Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 11 - Intermolecular Forces and Liquids and Solids - Questions & Problems: 11.47

Answer

Please see the work below.

Work Step by Step

We know that $n\lambda=2dsin\theta$ This can be rearranged as $d=\frac{n\lambda}{2sin\theta}$ $d=\frac{(1)(0.090nm)}{2sin15.2^{\circ}}=0.17nm$
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