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Photoelectron Spectroscopy

AP Chemistry· difficulty 4/5

Light of wavelength 200.0 nm strikes a metal surface. If the work function is 4.50 eV, what is the maximum kinetic energy of ejected electrons? (h = 6.626×10−34\times 10^{-34} J·s; c = 3.00×108\times 10^8 m/s; 1 eV = 1.602×10−19\times 10^{-19} J)

  • A

    4.504.50 eV

  • B

    6.206.20 eV

  • C

    0.500.50 eV

  • D

    1.701.70 eV

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Explanation

EphotonE_{photon} = hc/λ\lambda = (6.626e-34)(3.00e8)/(2.00e-7) = 9.94e-19 J = 6.20 eV. KEmax_{max} = EphotonE_{photon} - ϕ\phi = 6.20 - 4.50 = 1.70 eV.

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