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 936: 35


Scattering angle is approximately $66^{0}$ recoiling speed of the electron $v=3.1937\times10^{6}m/s$

Work Step by Step

Frequency of incident X-ray is given as $f_{0}=1.210\times10^{18}Hz$ since $\lambda=\frac{c}{f}$ , $\lambda_{0}=\frac{c}{f_{0}}=\frac{3\times10^{8}m/s}{1.210\times10^{18}Hz}=2.47933\times10^{-10}m=0.247933nm$ wavelength of incident X-ray light is $\lambda_{0}=0.247933nm$ Frequency of scattered X-ray is given as $f=1.203\times10^{18}Hz$ since $\lambda=\frac{c}{f}$ , $\lambda=\frac{c}{f}=\frac{3\times10^{8}m/s}{1.203\times10^{18}Hz}$=$2.49376\times10^{-10}m=0.249376nm$ wavelength of scattered X-ray light is $\lambda=0.249376nm$ for scattering by electron $\lambda_{C}=\frac{h}{m_{e}c}=2.43\times10^{-3}nm$ Wavelength shift or Compton shift is given by $\Delta \lambda= \lambda- \lambda_{0}=\lambda_{C} (1-cos\theta)$ $\lambda_{C} (1-cos\theta)= \lambda- \lambda_{0}=0.249376nm-0.247933nm$ $2.43\times10^{-3}nm (1-cos\theta)=1.443\times10^{-3}nm$ $1-cos\theta=0.5938$ $cos\theta=0.40617$ or $ \theta=cos^{-1}0.40617=66.035^{0}\approx66^{0}$ kinetic energy of recoiling electron will be difference in energy of incident X-ray and scattered X-ray. so $K=\frac{1}{2}mv^{2}=hf_{0}-hf$, here $v $ is the velocity of electron and $m$ is mass of electron $v= \sqrt\frac{2h(f-f_{0})}{m}$=$\sqrt \frac{2\times6.63\times10^{-34} J.s(1.210\times10^{18}Hz-1.203\times10^{18}Hz)}{9.1\times10^{-31}kg}$ $v=3.1937\times10^{6}m/s$
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