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Photoelectric effect and frequency of light

"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
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.

Solution Preview

Your setup is totally fine. Your equations are correct. Your problem lies in your algebra.

Recheck ...

Solution Summary

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