# How close to the wavelength of light have to be to inter-planar spacing for diffraction to occur?

I'm doing past papers for my exam on Wednesday, and I've come to a question I'm stuck on.

So, you have x rays of $\lambda = 0.25\,\mathrm{nm}$ incident on a cubic unit cell with lattice lattice constant $a = 0.286\,\mathrm{nm}$.

The miller indicies are (101), (022), (103) and (022).

We're supposed to say which of these will give Bragg diffraction.

So, I figure calculate inter-planar spacing for these, using $d=\frac{a}{\sqrt{(h^2+k^2+l^2)}}$ but this gives answers smaller than $\lambda$.

I had thought lambda would have to be smaller to give diffraction

Thanks

For Bragg scattering to occur, it has to satisfy n $\lambda$=2dsin$\theta$. n can be 1,2,3... For (101), 2d=0.404, For (022), 2d=0.202, For (103), 2d=0.181. So Bragg diffraction can occur on (101) plane but not the other two. The formula for calculating spacing is a/sqrt(h^2+k^2+l^2)