# Work function (photoelectric effect)

1. Sep 14, 2010

### fluidistic

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $$\lambda _1$$.
$$E_2=\phi + E_k_2$$

3. 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.

2. Sep 14, 2010

### 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.