Minimum wavelength of electrons ejected from metal

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Homework Statement



This problem has two parts.

A laser produces photons having an energy, E = 3.5eV.
a) What is the wavelength of photons produced, assuming that the index of refraction is 1.6? ->221nm
b) If this laser beam is focused on the clean surface of a metal having a work function of φ=1.5eV in vacuum, what is the minimum wavelength λmin of the electrons that will be ejected from the metal? Assume that the experiment is done in a vacuum. -> .867nm

Homework Equations



E = hv/λ
KEmax = hf - φ

The Attempt at a Solution


So I understand part a. It's simply λ = hv/E, where v is calculated using index of refraction. However I am stuck on part B, where the answer is .867 nm. At first I was trying to do λ = hc/φ, since v = c since it is in a vacuum, and since the minimum frequency will be when KE = 0 then hf = φ, however that gives me 826nm which is not correct, or even close to being correct.
 
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The question is asking for the minimum wavelength of the ejected electrons (de Broglie wavelength).