logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
$\;\;\;=204$pm
answered Jul 31, 2014 by sreemathi.v
 

Related questions

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