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Homework Help: Work function (photoelectric effect)

  1. Sep 14, 2010 #1


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    1. The problem statement, all variables and given/known data
    During successive illumination of a surface of a particular metal with radiation of wavelengths [tex]\lambda _1 =0.35 \mu m[/tex] and [tex]\lambda _2=0.54 \mu m[/tex], 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
    [tex]E_1=\phi + E_k_1[/tex] where [tex]E_1[/tex] is the energy of a photon whose wavelength is [tex]\lambda _1[/tex], [tex]\phi[/tex] is the work function and [tex]E_k_1[/tex] is the maximum kinetic energy of a photoelectron detached by a photon with wavelength [tex]\lambda _1[/tex].
    [tex]E_2=\phi + E_k_2[/tex]

    3. The attempt at a solution
    Using the equations, [tex]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}[/tex]. But I'm told that [tex]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)[/tex] and furthermore [tex]\phi =E_2 -E_k_2[/tex].
    Using [tex]E=h \nu[/tex], [tex]E_1 =5.67948768 \times 10 ^{-19}J[/tex], [tex]E_2=3.68114942 \times 10 ^{-19}J[/tex].
    [tex]m_e=9.10938215 \times 10 ^{-31}kg[/tex].
    I find that [tex]\phi =3.01503667 \times 10 ^{-19}J[/tex] 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 [tex]\phi =4.78 eV[/tex] 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. jcsd
  3. Sep 14, 2010 #2
    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.
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