College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 27 - Problems - Page 1042: 30


$\lambda = 2.485\times 10^{-11}~m$

Work Step by Step

We can find the energy of the scattered photon: $E_f = E_0-K$ $E_f = (240.0~keV)-(190.0~keV)$ $E_f = 50.0~keV$ We can find the wavelength of the scattered photon: $E_f = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E_f}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(50.0\times 10^3~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda = 2.485\times 10^{-11}~m$
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