Example:Light of frequency of 1.00 x 10¹5 Hz illuminates a sodium surface. The ejected photoelectrons are found to havea maximum kinetic energy of 1.78 eV. Find the threshold frequency for this metal.Given:Solution:KEmax = (1.78 eV)(1.60 x 10-19 J/eV)KEmax hf - hft ; ft=== 2.85 x 10-1⁹ Jhf-KEmaxh(6.63 x 10-34 Js) (1.00 x 10¹5 Hz)-(2.85 x 10-¹⁹ J)6.63 x 10-34 Jsf= 1.00 x 1015 Hzf₁= 5.70 X 1014 HzActivity:Solve the following problems. Use the rubrics as your guide in presenting your solution (refer to attachment A.1at the next page).1. In the photoelectric effect, it is found that incident photons with energy 5.00 eV will produce electronswith a maximum kinetic energy 3.00 eV. What is the threshold frequency of this material?

Given, Energy E = 5.00 eV = (5.00 eV)(1.60 × 10-19 J/eV)…

Answered: Activity: Solve the following problems.…

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