Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Using X-rays of wavelength 154.1pm and starting from the glancing angle,the reflection from silver crystal was found to occur at $\theta=22.20^{\large\circ}$.Calculate the spacing between the planes of Ag atoms that gave rise to the above reflection.($\sin 22.20^{\large\circ}=0.3778$)

$\begin{array}{1 1}204pm\\304pm\\604pm\\704pm\end{array} $

Can you answer this question?

1 Answer

0 votes
Here $\lambda=154.1pm,n=1,\theta=22.20^{\large\circ}$
According to Bragg's equation viz.,
$2d\sin \theta=n\lambda$
$d=\large\frac{n\lambda}{2\sin \theta}=\frac{1\times 154.1}{2\times 0.3778}$pm
answered Jul 31, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App