Question about Photon energy in Nuclear/Atomic Physics

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I picked up a book titled "Physics in Nuclear Medicine" that, while a slightly outdated edition, will presumably still have a bunch of valid fundamental information inside.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

[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 :^)

 

[AndyW]

Dear forum post,

Thank you for sharing your thoughts and questions about the relationship between photon energy and wavelength. As a scientist in the field of nuclear medicine, I can offer some insights into this topic.

Firstly, it is important to clarify that the equations you mentioned do represent the same thing – the energy of a photon. The equations E = hν and E = ħω both describe the energy of a single photon in terms of its frequency (ν) or angular frequency (ω), respectively. The equation E = ħω is the more general form, where ħ represents the reduced Planck constant (h/2π). On the other hand, the equation E = hν is a simplified version that only applies to electromagnetic radiation, where the frequency and angular frequency are related by ν = ω/2π.

Now, to address your confusion about the relationship between photon energy and wavelength, let me explain the equation you mentioned in the book: E(keV) = 12.4/λ(Å). This equation is derived from the relationship between the energy and frequency of a photon (E = hν), combined with the fact that the frequency of an electromagnetic wave is inversely proportional to its wavelength (ν = c/λ, where c is the speed of light). By substituting for ν in the first equation, we get E = hc/λ. Then, using the values of h and c in appropriate units, we get E(keV) = 12.4/λ(Å). So, this equation is essentially a conversion factor that allows us to calculate the energy of a photon in keV units when we know its wavelength in Ångstroms.

I hope this explanation helps to clarify your confusion. In short, the equations you mentioned do represent the same thing – the energy of a photon – and the equation from the book is simply a conversion factor derived from the relationship between energy and frequency of a photon, and the inverse relationship between frequency and wavelength of electromagnetic radiation.