mississippichem Posted November 22, 2011 Share Posted November 22, 2011 Yeah, I know, Swanson Isn't planck's constant proof that a photon has energy independent of it's frequency... do photons interact with electromagnetic radiation or are they electromagnetic radiation?* Got it backwards. A photon's energy is defined by its wavelength. A photon IS electromagnetic radiation also. Link to comment Share on other sites More sharing options...
Appolinaria Posted November 22, 2011 Share Posted November 22, 2011 Got it backwards. A photon's energy is defined by its wavelength. A photon IS electromagnetic radiation also. what is the purpose of h? why not just e=v Link to comment Share on other sites More sharing options...
A Tripolation Posted November 22, 2011 Share Posted November 22, 2011 what is the purpose of h? why not just e=v mississippichem was saying that a photon's energy is defined by its wavelength. That doesn't mean it's soley dependent on upon it. Think about units here. Wavelengths are defined by meters. Energy is joules. A statement that e=v just doesn't make any sense. Link to comment Share on other sites More sharing options...
Appolinaria Posted November 22, 2011 Share Posted November 22, 2011 ok i think i get it now. so wavelength is just how many photon packets there are. more packets = more energy of the photon Link to comment Share on other sites More sharing options...
derek w Posted November 23, 2011 Share Posted November 23, 2011 (edited) no.My understanding is that waves have frequency and amplitude,but in light the amplitude has a constant of "h",therefore the higher the frequency the more energy.The amplitude is always "h"? In musical instruments the amplitude of a wave depends on size of instrument,therefore the constant "h" is related to the the size of the big bang.In another universe with a bigger big bang,amplitude would be "h+"?Or am i misunderstanding something. Edited November 23, 2011 by derek w Link to comment Share on other sites More sharing options...
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