Another chem problem?

It is convenient to convert the work function to electron volts by dividing by the charge of the electron, 1.6E-19 coulombs. This gives an energy of 4.18 electron volts. Now we can use E = h/lambda (in appropriate units), where h is Planck's constant and lambda is the wavelength. Bottom line: it won't work; 4 ev puts the radiation in the ultraviolet.