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: 40

Answer

21 volts.

Work Step by Step

First assume that the electron is nonrelativistic. Find the wavelength using equation 27–8. $$\lambda=\frac{h}{p}=\frac{h}{mv}$$ $$v=\frac{h}{m\lambda}=\frac{6.626\times10^{-34}J \cdot s }{(9.11\times10^{-31}kg)(0.27\times10^{-9}m)}$$ $$=2.694\times10^{6}m/s=0.0090c$$ Our assumption is valid because v is a tiny fraction of c. Find the kinetic energy of the electron. $$KE=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31}kg)( 2.694\times10^{6}m/s)^2 $$ $$=3.31\times10^{-18}J\approx 21 eV$$ To give an electron a kinetic energy of 21 eV, it must be accelerated through a potential difference of 21 volts.
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