# energy of a photon

Discussion in 'Electronic Design' started by Jamie, Jul 2, 2011.

1. ### JamieGuest

Hi,

Ok so this formula E = h*c/Î» gives the energy of a "photon". To
simplify that formula, it could be thought of as E = amplitude /
wavelength, with the h constant scaling the amplitude correctly, and the
c constant scaling the wavelength constant correctly.

One interesting thing from this formula is it apparently shows that all
"photons" aka electromagnetic waves resulting from electron orbital
emissions, have the same amplitude! So this must mean that all electron
orbital emissions and absorptions (which are frequency keyed) must all
emit or absorb electromagnetic waves with the same fixed amplitude,
regardless what the orbital number is. I find that to be quite amazing
and hard to understand why that would occur.

cheers,
Jamie

2. ### JamieGuest

Ahh I think I figured it out, this amplitude is proportional to the the
electron's angular momentum (which is a constant I guess)

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydazi.html#c3

cheers,
Jamie

3. ### JamieGuest

Hi,

New idea: Planck's constant describes the constant angular momentum
transfer when an electron changes orbital states.

http://en.wikipedia.org/wiki/Angular_momentum#Angular_momentum_in_quantum_mechanics

"an electron standing at rest has an angular momentum of hbar/2."

This shows that it is the electron's angular momentum that is the
inherent source of Planck's constant, if the electron had inherently
more "at rest" angular momentum, then the photoelectric emission graph
of voltage over frequency would have a higher slope and Planck's
constant would be a proportionately different number.

http://en.wikipedia.org/wiki/Planck_constant

"The Planck constant has dimensions of physical action; these are the
same as those of angular momentum"

So right there the Planck units even match those of the electron's
angular momentum.

Einstein made up the photon to explain Planck's constant and the
photoelectric effect when really he should have picked the electron!

cheers,
Jamie

4. ### JamieGuest

Hi,

I don't typically think of photons rattling around, as I don't believe
in them either. However for the emission of electromagnetic waves from
an atom, each emission must have a fixed amplitude, as seen from the
photoelectric experiments. That amplitude is defined by Planck's
constant, which I think is defined itself by the electron angular
momentum, so that when an atom emits a "photon", the amplitude of the
photon is directly related to the electrons angular momentum. That
shows Planck's constant is derived from matter and not from quantized
electromagnetic fields aka "photons".

cheers,
Jamie

5. ### Rene TschaggelarGuest

How about having the amplitude as square root
of the power ? As just the energy is given,
we need a time. Here comes the interesting relation.
We'd naturally assign one wavelength giving
a time of Î»/c = 1/frequency. But then the
bandwidth of the photon would be huge and the
spectrum continuous. And a line spectrum requires
the photon to be infinitely long...
We can trade spatial uncertainty against frequency
uncertainty and arrive at gaussian pulses, the
eigenfunctions of the Fourier transform. Thus the
length of a photon is (low natural number)/frequency.

For further reading search for "squeezed photons"

Rene