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: 43

Answer

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Work Step by Step

All 3 energies are much less than the rest energy of an electron. We may use nonrelativistic relationships. $$KE=\frac{p^2}{2m}$$ $$p=\sqrt{2m(KE)}$$ Calculate the wavelength using equation 27–8. $$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$$ a. $$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(10eV)(1.60\times10^{-19}J/eV)}}$$ $$=3.88\times10^{-10}m\approx 4\times10^{-10}m $$ b. $$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(100eV)(1.60\times10^{-19}J/eV)}}$$ $$=1.23\times10^{-10}m\approx 1\times10^{-10}m $$ c. $$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(1.0\times10^3 eV)(1.60\times10^{-19}J/eV)}}$$ $$=3.88\times10^{-11}m\approx 3.9\times10^{-11}m $$
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