Fill in the blanks about blackbody radiation?
1 Answer
Well, here's my attempt... I could explain it better than I can fill in the blanks, though...
A paraphrase that summarizes the main idea is at the bottom of the answer.
An
The photoelectric effect occurs when light shining on a metal creates an
#color(red)"Red"# = I'm not sure."Atom" could be "metal", and "electric current" could be "photoelectron", for example.
#" "# #" "# #" "# APPENDIX
Shine light on a metal with enough energy, and it ejects an electron. Classical and quantum physicists argued whether it was due to higher intensity light (classical) or higher frequency light (quantum).
#E prop "intensity"# , or#E prop nu# ?Classical physics predicted that light frequency simply increased with temperature because the particles moved faster and had higher average kinetic energy, thereby having higher frequencies (lower wavelengths
#lambda# ) of oscillation:
#B_(lambda)(T) = (2ck_BT)/(lambda^4)# ,#" ""Rayleigh-Jeans Law"# Thus, it was predicted that black things emit an infinite amount of energy, which is clearly illogical. We know that perfect blackness (very small wavelength) absorbs all light, rather than emitting all of it.
Turns out, there is a more accurate description coming from Planck as to how blackbodies behave at high temperatures:
#B_(lambda)(T) = (2hc^2)/(lambda^5) 1/(e^(hc//lambdak_BT)-1)# ,#" ""Planck Radiation Law"#
Indeed, it shows that black things absorb all light, instead of emitting all light. Makes sense...