Answer
$ 0.22 nm$
Work Step by Step
Given, $n = 1$
$\theta = 23^{\circ}$
$d = 282 pm = 0.282 nm$
The wavelength of X-rays is
$2dsin\theta = n\lambda$
$\lambda = \frac{2dsin\theta}{n}= \frac{2 * 0.282nm (sin 23^{\circ})}{1} = 0.22 nm$
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: 11.48
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