A photocathode having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 3.4 eV. What is maximum kinetic energy of the photoelectrons produced?(c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 1.6 × 10^-19_J B) 3.8 × 10^-19_J C) 4.4 × 10^-19_J D) 4.9 × 10^-19_J E) 5.4 × 10^-19_J

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Quizplus · Accepted Answer

The maximum kinetic energy of the photoelectrons is given by the equation: KE_max = photon energy - work function. Converting the work function and photon energy from eV to joules: Work function = 2.4 eV = 2.4 × 1.60 × 10^-19 J = 3.84 × 10^-19 J. Photon energy = 3.4 eV = 3.4 × 1.60 × 10^-19 J = 5.44 × 10^-19 J. Therefore, KE_max = 5.44 × 10^-19 J - 3.84 × 10^-19 J = 1.6 × 10^-19 J.

A Photocathode Having a Work Function of 2