Calculating Work Function of Metal in a Photocell

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SUMMARY

The work function of the metal in a photocell can be calculated using the stopping potential and the energy of the incoming photons. Given a light frequency of 8×1014 Hz and a stopping potential of 1.2V, the energy of the photons is determined to be 5.304×10-19 J. By applying the equation hf = EK + φ, where EK is the kinetic energy of the emitted electrons (EK = eVs), the work function φ can be calculated definitively.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the equations E=hf and hf=φ + EK
  • Knowledge of stopping potential in photocells
  • Basic concepts of photon energy and wavelength
NEXT STEPS
  • Calculate the work function of various metals using different frequencies of light
  • Explore the relationship between stopping potential and kinetic energy in photocells
  • Investigate the experimental setups that demonstrate the photoelectric effect
  • Learn about the implications of work function in photovoltaic cells
USEFUL FOR

Students studying physics, particularly those focusing on quantum mechanics and the photoelectric effect, as well as educators looking to explain the principles of photocells and work functions in metals.

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Homework Statement


In a particular photocell, light with a frequency of 8×10^14Hz is directed onto the cell and it is found that a stopping potential of 1.2V is needed to reduce the photoelectric current to zero.
Calculate the work function of the metal in the cell.


Homework Equations


E=hf
E=hc/λ
hf=\phi + Ek

The Attempt at a Solution


Not much luck with this. Most sources that calculate the work function require the threshold frequency first, but we don't have that. Using E=hf and E=hc/λ I managed to find the energy of the light/photon as 5.304×10^-19 and the wavelength to be 3.725×10^-7 but now I'm stuck.
 
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I think you need to look at the experimental setup and then consider what is necessary to have the experiment exhibit the following behaviour.

A photon smashes into the surface of the metal and causes an electron to be ejected from the surface. Change the experiment parameters a bit and now the incoming photon doesn't manage to knock an electron off the surface.
 


The maximum kinetic energy carried by an electron is related to the stopping potential. To reduce the photo current to zero, the stopping potential has to overcome the kinetic energy of the electrons. Therefore, E_K = e V_s . But you also know, from equation 3 in your list that hf = E_K + \phi. You can now solve for \phi.
 

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