Work function of cathode, visible light

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The maximum possible work function for a photodetector responding to the entire range of visible light is 1.78 eV or less. At this work function, 700 nm light can just barely release electrons, while 400 nm light can easily do so. A work function greater than 3.11 eV would prevent the emission of electrons from visible light entirely. The confusion arises from the energy calculations, where 3.11 eV is the threshold for 400 nm light, but it does not allow for effective electron emission across the visible spectrum. Therefore, the work function must be optimized to ensure electron emission from both ends of the visible light spectrum.
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Homework Statement


You need to design a photodetector that can respond to the entire range of visible light (400 nm - 700 nm). What is the maximum possible work function?

Homework Equations


E = hf = hc/lambda

The Attempt at a Solution


For 400nm, I got E = 3.11 eV
and for 700nm I got 1.78 eV.

The solution is that the maximum work function is 1.78eV or less, and I do not understand why because 3.11eV is beyond the threshold for 400nm light to emit right? So shouldn't the answer be 3.11eV or greater?
 
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If your material has a work function of 3.11 eV, then 400 nm light is barely able to knock out electrons, the 700 nm light at 1.78 eV/photon cannot knock out any electrons.
If your material has a work function of more than 3.11 eV, you cannot release any electrons with visible light.

If your work function is 1.78 eV/photon, then 700 nm light barely manages to release electrons, 400 nm light can do so easily.
 
What is the energy of the emitted electrons for a work function of 3.11 eV with incident light of 700 nm?
 
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