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: 32


Wavelength of scattered radiation will be $\lambda=0.005211nm$

Work Step by Step

Wavelength of incident X-ray $\lambda_{0}=0.0045nm$ suppose wavelength of X-ray after scattering is $\lambda$ Angle of scattering is given as $\theta= 45^{0}$ for scattering by electron $\lambda_{C}=\frac{h}{m_{e}c}=2.43\times10^{-3}nm$ Wavelength shift or Compton shift is given by $\Delta \lambda= \lambda- \lambda_{0}=\lambda_{C} (1-cos\theta)$ putting the values in above equation we will get $\lambda- \lambda_{0}=2.43\times10^{-3}nm(1-cos45^{0})$ $\lambda- \lambda_{0}=2.43\times10^{-3}nm(1-0.7071)=0.7117\times10^{-3}nm$ $\lambda- \lambda_{0}=0.0007117nm$ $\lambda=0.0007117nm+\lambda_{0}=0.0007117nm+0.0045nm$ $\lambda=0.005211nm$
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