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### Author Topic: The photoelectric effect  (Read 5217 times)

#### McQueen

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##### The photoelectric effect
« on: 29/08/2004 22:50:18 »

I would like your opinion on this experiment and the conclusions drawn from it. A 4 W. ultraviolet LED shining on a light sensitive surface from a distance of 10cms will result in the ejection of electrons from the surface it shines on. Yet if the light sensitive surface is moved to a distance of 10m from the source no photoelectric effects are detected although the ultraviolet rays may still cause objects to fluoresce at that distance.  What are the conclusions to be drawn from this ? Suppose the ultraviolet is in the 300 nM range , this gives an energy of  approx 4 electron volts or 6.4 x 10 ^^-19 J .  At such  small wave-lengths high directivity is possible and even a simple reflector results in a directivity of  about 10:1 in other words at 10m the beam will spread out over an area of 1 m. Then 4/6.4 x 10^^-19 = 6.25^^18 photons per sec. At a distance of 10 cms. there would be approx  10 x ^^18 photons striking the surface  , at a distance of 10 m there would still be  10 x ^^17 photons striking the surface. This works out to 10^^13 photons striking every cm sq of the surface , how is it that no photoelectric effects are detected. The only answer would seem to be that our interpretation of the term intensity is wrong. At present we tend to think of the photoelectric event as being due to a single event phenomenon. In other words the photoelectric effect is taken as being the result of  the eigen energy value of a single photon being transferred to an electron. Yet  if we examine the formation of x-rays using the bremmstrahlung effect it is  possible to see that the effect of a single electron passing near the nucleus and being slowed down is not the emission of just one photon but a whole series of photons whose wave lengths and frequencies might vary over a small range , in a very small period of time 10^^-12secs possibly. Similarly x-rays produced by the k-shell process result in multiple photons of a single frequency being emitted.  K-shell photons although they do not pass so close to the nucleus are more intense and have higher energy than x –rays produced by the bremmstrahlung effect. An inevitable conclusion is that the intensity of any electromagnetic radiation is due to both the eigen value of the photon and to its frequency (i.e to the number of times a second it is able to deliver that energy to an electron) in other words multiple photons striking at the same location  over a fixed period of time , translate into intensity.  This would explain why at close range the UV light has enough energy to eject electrons from the surface it is  striking , while at further distances although fluorescence is still observed due to the individual eigen values of the photons the photoelectric effect is absent.

#### gsmollin

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##### Re: The photoelectric effect
« Reply #1 on: 30/08/2004 20:33:40 »
The photoelectric effect is indeed the result of a single photon being absorbed by the outer shell of an atom and ejecting a single electron. This is not the same as bremmstrahlung, and the photon in question does not have X-ray energy, 100's of keV, but UV, ~4 eV. Let's keep these separate.

If you are not measuring any photo-electric current at 10 m, but are at 10 cm, then there can be a couple of problems with your experiment. 1) You need more sensitivity in your detector. You should measure something, but you may lack the sensitivity. 2) The atmosphere is absorbing the UV light, so the intensity of the UV beam is less than your sensitivity, see 1).

You should do a series of measurements, as a function of distance from the UV source. This would show you the dependence on distance, and what your maximum range is. You should take a complete set of readings at each distance point, showing the extinction voltage of the photo-electric effect, and the amount of photo current flowing in the detector tube.

I don't know what significance the fluorescing materials has, except to suggest it shows the UV is able to excite materials. Remember that fluorescence and photo-electric effect are closely related phenomena. However, unless you know the work function of the fluorescent material, you can't say much about the eV level of the UV.

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##### Re: The photoelectric effect
« Reply #1 on: 30/08/2004 20:33:40 »