a.) A surface whose work function is = 4 eV is illuminated by a light whose wave
ID: 1455723 • Letter: A
Question
a.) A surface whose work function is = 4 eV is illuminated by a light whose wavelength is 85.3 nm.
What is the maximum kinetic energy of a photoelectron emitted from the surface? The speed of light is 3 × 108 m/s and Planck’s constant is 6.63 × 1034 J · s .
Answer in units of eV.
b.) The light intensity incident on a metallic surface produces photoeletrons with a maximum kinetic energy of 4.78 eV. If the light intensity is doubled, what is the maximum kinetic energy of the pho- toeletrons? Answer in units of eV.
Explanation / Answer
From the photo-electric equation:
hf - = max. KE
= 4 eV*(1.60E-19 C) = 6.40 E-19 J
f = (3.0 E8)/(85.3 E-9) = 3.516 E15 Hz
(6.63E-34)*((3.516 E15) - (6.40E-19) = 2.33 E-18 J
2.33 E-18 J/ 1.60 E-19 C ... = 14.56 eV
Please post the other question separately.
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