University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 38 - Photons: Light Waves Behaving as Particles - Problems - Exercises - Page 1275: 38.6


1.45 eV.

Work Step by Step

The threshold frequency is $f=\frac{c}{\lambda}=1.10\times10^{15}Hz $. Now find the work function. It is the energy of this threshold photon. $\phi =(6.626\times10^{-34}J \cdot s)( 1.10\times10^{15}Hz)=7.29\times10^{-19}J$ Find the energy of the uv photon. $hf=(6.626\times10^{-34}J \cdot s)( 1.45\times10^{15}Hz)=9.61\times10^{-19}J$ The maximum KE is the uv photon’s energy, minus the work function. $$K_{max}=hf-\phi=2.32\times10^{-19}J =1.45 eV$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.