Physics: Principles with Applications (7th Edition)

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

Chapter 17 - Electric Potential - Problems - Page 496: 16

Answer

a. 4.8 keV b. 42.8

Work Step by Step

a. The proton is accelerated through the opposite potential difference as the electron, and its charge is opposite to that of the electron. Considering equation 17-2 or equation 17-3, we see that the proton gains the same kinetic energy as the electron does, 4.8 keV. b. They have the same kinetic energy. Find the ratio of the speeds. $$\frac{1}{2}m_ev_e^2=\frac{1}{2}m_pv_p^2$$ $$\frac{v_e}{v_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{\frac{1.67\times10^{-27}kg}{9.11\times10^{-31}kg }}$$ $$=42.8$$ The electron’s speed is 42.8 times greater than that of the proton.
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