University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 39 - Particles Behaving as Waves - Problems - Exercises - Page 1313: 39.2


See explanation.

Work Step by Step

a. $E=\frac{hc}{\lambda}=\frac{(4.136\times10^{-15}eV)(3.00\times10^8 m/s)}{0.20\times10^{-9}m}=6.2keV$ b. $E=\frac{p^2}{2m}=(\frac{h}{\lambda})^2 \frac{1}{2m}=\frac{h^2}{2m \lambda^2}$ Evaluate using the electron’s mass to find 38 eV. c. Evaluate the expression using the alpha particle’s mass to find an energy of $5.2\times10^{-3}eV$. For a given wavelength, a photon has much more energy than an electron. The electron has more energy than an alpha particle.
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