## Photoelectric Effect

$c=\lambda v$

Dustin Shin 2I
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

### Photoelectric Effect

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?

katherinemurk 2B
Posts: 68
Joined: Wed Nov 15, 2017 3:02 am

### Re: Photoelectric Effect

Intensity is what effects the amount of electrons. So as intensity increases, quantity of electrons being removed increases

Jessica Dharmawan 1G
Posts: 32
Joined: Fri Sep 28, 2018 12:15 am

### Re: Photoelectric Effect

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.

clamond3F
Posts: 8
Joined: Fri Sep 28, 2018 12:23 am

### Re: Photoelectric Effect

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.

Andreana Vetus 1A
Posts: 28
Joined: Fri Sep 28, 2018 12:17 am

### Re: Photoelectric Effect

Yes because intensity is what controls the amount of electrons being ejected. Remember that intensity however does not increase the energy of the light.

Luis_Yepez_1F
Posts: 50
Joined: Fri Sep 28, 2018 12:18 am

### Re: Photoelectric Effect

Yes, because intensity is proportional to electrons ejected.

taryn_baldus2E
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

### Re: Photoelectric Effect

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.

Alli Hinmon 3E
Posts: 30
Joined: Wed Nov 08, 2017 3:00 am

### Re: Photoelectric Effect

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.

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