Light with a frequency of 8.70 × 10¹⁴ Hz is incident on a metal that has a work function of 2.8 eV. What is the maximum kinetic energy that a photoelectron ejected in this process can have? A)8.71 × 10^-1⁹ J B)3.11 × 10^-19 J C)1.28 × 10^-19 J D)2.41 × 10^-19 J E)4.52 × 10^-19 J

Question

Quizplus · Accepted Answer

The maximum kinetic energy of a photoelectron is given by the equation:KE = hf - Φwhere h is Planck's constant, f is the frequency of the light, and Φ is the work function of the metal.Plugging in the given values, we get:KE = (6.626 × 10^-34 J s)(8.70 × 10^14 Hz) - (2.8 eV)(1.602 × 10^-19 J/eV)KE = 1.28 × 10^-19 JTherefore, the maximum kinetic energy that a photoelectron ejected in this process can have is 1.28 × 10^-19 J.

Light with a Frequency of 8