Find current generated from photoelectric effect.

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SUMMARY

The discussion focuses on calculating the current generated from the photoelectric effect when a beam of light with a wavelength of 300nm strikes a metal surface with a work function of 2.25 eV. The power of the light is 1 microW per square centimeter, illuminating a total area of 3 cm². The correct approach to find the current involves using the number of emitted electrons per second, calculated from the photon flux, rather than directly applying power and voltage. The final current produced is determined to be 0.726 microamps, as opposed to the incorrect calculation of 1.59 microamps.

PREREQUISITES
  • Understanding of the photoelectric effect and its principles
  • Familiarity with the concepts of energy (E = hf) and work function in electron emission
  • Knowledge of current calculation using charge and time (I = Q/t)
  • Basic understanding of power (P = IV) and its application in electrical circuits
NEXT STEPS
  • Study the relationship between photon energy and electron emission in the photoelectric effect
  • Learn about the calculation of current from charge flow in photoelectric devices
  • Explore the concept of work function and its impact on electron emission efficiency
  • Investigate the effects of light intensity and wavelength on the photoelectric current
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and quantum mechanics, as well as educators and anyone interested in the practical applications of the photoelectric effect in technology.

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Homework Statement


A beam of light of 300nm is shone at a metal surface of work function 2.25eV. The power of the light striking each square centimeter is 1 microW. A total of 3cm^2 is illuminated by the beam. If each photon caused an electron to be ejected, what current is produced?




Homework Equations



E = hf - (work function)]
P/V = I
E = hf



The Attempt at a Solution



I calculated each electron emitted to have 1.89eV, and for each square centimeter, 1.51x10^12 photons strike per second, So, I try to use P/V = I, using P = 3microwatts and V = 1.89eV, which gives me an answer of 1.59microamps. However, the answer given in the book is 0.726microamps. Can anyone see where I'm going wrong, or any concepts I'm not understanding correctly? Thanks for any help!
 
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The I = P/V doesn't make sense to me.
You get one electron for each photon, so you have 3*1.51x10^12 electrons flowing per second. Use I = Q/t to figure out the resulting current.

It is incorrect to use the 3 microwatts on the electron side because much of the power is lost in knocking those electrons loose.
 
That 1.89 eV you calculated, is the kinetic energy of the emitted electron- it is not voltage. (voltage is work/charge).You need the current- the charge flowing through a cross section in one second. You got 1.51x10^12 photons/cm2, they will kick out the same number of electrons in one second and three times that from 3 cm2. The out-flowing electrons carry e=1.6x10^19 charge each what is the total charge supplied by the photocell in one second?

ehild
 

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