Question

The corresponding interplanar spacing is ($inpm$)?

Answer 1

$Wavelength\left(\lambda \right)=229pm\alpha ={23}^0{20}^{\prime }Brag{g}^{\prime }sequation=n\lambda =2d\sin Qn=orderofdefraction=1d=interplanarspacingd=\frac{\lambda }{2\sin Q}=\frac{229}{2\sin {(23}^0{20}^{\prime })}=\frac{229}{2\left(0.396\right)}=289.14pm$