Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 37 - The Foundation of Modern Physics - Exercises and Problems - Page 1082: 14

Answer

(a) $v = 3.66\times 10^7~m/s$ (b) $v = 2.68\times 10^7~m/s$ (c) The particle is an electron.

Work Step by Step

(a) We can convert the energy to units of joules: $7.0~MeV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 1.12\times 10^{-12}~J$ We can find the speed: $\frac{1}{2}mv^2 = 1.12\times 10^{-12}~J$ $v^2 = \frac{(2)(1.12\times 10^{-12}~J)}{m}$ $v = \sqrt{\frac{(2)(1.12\times 10^{-12}~J)}{m}}$ $v = \sqrt{\frac{(2)(1.12\times 10^{-12}~J)}{1.67\times 10^{-27}~kg}}$ $v = 3.66\times 10^7~m/s$ (b) We can convert the energy to units of joules: $15~MeV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 2.4\times 10^{-12}~J$ We can find the speed: $\frac{1}{2}mv^2 = 2.4\times 10^{-12}~J$ $v^2 = \frac{(2)(2.4\times 10^{-12}~J)}{m}$ $v = \sqrt{\frac{(2)(2.4\times 10^{-12}~J)}{m}}$ $v = \sqrt{\frac{(2)(2.4\times 10^{-12}~J)}{(4)(1.67\times 10^{-27}~kg)}}$ $v = 2.68\times 10^7~m/s$ (c) We can convert the energy to units of joules: $1.14~keV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 1.824\times 10^{-16}~J$ We can find the mass: $\frac{1}{2}mv^2 = 1.824\times 10^{-16}~J$ $m = \frac{(2)(1.824\times 10^{-16}~J)}{v^2}$ $m = \frac{(2)(1.824\times 10^{-16}~J)}{(2.0\times 10^{7}~m/s)^2}$ $m = 9.12\times 10^{-31}~kg$ We can see that this mass is very close to the mass of an electron, so we can assume that the particle is an electron.
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