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: 48

Answer

23.0 pm.

Work Step by Step

The KE is much less than the rest energy of an electron. We may use nonrelativistic relationships. Calculate the wavelength using equation 27–8. $$KE=\frac{p^2}{2m}$$ $$p=\sqrt{2m(KE)}$$ $$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$$ $$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(2850eV)(1.60\times10^{-19}J/eV)}}$$ $$=2.30\times10^{-11}m= 23.0\times10^{-12}m $$ As discussed in sections 25-7 and 25-8, the microscope's resolution is about the size of the wavelength of the electron/radiation used to view an object.
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