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: 68

Answer

$1.2\times10^{-10}m$.

Work Step by Step

The spacing between planes, d, for the first-order peaks is given in equation 25–10, $\lambda=2d sin\theta$. Calculate the wavelength of the electrons from the kinetic energy. The KE is much less than the rest energy of an electron. We may use nonrelativistic relationships. Use equation 27–8. $$KE=\frac{p^2}{2m}$$ $$p=\sqrt{2m(KE)}$$ $$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$$ $$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(72eV)(1.60\times10^{-19}J/eV)}}$$ $$=1.446\times10^{-10}m $$ Now find the spacing. $$d=\frac{\lambda}{2 sin \theta}=\frac{1.446\times10^{-10}m }{2sin38^{\circ}}=1.2\times10^{-10}m $$
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