Question about Photon energy in Nuclear/Atomic Physics

  • #1

Main Question or Discussion Point

I picked up a book titled "Physics in Nuclear Medicine" that, while a slightly outdated edition, will presumably still have a bunch of valid fundamental information inside. One thing I noticed in it talking about EM radiation behaving as packets of energy was that it states: "The energy of the photon E and the wavelength lambda of its associated electromagnetic field are related by:"

[tex] \text{E(keV)} = 12.4/ \lambda (\text{\AA}) [/tex]

(On another note, how the heck do I get the Angstrom symbol to show up?)

But my confusion was when I think of the energy of a photon or a collection of photons, I think of the formulas [tex] E = h \nu = \hbar \omega [/tex]
or just the De Broglie relations in general where nu is the frequency, omega is the angular frequency, hbar is h/2pi, h is the normal Planck constant, etc; is this just a case of dimensional analysis where these are the same things but units need to be moved around? I can't see how that would be the case. Sure, momentum in the form [tex] P = \frac{h}{\lambda} [/tex] has frequency in the denominator...but energy for a photon doesn't. Do these equations represent the same things?

EDIT: NEVERMIND I JUST REALIZED I FORGOT WHAT NU REPRESENTS :^)
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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