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Photoelectric Effect - Intensity in Phototube

  1. May 18, 2006 #1
    We have been doing some experiments using a mercury light source and filters for each wavelenght and a Detector Box (phototube) to measure the output voltage, to investigate planks constant, stopping voltage, intensity's effect..... the typical - "energy is quantised" idea. :zzz:

    There is one part of it i'm not too clear about and not sure how to explain it correctly.

    I know intensity should not affect the output voltage. But as we lower the intensity of the light/wavelength the voltage drops slightly.

    Here is the hint my lecturer gave:

    "Think of the phototube as a capacitor. The capacitance C is tiny (~pF) but as long as the voltmeter resistance R is very large (in this case ~ 10^13 ohms) the RC time constant will be long enough to ensure that the recorded voltage is accurate. However this may not apply when the incident intensity is low..." WHY? :uhh:
     
  2. jcsd
  3. May 18, 2006 #2
    Well there is most likely a leakage of charge in your device. So as the apparatus is charging up, a small amount of charge is also leaving. Obviously, if the photocurrent is very small (which is expected if the light intensity is low, following the quantum model), the rate of charge build-up will be correspondingly small. If the current is small enough, the rate of charge leakage will no longer be insignificant in comparison to the rate of charge build-up, and the apparent stopping potential will be lower than expected.
     
  4. May 21, 2006 #3
    Thanks for that explaination.

    How would i go more in to depth and integrate "time-constant" into the explaination?

    Because in the experiment we were initially ask to find out the detector's time constant, by timing it until the voltage dropped to e^-1 of the initial E.
    and i figured somehow it relates to the whole "intensity" thing
     
  5. May 22, 2006 #4
    Anyone enlighten me about how a low intensity relates to the RC time constant??? :frown:

     
  6. May 22, 2006 #5
    Well actually I didn't really look at the hint your lecturer gave you when I first read your post, so I just assumed your device had a leakage of charge.

    However, from reading that hint he gave you I'd expect that my initial response isn't what he is looking for.

    The only reason I can really think of at this point is that because the RC time constant is so large due to the high impedance, it will take longer for the capacitor to charge. That is, when you reduce the number of photons incident on the phototube you reduce the photocurrent, and therefore, it will take the capacitor that much longer to charge. What this means is that you might prematurely take the reading of the stopping potential without allowing enough time for the capacitor to charge to the same point as before (with the greater photocurrent).

    Like I said this might not be what he is wanting either, but this is what I came up with based on how your lecturer worded the hint.
     
  7. May 24, 2006 #6

    ZapperZ

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    Your voltmeter is typically place parallel to the cathode-anode assembly (phototube) since you want to measure the potential difference across it. If you recall from elementary physics, a voltmeter is suppose to have an intrisically high resistance, very much larger than the resistance of the resistor that you want to meausre the pd of. This is to ensure that practically all of the current is flowing through the resistor, and very, very little goes through the voltmeter.

    Now when you have a lot of intensity (photons), your cathode is producing lots of electrons, and thus, the resistance of the phototube is small. But as you decrease the intensity, less electrons are being produced. You have smaller current, and this results in a resistance of your phototube being very high and start to be appreciable to the internal resistance of your voltmeter. So now the voltmeter is no longer accurately measuring the pd across your phototube.

    Zz.
     
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