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.($\sin 22.20^{\large\circ}=0.3778$)

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

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
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