Physics: Principles with Applications (7th Edition)

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

Chapter 27 - Early Quantum Theory and Models of the Atom - General Problems - Page 802: 93

Answer

$1.8\times10^{11}C/kg$.

Work Step by Step

After the acceleration, the electron has a kinetic energy of 96 eV. Find the speed. Assume that the electron is nonrelativistic. $$KE=\frac{1}{2}mv^2$$ $$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(96eV)(1.60\times10^{-19}J/eV)}{9.11\times10^{-31}kg}}$$ $$=5.807\times10^6 m/s$$ The electron is not relativistic. In Example 20-6, the radius of the circular path is related to the mass, speed, charge, and the magnetic field strength. Solve for the charge-to-mass ratio. $$r=\frac{mv}{qB}$$ $$\frac{q}{m}=\frac{v}{Br}$$ $$=\frac{5.807\times10^6 m/s }{(3.67\times10^{-4}T)(0.09m)}$$ $$= 1.8\times10^{11}C/kg $$
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