## photoelectric effect

$c=\lambda v$

Savannah Mance 4G
Posts: 107
Joined: Fri Aug 30, 2019 12:17 am

### photoelectric effect

In the photoelectric experiment when they doubled the intensity of light, electrons were still not ejected. What is intensity? Is the intensity the frequency or the wavelength?

Camellia Liu 1J
Posts: 51
Joined: Sat Aug 24, 2019 12:15 am

### Re: photoelectric effect

I don't think intensity is related to either the frequency or the wavelength. When you double the intensity of light, you increase the number of photons emitted per second by the light source.

AngieGarcia_4F
Posts: 120
Joined: Thu Jul 25, 2019 12:17 am

### Re: photoelectric effect

When you increase intensity of light, you are increasing the number of photons emitted by the light, not to be confused with the energy per photon emitted by the light. To increase the energy per photon of the light and therefore eject more electrons from the metal, you'd increase the frequency (decrease the wavelength) of the light.

005206171
Posts: 107
Joined: Wed Sep 18, 2019 12:20 am

### Re: photoelectric effect

I like to think of light intensity a how strong a light shines - the number of photons a light emits. If it changed frequency, i'd see a different color light. But Einstein's experiment shows that's what you need to do to knock off more electrons because colors with higher frequencies (lower wavelengths) have more energy in the photons they emit. This is also why increasing frequency also increases the KEmax of the ejected e-. All 3 terms describe light differently.

Haley Chun 4H
Posts: 54
Joined: Sat Aug 17, 2019 12:18 am

### Re: photoelectric effect

When light acts as a particle (as in the photoelectric effect), increasing intensity means increasing the # of photons.