If the work function of a metal surface is 2.20 eV,what frequency of incident light would give a maximum kinetic energy of 0.25 eV to the photoelectrons ejected from this surface? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s) A)2.05 × 10¹⁴ Hz B)1.02 × 10¹⁴ Hz C)2.50 × 10¹⁴ Hz D)3.53 × 10¹⁴ Hz E)5.92 × 10¹⁴ Hz

Question

Quizplus · Accepted Answer

The energy of the incident light must be equal to the work function plus the maximum kinetic energy of the ejected photoelectrons. Using the equation E = hf, where E is the total energy (2.20 eV + 0.25 eV = 2.45 eV), and converting eV to joules, we find the frequency f = (2.45 eV * 1.60 × 10^-19 J/eV) / (6.626 × 10^-34 J∙s) = 5.92 × 10^14 Hz.

If the Work Function of a Metal Surface Is 2