How Many Photoelectrons Are Ejected Per Second in This Experiment?

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SUMMARY

The discussion focuses on calculating the number of photoelectrons ejected per second in a specific experiment, as represented by a provided graph. The participant has identified that the maximum kinetic energy (K[max]) is 10 eV and is attempting to relate this to the current measured in coulombs per second (C/s). The key insight is that to find the number of photoelectrons, one must divide the maximum current by the charge of a single electron (1.60E-19 C). This method is essential for accurately determining the photoelectron emission rate.

PREREQUISITES
  • Understanding of photoelectric effect principles
  • Familiarity with kinetic energy calculations
  • Knowledge of electric current measurement in coulombs
  • Basic grasp of electron charge (1.60E-19 C)
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  • Research the photoelectric effect and its applications in experiments
  • Learn about calculating kinetic energy in electron volts (eV)
  • Explore methods for measuring electric current in physics experiments
  • Study the relationship between current and electron emission rates
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This discussion is beneficial for physics students, educators, and researchers interested in the photoelectric effect, as well as anyone involved in experimental physics and electron emission analysis.

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

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