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 - Problems - Page 800: 42

Answer

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Work Step by Step

a. Calculate the momentum using equation 27–8. $$\lambda=\frac{h}{p}$$ $$p=\frac{h}{\lambda}=\frac{6.626\times10^{-34}J \cdot s }{4.5\times10^{-10}m}$$ $$=1.5\times10^{-24}kg \cdot m/s$$ b. Assume that the electron is nonrelativistic. Find the speed using equation 27–8. $$\lambda=\frac{h}{p}=\frac{h}{mv}$$ $$v=\frac{h}{m\lambda}=\frac{6.626\times10^{-34}J \cdot s }{(9.11\times10^{-31}kg)(4.5\times10^{-10}m)}$$ $$=1.6\times10^{6}m/s$$ Our assumption is valid because v is less than 1 percent of c. c. First, calculate the kinetic energy of the electron. $$KE=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31}kg)( 1.616\times10^{6}m/s)^2 =1.19\times10^{-18}J\approx 7.4 eV$$ To give an electron 7.4 eV of kinetic energy, the electron at rest must be accelerated through a potential difference of 7.4 volts.
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