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    Photoelectric effect and frequency of light

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    "The work function of gold is 4.58 eV. What frequency of light must be used to eject electrons from a gold surface with a maximum kinetic energy of 6.48e-19?"

    Here's what I did:

    W0 = (4.58 eV) (1.6e-19) = 7.33e-19 J

    Kmax = hf - W0
    so
    f = (Kmax / h) + W0
    f = (6.48e-19 / 6.626e-34) + 7.33e-19 = 9.78e14 Hz

    But that's not the right answer. The right answer is supposedly 2.08e15.

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    https://brainmass.com/physics/photoelectric-effect/photoelectric-effect-frequency-light-41906

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    Your setup is totally fine. Your equations are correct. Your ...

    Solution Summary

    This solution provides a brief outline on the mistake in the problem, involving frequency calculations.

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