A photon of energy 8 eV strikes a metal surface whose threshold frequency is 1.6 * 10¹⁵ Hz. Find the maximum kinetic energy of emitted photoelectrons. (h = 6.6 * 10⁻³⁴ J s, 1 eV = 1.6 * 10⁻¹⁹ J)

1.4 eV
0.8 eV
4.2 eV
2.8 eV

Correct answer: 1.4 eV

Solution

Work function phi = h * nu_threshold = 6.6e-34 * 1.6e15 = 10.56e-19 J = 10.56e-19/1.6e-19 eV = 6.6 eV. KE_max = E_photon - phi = 8 eV - 6.6 eV = 1.4 eV.

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