## Photoelectric effect

Posts: 32
Joined: Fri Sep 28, 2018 12:23 am

### Photoelectric effect

In photoelectric experiments, typically what part of the electromagnetic spectrum is the incoming light?
I thought the incoming light was visible light. Could someone explain why this is not the case/what part of the EM the incoming light is from?

Alexandra Ortega 4D
Posts: 31
Joined: Wed Oct 03, 2018 12:17 am

### Re: Photoelectric effect

Typically, ultraviolet radiation is used to perform the photoelectric experiments. UV light has shorter wavelengths than other electromagnetic radiation, like visible light, which is why it is used. The electromagnetic radiation being used in these experiments must have enough energy for one photon to be emitted from the metal.

juleschang16
Posts: 55
Joined: Fri Sep 28, 2018 12:19 am

### Re: Photoelectric effect

In the photoelectric experiment, an ultraviolet radiation source is shined on a metal and UV light is used over visible light because it has a higher intensity and therefore, more electrons can be emitted.

Dong Hyun Lee 4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

### Re: Photoelectric effect

Using the equation v = c / lambda where lambda = the wavelength. When wavelength is shorter v is greater (as for UV lights) and using the second equation of E = hv, v being greater would result in more energy.

Yukta Italia 3I
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

### Re: Photoelectric effect

I have a question from Chapter 1 in the 6th edition of the textbook, regarding question 22. Why does the photoelectric effect best support the idea that electromagnetic radiation has the properties of particles?

105002507
Posts: 30
Joined: Fri Sep 28, 2018 12:15 am

### Re: Photoelectric effect

It is because EMR consists of particles of energy called photos and each of those photos has a fixed energy of hV. This is according to Einstein.