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