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: 19

Answer

wavelength of light is $\lambda=2.53698\times10^{-7}m=253.69nm$ approximately $254nm$

Work Step by Step

Work function of the metal is $\phi_{0} =2.4eV$ stopping potential is given as $V_{0} =2.5V$ from equation 27.5 maximum kinetic energy $K_{max}$of photoelectron is defined as $K_{max}=eV_{0}$ putting the value of $V_{0} =2.5V$ we will get $K_{max}=e2.5V=2.5eV$ From equation 27.7 $E=hf= K_{max} + \phi_{0}$ where $hf$ is energy of incident photon, $K_{max}$ is maximium kinetic energy of photelectron, $\phi_{0}$ is work function putting the values of $K_{max}=2.5eV$ and $\phi_{0} =2.4eV$ we will get $E=hf=2.5eV+2.4eV=4.9 eV$ $1eV$ is equal to $1.6\times10^{-19}J$ so $ 4.9 eV$ is equal to $ 4.9\times1.6\times10^{-19}J=7.84\times10^{-19}J$ $E=hf=\frac{hc}{\lambda}$ so $\lambda= \frac{hc}{E}$ $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$ and $E=7.84\times10^{-19}J$ we will get $\lambda= \frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{7.84\times10^{-19}J}$ $\lambda=2.53698\times10^{-7}m=253.69nm$ approximately $254nm$
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