L.N.S. Posted May 15, 2016 Posted May 15, 2016 While reading about Ultraviolet Catastrophe from a book its confusing me with Black Body radiation.. Can anyone help me out explaining this topic in an easy way?
Sensei Posted May 15, 2016 Posted May 15, 2016 (edited) Objects that are hot are emitting photons, Photon has energy [math]E=h*f[/math] h - Planck constant = 6.62607004*10^-34 J*s f - frequency in Hz=1/s [math]f=\frac{c}{\lambda}[/math] so [math]E=\frac{h*c}{\lambda}[/math] The hotter object (the larger temperature), the more energetic photons are emitted by object. The more energy they have, the larger frequency f, and shorter wavelength. Black body radiation is summing the all photons energies emitted per second together (Joules/second=Watt), and assigns to it appropriate temperature in Kelvins. Ultraviolet photons (the one with wavelength smaller than 400 nm) have enough energy to ionize matter, and make plasma. Therefore they are rarely encountered by now, otherwise everything would be in plasma state (like in star). Edited May 15, 2016 by Sensei
swansont Posted May 15, 2016 Posted May 15, 2016 A blackbody is the most efficient absorber and emitter of radiation, so it is the system that is being described by the Rayleigh-Jeans law that quantifies/predicts the "catastrophe".
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now