# I Question about Photon energy in Nuclear/Atomic Physics

1. Mar 25, 2016

### Zacarias Nason

I picked up a book titled "Physics in Nuclear Medicine" that, while a slightly outdated edition, will presumably still have a bunch of valid fundamental information inside.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:"

$$\text{E(keV)} = 12.4/ \lambda (\text{\AA})$$

(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 $$E = h \nu = \hbar \omega$$
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 $$P = \frac{h}{\lambda}$$ 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 :^)