Daniel's Inquiry into Electron Position Energy and EM Wave Frequency

[danne89]

Hi,
I wonder how the electron's position energy reflects the electro magnetic wave's frequency.
Daniel

 

[Integral]

When an electron transitions from a higher to lower energy level a photon of electromagnetic energy is generated. The Energy of that photon is equivalent to the energy change of the electron and is given by $E= h \nu$ Where E is the energy of the Photon and $\nu$ is its frequency.

 

[danne89]

Hmm. I don't understand. So the wave length is arbitary, right? That don't make seens to me. Or is it always gamma rays?

 

[tfleming]

think of the maths; there's a eigenvalue solution; iff (if and only if) there's enough energy between these postional and motional energy states, THEN there's a certain frequency of any gamma particles emitted; there could even be more than one particle if the energy state change is big enough. depending on how many atoms are in that state then there will be so many gamma particles emitted; so in a population of atoms, there will be range of different energies involved and hence a range of gamma particles of discrete frequencies. this is where statistical mechanics comes in

 

[tfleming]

the energy is a solution to the eigenvalue problem so it is NOT arbitrary; the high-energy photon (gamma particle) has become EXCESS to the atom's needs and is expelled from the atom as it shifts to a lower energy state.

 

[danne89]

I know understand it. The problem was that I lacked knowleage about waves. Thanks!

 

[Integral]

Hmm. I don't understand. So the wave length is arbitary, right? That don't make seens to me. Or is it always gamma rays?

No the wavelength is not arbitrary it is determined by the enegry of the photon and is related to $\nu$ by

$$\lambda = \frac c \nu$$

 

[danne89]

So
$\lambda = \frac{hc}{e}$

 

[Integral]

that is correct,

I made an errorI my last post, now corrected.

$$\lambda = \frac c \nu$$

 

[Astronuc]

By convention, gamma rays originate in the nucleus of an atom (or from the decay of some subatomic particles), as distinguished from X-rays which involve the electrons in the electron field surrounding the atom.

However, an X-ray or gamma-ray of the same energy is the same photon as far as interactions with other matter is concerned.

The gamma-rays reflect the differences in energy states in an excited nucleus, just as X-rays, UV and photons of optical (visible) wavelengths reflect characteristic differences in energy states within the atomic electron field.

As for interactions between gamma or X-rays and electrons, see the Compton - http://scienceworld.wolfram.com/physics/ComptonEffect.html, and also

http://scienceworld.wolfram.com/physics/PhotoelectricEffect.html as it applies photoelectric absorption of the incident photon.

 

[danne89]

Thanks. Good links.

 

[FotonSingularity]

This misunderstanding seems to be due to the way the formula was shown E=hv. It’s easier to recognize if you write E=hf where E is the energy in joules, h is hertz constant, and f is frequency cycles per second. Hope this helps, I suggest you read on valence jumps. It will make things more clearer.

Just wondering Daniell but did you want to know the polerization axis due to stimulated emisson?