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

Question

Quizplus · Accepted Answer

The maximum kinetic energy of the photoelectrons is given by the difference between the photon energy and the work function: $3.4 \, \text{eV} - 2.4 \, \text{eV} = 1.0 \, \text{eV}$. Converting this to joules using $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$, we get $1.0 \times 1.60 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-19} \, \text{J}$.

A Photocathode Having a Work Function of 2