Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - General Problems - Page 769: 62

Answer

$v = 0.822c$

Work Step by Step

The electron’s KE comes from the decrease in potential energy, so it has a kinetic energy of $6.20\times10^{-14}J$. Use equation 26–5b. $$KE=6.20\times10^{-14}J =(\gamma-1)mc^2=$$ $$\frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{6.20\times10^{-14}J }{(9.11\times10^{-31}kg)(3.00\times10^8m/s)^2}$$ $$v=c\sqrt{1-\frac{1}{(\frac{6.20\times10^{-14}J }{(9.11\times10^{-31}kg)(3.00\times10^8m/s)^2}+1)^2}}$$ $$v=0.822c$$
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