How Many Photoelectrons Are Ejected Per Second in This Experiment?

[elephantorz]

1. To make it easier I just scanned it

http://farm3.static.flickr.com/2062/2457086181_c2a13b6077_o.jpg"

I am supposed to find how many photoelectrons are ejected per second in the experiment represented by the graph.

2. See link



3. Also see link, all of my work so far is there [not all of it, my tutor took other figurings with her, so excuse the lack of lots of writing], I figured since Kinetic energy and potential difference were directly proportional K[max] is 10eV, but as you can see my answer is way off, anyway, any help would be greatly appreciated, and yes, I am female, I write in pink =D.

 

[elephantorz]

Nm, you just have to divide max current since it's C/s by 1.60E-19 C.

 

[Moetasim]

Thank you for providing the graph and your work so far. It is clear that you have put effort into understanding the experiment and calculating the number of photoelectrons ejected per second. However, your answer may be off because there are a few key factors that need to be considered in this experiment.

Firstly, the number of photoelectrons ejected per second depends on the intensity of the incident light. The graph you provided does not show the intensity of the light used in the experiment, which could affect the number of photoelectrons ejected. Additionally, the work function of the material being used also plays a role in determining the number of photoelectrons ejected. The work function is the minimum amount of energy required to remove an electron from the surface of a material.

Furthermore, the maximum kinetic energy of the photoelectrons is determined by the potential difference applied in the experiment. In your calculations, you have correctly identified that the maximum kinetic energy is 10eV, but it is important to note that this energy is only applicable to the fastest photoelectrons. There will be a range of kinetic energies for the ejected photoelectrons, with some having lower energies due to the varying work function of the material.

In order to accurately calculate the number of photoelectrons ejected per second, you will need to have more information about the experiment, such as the intensity of the light and the work function of the material. Additionally, it would be helpful to have the potential difference used in the experiment.

I hope this helps and good luck with your further calculations!