Photoelectric Effect

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1.As the transmission is decreased by percentages, the stopping voltage required for the circuit is dramatically decreased as well as taking a longer charge time.

2.  Higher energy light (Green vs. Yellow) required a much larger stopping voltage in order to hold back all of the electrons given off.


1. Yes, since the frequency would result in an increase or decrease of light entering the tube over time, the stopping voltage would be affected by that amount per unit time.

2.  Yes because if the wave had a higher amplitude, it would increase the energy of the wave resulting in a greater energy transfer to the knocked off electrons

3.  No because it would constantly be being received from by the phototube which would have a constant emission rate, unless the wave is changed.

4.  No because it doesn't matter how many strike the emitter in a given time because they will have the same amount of energy to transfer to the electrons

5.  No, because as in number 4 each photon would still have the same energy which would result in a constant energy transfer at the same level.  No increase or decrease of voltage would be needed.

6.  Yes because there is space in between each photon striking the emitter, even if it is very small.  This causes a delay between energy transfers from photons to electron.

7.  It acts as a wave because as we varied the intensity of the light allowed into the phototube, the stopping voltage at that time greatly varied in time to reach the level as well where as a particle wouldn't have the same effect.

             Green Light:

Transmission Stopping Voltage Approx Charge Time
100% .974 18.5s
80% .974 24.9s
60% .974 63.4s
40% .974 180.4s
20% .92 164.2s

           Yellow Light:

Transmission Stopping Voltage Approx Charge Time
100% .83 11.9s
80% .83 14.4s
60% .83 18.5s
40% .83 37.2s
20% .826 112.5s
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