Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 37 - Relativity - Problems - Page 1149: 58f

Answer

$\beta = 0.99999987$

Work Step by Step

We can find $\gamma$: $K = (\gamma-1)~mc^2$ $\gamma-1 = \frac{K}{mc^2}$ $\gamma = 1+\frac{K}{mc^2}$ $\gamma = 1+\frac{(1.0000000\times 10^9~eV)(1.602 176 462\times 10^{-19}~J/eV)}{(9.109 381 88\times 10^{-31}~kg)(2.998\times 10^8~m/s)^2}$ $\gamma = 1+1956.8529$ $\gamma = 1957.8529$ We can find $\beta$: $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ $1-\beta^2 = \frac{1}{\gamma^2}$ $\beta^2 = 1-\frac{1}{\gamma^2}$ $\beta = \sqrt{1-\frac{1}{\gamma^2}}$ $\beta = \sqrt{1-\frac{1}{(1957.8529)^2}}$ $\beta = 0.99999987$
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