College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1043: 57


$\lambda = 3.64\times 10^{-12}~m$

Work Step by Step

We can find the minimum energy of the photon: $E = m_e~c^2+m_p~c^2$ $E = (9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2+(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $E = 5.46\times 10^{-14}~J$ We can find the maximum wavelength of the photon: $E = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(5.46\times 10^{-14}~J)}$ $\lambda = 3.64\times 10^{-12}~m$
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