Saturation voltage for photoelectric current physics forums gas leak los angeles

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I was wondering about saturation current in the photoelectric effect. It is clear to me that for a sufficiently large accelerating potential all of the electrons are gathered by the collecting electrode. Since it is all of them, there cannot be more, and the current won’t change if the accelerating potential is increased further.

What I’m not entirely clear about is what happens when the current is not saturated. My intuition is that some of the electrons simply won’t make to the collecting electrode, but I find answers to related questions that invoke the speed of the electron (kinetic energy) and the fact that the current is the quantity of charge per unit time.

I guess that not all electrons are necessarily emitted towards the collecting electrode, so some of them could simply "miss the target". If an accelerating potential is switched on, the electrons trajectories are bent towards the electrode. If the potential is too weak some electrons might still escape, though.

Does this make sense? Or else one should really take into account the different initial velocities of the electrons, and the current being "quantity of charge per unit time"? In that case the saturation would come about because the final velocities of the accelerated electrons would be much larger than any of their initial values, and so the latter would not matter. But for some reason this does not feel right.

I expect that the "slope" of the ramp towards saturation is affected by geometric factors, such as the area and shape of the collector and its distance from the target. For instance, the farther apart the electrodes are, the more "stretched" the saturation is, because only electrons emitted within a small solid angle would reach the collector.

I was wondering about saturation current in the photoelectric effect. It is clear to me that for a sufficiently large accelerating potential all of the electrons are gathered by the collecting electrode. Since it is all of them, there cannot be more, and the current won’t change if the accelerating potential is increased further.

What I’m not entirely clear about is what happens when the current is not saturated. My intuition is that some of the electrons simply won’t make to the collecting electrode, but I find answers to related questions that invoke the speed of the electron (kinetic energy) and the fact that the current is the quantity of charge per unit time.

I guess that not all electrons are necessarily emitted towards the collecting electrode, so some of them could simply "miss the target". If an accelerating potential is switched on, the electrons trajectories are bent towards the electrode. If the potential is too weak some electrons might still escape, though.

Does this make sense? Or else one should really take into account the different initial velocities of the electrons, and the current being "quantity of charge per unit time"? In that case the saturation would come about because the final velocities of the accelerated electrons would be much larger than any of their initial values, and so the latter would not matter. But for some reason this does not feel right.

I expect that the "slope" of the ramp towards saturation is affected by geometric factors, such as the area and shape of the collector and its distance from the target. For instance, the farther apart the electrodes are, the more "stretched" the saturation is, because only electrons emitted within a small solid angle would reach the collector.

2. However, the range of energy or speed depends very much on the energy of the incident photons. If the photons have energy just barely above the work function, then the range of energy of the photoelectrons may not matter that much. If that is the case, then the difference in KE plays a very small role in the current below saturation.

4. The caveat in all of this is the geometry and location of the anode. If the anode is big and encompassing the cathode, then the "missed electrons" no longer matter. I’ve done this when I reverse-biased the cathode and ground the vacuum chamber that the whole contraption sat in. Essentially, the stainless steel chamber is the anode, meaning I captured all the emitted photoelectrons. So in that case, clearly, below saturation, the variation in current is due to photoelectrons being emitted at different KE.

Also, sometimes I’m having a hard time following your explanations, but perhaps that has to do with me not being a native English speaker.Actually, jartsa did bring in a possibly relevant point, and again, it depends on how accurately you measure your photocurrent and how sensitive your equipment are at detecting these various mechanisms. Certainly, if you do this with a light source with such a low energy above threshold and at a very low forward bias, the photoelectrons will come out "lazily" and build up space-charge above the surface and will cause subsequent electrons to be pushed back into the cathode. But as I’ve said, typically one uses a He or Hg discharge lamp for basic lab experiment, and the photon energy is usually several times higher than the work function.

Also note that this typically applies to cathode that has been grounded (as is the rest of the containment vessel), while only the anode is forward biased. This is NOT the only setup that is done. I’ve mentioned earlier of reverse biasing just the cathode, while keeping the rest of the vessel grounded. You might possibly notice a slightly different photocurrent curve with this one.

Oh, btw, this is off topic, but if you can crank up the forward bias so that the E-field reaches to about 1 kV/m, you might notice that the "saturation" curve is not perfectly horizontal! The value of the "saturation current" increases with increasing forward bias. 🙂