# Planck's Law for Blackbody Radiation vs. The Rayleigh-Jeans Law

Given: f(lambda) = 8pi*kt(lambda^4)

Where lambda is measured in meters, T is the temperature in kelvins, k is Boltzmann's constant. The Rayleigh-Jeans Law agrees with experimental measurements for long wavelengths but disagrees drastically for short wavelengths. [The law predicts that f(lambda) -> 0 as f(lambda) --> infinity but experiments have shown that f(lambda) --> 0.] This fact is known as the ultraviolet catastrophe.

In 1900 Max Planck found a better model (known as Planck's Law) for blackbody radiation:

f(lambda) = 8pihc(lambda)^05 / [e^hc / (lamdba*kT) - 1]

where lambda is measures 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 constrant = 1.3807 * 10^-23 J/K

1. Use L'hopitals rule to show that the lim f(lambda) = 0 and lambda goes to 0 and that the lim f(lambda) = 0 as lambda goes to 0 for Planck's Law. So this law models blackbody radioation better than Rayleigth-Jeans Law for short wavelengths.

Â© BrainMass Inc. brainmass.com November 24, 2022, 11:30 am ad1c9bdddfhttps://brainmass.com/physics/blackbody-radiation/plancks-law-blackbody-radiation-vs-rayleigh-jeans-law-2725

#### Solution Summary

This shows that Planck's law models blackbody radiation better than Rayleigh-Jeans Law for short wavelengths in an attached Word document.