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### Photoelectric Effect

Posted: Mon Oct 29, 2018 12:08 am
In terms of the photoelectric effect, just to make sure I fully understand this concept, if the quantity of electrons increases, would the intensity of the light also increase due to the photoelectric effect?

### Re: Photoelectric Effect

Posted: Mon Oct 29, 2018 12:14 am
Intensity is what effects the amount of electrons. So as intensity increases, quantity of electrons being removed increases

### Re: Photoelectric Effect

Posted: Mon Oct 29, 2018 9:12 am
As the light source is increased in intensity, then the number of electrons ejected increases. Light intensity is proportional to the number of electrons ejected.

### Re: Photoelectric Effect

Posted: Sun Nov 04, 2018 9:42 pm
As long as the wavelength of light being emitted from the source is short enough to eject electrons from the surface then increasing the intensity (number of photons) of the light would increase the number of electrons being ejected, because each photon has the potential to dislodge one electron.

### Re: Photoelectric Effect

Posted: Sun Nov 04, 2018 9:51 pm
Yes because intensity is what controls the amount of electrons being ejected. Remember that intensity however does not increase the energy of the light.

### Re: Photoelectric Effect

Posted: Sun Nov 04, 2018 10:13 pm
Yes, because intensity is proportional to electrons ejected.

### Re: Photoelectric Effect

Posted: Sun Nov 04, 2018 10:23 pm
Yes, the intensity of the light will influence the number of electrons released, but first, the light must reach the needed frequency to eject any electrons.

### Re: Photoelectric Effect

Posted: Sun Nov 04, 2018 10:41 pm
I am having trouble with when to use the equation E=hv(frequency) and when to use E=hc/Wavelength
I understand that the 1s is used for "electromagnetic radiation" and the second is "photons and light" but in an example it asked for frequency of the lazerpointer and the correct answer used E=hv.