Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 28 - Quantum Mechanics of Atoms - Problems - Page 826: 1

Answer

$x=0.45\mu m$

Work Step by Step

$\lambda=\frac{h}{p}$ $KE=\frac{m_nv^2}{2}=\frac{p^2}{2m_n}$ $p=\sqrt{2m_nKE}$ $\lambda=\frac{h}{\sqrt{2m_nKE}}$ $d\frac{x}{l}=m\lambda$ $x=\frac{l\lambda}{d}=\frac{lh}{d\sqrt{2m_nKE}}=0.45\mu m$
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