Photo-electric effect-calculating the saturation current

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K29
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Got an exam soon, and a question came up in a past paper that I'm not too sure if my reasoning is right.

If we have a lamp outputting 10W or 10J.s-1 and we do one photo-electric experiment with wavelength 150 nm and one with 300nm light, what will be the saturation current for each experiment. Oh and we assume ALL light from the lamp is incident on the photo-cell.

Now if you believe that saturation current is "only dependant on intensity" then you have been subjected to a very HORRID ambiguity and should leave now and go do some reading. (The physics-forums archive has some nice threads on this) btw higher frequency implies lower current. Its in my physics lab notes and I've done some lab work on this and confirmed it :O.

SOLUTION ATTEMPT

Well I can get the number of photon's per second. Its simple:
[tex]\frac{EnergyPerSecond}{EnergyPerPhoton}= PhotonsPerSecond[/tex]
So using the info we have we use[tex]E=hf[/tex] to get the energy per photon in Joules.
And of course Energy per second is 10J.s-1.

Now for a saturating current we know that all photo-electrons make it to the collector plate. So the number of photons from the light source per second are equal to the number of photo-electrons per second. So since an amp is equal to 6.241 × 1018 electrons per second. We can easily get the current. yay!! right?
 

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