College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 27 - Quantum Physics - Learning Path Questions and Exercises - Exercises - Page 935: 14

Answer

Maximum kinetic energy of the photoelectrons ejected from the surface is $K_{max}=1.63\times10^{-19}J$

Work Step by Step

given that work function of the surface $\phi_{0}$ =$5.0\times10^{-19}J$ wavelength of incident light is $\lambda=300nm=300\times10^{-9}m=3\times10^{-7}m$ $E=hf$, since $f=\frac{c}{\lambda}$ Energy of a Quanta of light is $E=hf=\frac{hc}{\lambda}$ $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$, $\lambda=3\times10^{-7}m$ putting these values we will get $E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{3\times10^{-7}m}$ $E=hf=6.63\times10^{-19}J$ From equation 27.7 $hf= K_{max} + \phi_{0}$ where $hf$ is energy of incident photon, $K_{max}$ is maximium kinetic energy of photelectron, $\phi_{0}$ is work function of the surface. $K_{max} =hf-\phi_{0}$ putting values of $hf$ and $\phi_{0}$ in this equation we will get $K_{max}=6.63\times10^{-19}J-5.0\times10^{-19}J$ $K_{max}=1.63\times10^{-19}J$
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