Answer
Please see the work below.
Work Step by Step
We know that
$n\lambda=2dsin\theta$
This can be rearranged as
$d=\frac{n\lambda}{2sin\theta}$
$d=\frac{(1)(0.090nm)}{2sin15.2^{\circ}}=0.17nm$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
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 - Page 511: 11.47
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