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 801: 85

Answer

0.82 nm.

Work Step by Step

At such a low energy, the electrons are nonrelativistic. The maximum kinetic energy of the photoelectrons is given by Eq. 27–5b. Use the nonrelativistic equation for momentum and KE. $$KE_{max}= \frac{hc}{\lambda}-W_0=\frac{p^2}{2m}$$ Now use the de Broglie relation. $$ \frac{hc}{\lambda}-W_0=\frac{h^2}{2m\lambda_e^2}$$ The shortest electron wavelength corresponds to the largest momentum and kinetic energy. $$\lambda_e=\frac{h}{\sqrt{2m(\frac{hc}{\lambda}-W_0)}}$$ $$=\frac{6.626 \times10^{-34}J s}{\sqrt{2(9.11\times10^{-31}kg)(\frac{1240 eV nm}{280nm}-2.2eV)(1.60\times10^{-19}J/eV)}}$$ $$=8.2\times10^{-10}m$$
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