Calculating Work Function of Metal in a Photocell

AI Thread Summary
To calculate the work function of the metal in a photocell, the stopping potential of 1.2V indicates the maximum kinetic energy of the emitted electrons. The energy of the incoming photons, calculated using the frequency of 8×10^14 Hz, is approximately 5.304×10^-19 J. The relationship between photon energy, kinetic energy, and work function is given by the equation hf = Ek + φ. By substituting the stopping potential into the kinetic energy equation (Ek = eVs), the work function can be determined. The solution involves rearranging the equations to isolate φ, leading to the final calculation of the work function.
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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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