# Current in photoelectric effect

In the first lecture about photoelectric effect in my physics course my teacher proposed this device:

Then he started to say why photoelectric was not classical and made this statements:

1) Increasing on radiation intensity only increases the number of electrons being emitted and intensity of the current measured.

2) Increasing the frequency only leads to faster electrons not to current's intensity.

My question was: But if more electrons are released in a same amount of time (faster electrons) i.e. They will travel the space between the plates faster that would not lead to a increasing the current? (He said no, and I still can't make out why.)

Edit: What confused me more was this equation from my electrodynamics course: $i=NAve$ Where $N$ stands for number of electrons per unit of volume, $A$ is cross-sectional area, $v$ is velocity of the electron and $e$ is the fundamental charge.

• "faster electrons" is not the same as "more electrons released in the same amount of time", so I'm wondering why you've phrased your question that way. Is your question asking only about situation 2? Dec 15, 2016 at 23:20
• If you are watching a highway, and one car per minute passes your location at 10 mph, and an hour later one car per minute passes your location at 100 mph, how many cars pass your location per minute? Is there a difference? Now pretend the cars are electrons. Does the current change between the two situations? Dec 16, 2016 at 0:08
• There is a transient effect when the frequency is changed. Then it returns to steady state. Dec 16, 2016 at 3:02
• @BowlOfRed I used more because on my understanding when electrons move faster they open space for the next ones behind them, like in a pipe of water. Dec 16, 2016 at 3:32
• Did your instructor refer to 'increasing the frequency' meaning the light wavelength being decreased? That doesn't change the number of photons per second, it just lessens the oscillation period of the electric waves. Dec 16, 2016 at 8:53