Electron/Photon locations
1) Is there any way that we can trace the path of an electron in an electron cloud (surrounding an atom). I know we can calculate the probability of an electron at any point, but can we actually map the path?
2) How 'localised' is a photon? Can we actually pin point...
hi guys thanks for ur replies!
What i am trying to do is to produce a smooth voltage to power an xray tube. I am trying to convert the AC currect into a DC current then smooth the voltage out so that the peak voltage is effectively the average voltage.
Exequor: What is L1? I'm not very...
Hi all!
Is it possible to produce a smooth(or almost smooth) voltage through full wave rectification? I know for half-wave, you can use a capacitor, but this doesn't seem to work for full wave.
Cheers
Can someone explain to me how the photon energy can heat up a material. Shouldn't the photon energy cause interactions(photoelectric, compton etc.) How can this energy be "absorbed" to make atoms move faster?
I'm currently in college at the moment and of course in high school we did all these stuff about electricity and I got through that alright.
But now I'm getting bugged: what exactly is voltage?
Sure V = Joules/coulomb so can one treat it as the energy "carried" by an electron? If so...
thank you so much. my lecturer didn't really explain how scintillators work (more on the lines of xray photon in, visible photon out in a nutshell) and this clarified that perfectly.
by any chance, do you know the answers to my 2nd question, the energy shells questions?
the intersection of y=1/(x^2) and y=x is at x=1
so the area is the area under the top curve - area under the bottom curve.
A = [I(1 to 2)(x dx)] - [I(1 to 2)((dx)/(x^2))]
I = integral sign
I believe that a scintillator converts xray photons into visible photons through the process of exciting electrons, then allowing it to fall to ground state.
My q is: shouldn't the energy of the photon released be the same as the photon energy that is used to excite the electron?
Just on...
This may sound dumb but...
since light is classified as an electromagnetic wave, it has an E field and a B field. But doesn't this mean the E field will attract metal, and the B field will cause currents to flow?
And since E = F/c = V/m does that mean the emr has a "voltage" or a "force"?
Use the lens formula:
1/f = 1/u + 1/v
where f = focus, u = object distance from lens, and v = image distance from lens
So:
1/20 = 1/10 + 1/v
Solve for v
v = -20cm
So the image is located at 20cm