}kT}{\lambda^{4}e^{\frac{hc}{kt\lambda}}} \approx \lim_{\lambda \rightarrow \infty} \frac{1}{\lambda^{4}e^{\frac{1}{\lambda}}} \approx \lim_{\lambda \rightarrow \infty} \frac{1}{\infty*1} = 0$$

Continue L.H further to bring $\lambda^{x} \rightarrow \lambda^{0}$ I'd eventually get:
$$\frac {960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{kt\lambda}}}$$
$$\lim_{\lambda \rightarrow 0^{+}}\frac {960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{kt\lambda}}} \approx \lim_{\lambda \rightarrow 0^{+}}\frac {1}{e^{\frac{1}{\lambda}}} \approx \frac {1}{\infty} = 0$$

12. Apr 21, 2015

### Staff: Mentor

This looks fine except for two things. First, in your next-to-last expression, you have already taken the limit, so "lim" should not be included. Second, we never use $\infty$ in arithmetic expressions, so don't write $\frac 1 {\infty * 1}$ -- just go directly from the previous expression to the value of the limit.

Also, instead of saying that the second expression is approximately equal to the third, just bring the constant $40\pi kT$ out, and write the exponential factor as $e^{M/\lambda}$.
What you have here looks OK although I didn't check your work (all I did was what I wrote in my earlier post). Same comments as before -- don't write $\frac 1 {\infty}$ and don't use $\approx$ - just bring out the constants.

13. Apr 21, 2015

### WyzZero

Thank you very much for all your help.