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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.

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.

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##

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.

## 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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