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The Photoelectric Effect

  1. Sep 28, 2014 #1
    I want to understand how this circuit demonstrating the photoelectric effect works. Please see attached picture.
    ?temp_hash=37e06220550d4795670c14017d0d3890.png
    On the picture we see light entering the cathode C, which, if the light is above the threshold frequency can eject electrons from the cathode, which hits the anode A, creating a (conventional) current flowing from A to C. This current can be measured in the Ammeter and the voltage can be measured in the Voltmeter. Then we can connect a battery, where we can put in a "retarding voltage". The retarding voltage can then be increased until the current between A and C becomes zero. From this point on and for higher retarding voltage the electrons with the most kinetic energy don't have enough energy to make it through from C to A anymore. So if we for example know that the maximum kinetic energy of the electrons is 1.5 eV, then this will happen when the retarding voltage is increased to 1,5 V. Correct so far?

    So then, I think the only thing I don't understand about this circuit is why there is a potentiometer in it. I understand it's to regulate the voltage from the battery, as I described above, but it seems unnecessary to use a potentiometer for that. Wouldn't it make more sense to just have a variable resistor?
     

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  3. Sep 28, 2014 #2
    The battery and potentiometer are used to set the retarding voltage that you read on the voltmeter. The potentiometer is nothing else than two variable resistor used as a voltage divider. The photocurrent flows through the Ampmeter and potentiometer back to the cathode C and never builds up much of a potential of its own.

    This is a bit of a schematic circuit diagram. To make accurate measurements you would have to make it a bit more complicated.
     
  4. Sep 28, 2014 #3

    sophiecentaur

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    If there were no 'lower half' to the potential divider then what potential would appear at the point A and what current would flow? Would things change if the 'top half' resistor value changed? If you answer that, you should see why you need the lower half of the potential controller.
     
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