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: 44.19

Answer

$1.62\times10^{-25}kg$. This is about 97 proton masses.

Work Step by Step

Table 44.1 gives the mass of the $Z^0$ in energy units, $91.2 GeV/c^2$. $$91.2\times10^9\;eV \frac{1.602\times10^{-19}J}{eV}=1.461\times10^{-8}J$$ This is the value of $mc^2$, where m is the mass in kilograms. Divide by the mass of the proton to find the mass ratio of 97.
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