Planck's Law vs Rayleigh-Jeans Law (Blackbody radiation)

In summary: I think you are getting a little mixed up here. Let's back up. The problem asks you to use L'Hopital's rule to show that ##\lim_{\lambda \rightarrow 0^+} f(\lambda) = 0## and ##\lim_{\lambda \rightarrow \infty} f(\lambda) = 0##. What is ##f(\lambda)##? Remember that Planck's Law is$$f(\lambda) =\frac{8{\pi}hc{\lambda}^{-5}}{e^{\frac{hc}{{\lambda}kT}}-1}$$and the problem is to show that the two limits above are both zero. What you need
  • #1
WyzZero
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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.

Homework Statement



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.

Homework 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##

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.

Thanks in advance for any help you can provide.
 
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  • #2
WyzZero said:
Using $${\lim_{\lambda\rightarrow \infty}} \frac{960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{{\lambda}kT}}}$$ I compare this to ##\lim_{\lambda \rightarrow \infty} \frac{1}{e^{\frac{1}{\lambda}}}## and ##e^{\frac{1}{\lambda}}## approaches ##\infty## so this goes to 0, but how do I show the same, as the limit of ##\lambda## goes to 0?

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:
  • #3
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
WyzZero said:
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.

Homework Statement



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.

Homework 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##

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
$$\frac{40{\pi}kt{\lambda}^{-4}}{e^{\frac{hc}{{\lambda}kT}}}$$
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.
WyzZero said:
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.

Thanks in advance for any help you can provide.
 
  • #5
Mark44 said:
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.

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
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
WyzZero said:
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
$$\frac{40{\pi}kt}{\lambda^{4}e^{\frac{hc}{kt\lambda}}}$$
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##.
WyzZero said:
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##
 
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  • #8
Mark44 said:
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##.

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
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.

WyzZero said:
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##
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
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
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
WyzZero said:
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.:
$$\lim_{\lambda \rightarrow \infty} \frac{40{\pi}kT\lambda^{-4}}{e^{\frac{hc}{kt\lambda}}} = \lim_{\lambda \rightarrow \infty} \frac{40{\pi}kT}{\lambda^{4}e^{\frac{hc}{kt\lambda}}} \approx \lim_{\lambda \rightarrow \infty} \frac{1}{\lambda^{4}e^{\frac{1}{\lambda}}} \approx \lim_{\lambda \rightarrow \infty} \frac{1}{\infty*1} = 0$$
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}##.
WyzZero said:
Continue L.H further to bring ##\lambda^{x} \rightarrow \lambda^{0}## I'd eventually get:
$$\frac {960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{kt\lambda}}}$$
$$\lim_{\lambda \rightarrow 0^{+}}\frac {960{\pi}k^{5}T^{5}}{h^{4}c^{4}e^{\frac{hc}{kt\lambda}}} \approx \lim_{\lambda \rightarrow 0^{+}}\frac {1}{e^{\frac{1}{\lambda}}} \approx \frac {1}{\infty} = 0$$
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.
 
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  • #13
Mark44 said:
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.
Thank you very much for all your help.
 

What is Planck's Law?

Planck's Law, also known as the Planck radiation law, is a mathematical formula that describes the spectral density of electromagnetic radiation emitted by a blackbody at a certain temperature. It was developed by German physicist Max Planck in 1900 and is considered to be one of the cornerstones of modern physics.

What is the Rayleigh-Jeans Law?

The Rayleigh-Jeans Law, also known as the Rayleigh-Jeans formula, is a mathematical formula that describes the spectral density of electromagnetic radiation emitted by a blackbody at a certain temperature. It was developed independently by British physicist Lord Rayleigh and Dutch physicist Hendrik Lorentz in the late 19th century. However, it was later found to be incorrect at high frequencies and was replaced by Planck's Law.

What is the difference between Planck's Law and the Rayleigh-Jeans Law?

The main difference between these two laws is that Planck's Law takes into account the quantum nature of radiation, while the Rayleigh-Jeans Law does not. This means that Planck's Law accurately describes the emission of radiation at all frequencies, while the Rayleigh-Jeans Law fails at high frequencies (known as the ultraviolet catastrophe).

Which law is more accurate?

Planck's Law is considered to be more accurate because it takes into account the quantum nature of radiation, which is essential for describing the emission of radiation at all frequencies. The Rayleigh-Jeans Law, on the other hand, was found to be incorrect at high frequencies and was replaced by Planck's Law.

What are the practical applications of Planck's Law and the Rayleigh-Jeans Law?

Both laws have numerous practical applications in fields such as astrophysics, thermodynamics, and spectroscopy. They are used to study the properties of blackbodies and to understand the emission and absorption of radiation by various materials. They also play a crucial role in the development of technologies such as lasers, solar panels, and infrared cameras.

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