# Calculate the wavelength of X-rays which produces a diffraction angle $2\theta$ equal to $16.80^{\large\circ}$ for a crystal.Assume first order diffraction with inter particle distance in crystal of 0.2 nm

$\begin{array}{1 1}5.84\times 10^{-11}m\\4.84\times 10^{-12}m\\3.84\times 10^{-11}m\\6.84\times 10^{-11}m\end{array}$

$n\lambda =2d\sin \theta$
Given $n=1$
$d=0.2\times 10^{-9}m$
$\theta=\large\frac{16.80}{2}$$=8.40^{\large\circ}$
Thus $\lambda=\large\frac{2\times 0.2\times 10^{-9}\sin 8.40^{\Large\circ}}{1}$
$\Rightarrow 5.84\times 10^{-11}m$