Work function (photoelectric effect)

[fluidistic]

Homework Statement

During successive illumination of a surface of a particular metal with radiation of wavelengths \lambda _1 =0.35 \mu m and \lambda _2=0.54 \mu m, we find that the respective maximum velocities of the photoelectrons have a difference between each other of a factor 2. Calculate the work function of the surface of the metal.

Homework Equations

E_1=\phi + E_k_1 where E_1 is the energy of a photon whose wavelength is \lambda _1, \phi is the work function and E_k_1 is the maximum kinetic energy of a photoelectron detached by a photon with wavelength \lambda _1.
E_2=\phi + E_k_2

The Attempt at a Solution

Using the equations, E_k_2 -E_k_1 = E_2-E_1 \Rightarrow \frac{m_e v_2 ^2}{2}-\frac{m_e v_1 ^2}{2}. But I'm told that v_1=2 v_2 \Rightarrow -\frac{3 m_e v_2 ^2}{2}=E_2-E_1 \Rightarrow v_2 ^2 =\frac{2}{3 m_e} (E_1 - E_2) and furthermore \phi =E_2 -E_k_2.
Using E=h \nu, E_1 =5.67948768 \times 10 ^{-19}J, E_2=3.68114942 \times 10 ^{-19}J.
m_e=9.10938215 \times 10 ^{-31}kg.
I find that \phi =3.01503667 \times 10 ^{-19}J which is less than 2eV.
However a friend told me he solved the exercise without having to plug the mass of the electron (I think it cancels out in his arithmetics) and he found out \phi =4.78 eV and according to him it corresponds to a copper surface (according to him but I see that silver and carbon have a closer work function to his value than copper has), which makes his result much more credible than mine.
Where did I go wrong? I'm really clueless.

 

[Mindscrape]

What you've done looks fine to me, as long as your numbers are right, but let me mull it over in the back of my mind for a bit.