University Physics with Modern Physics (14th Edition)

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

Chapter 44 - Particle Physics and Cosmology - Problems - Exercises - Page 1520: 44.20


See explanation.

Work Step by Step

a. From Table 44.3, the rest energy of the $\Sigma^0$ is 1193 MeV. From Table 44.3, the rest mass of the $\Lambda^0$ is 1116 MeV. Assuming the $\Lambda^0$ is produced at rest, the energy of the photon is the difference, 77 MeV. b. $p=\frac{E}{c}=4.1\times10^{-20}kg\cdot m/s$. If the $\Lambda^0$ had this much momentum, its KE would be $\frac{p^2}{2m}$, which is only about 2.8 MeV. This is much less than the 77 MeV of the photon’s energy, justifying the original assumption.
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