X-rays of wavelength 0.14 nm are directed toward a crystal.The separation between the planes is 0.512 nm.At what angle,relative to the crystallographic planes does the first order diffraction peak appear?

Question

Quizplus · Accepted Answer

The angle for the first order diffraction peak can be found using Bragg's law, $n\lambda = 2d\sin\theta$, where $n=1$ for first order, $\lambda=0.14$ nm is the wavelength, and $d=0.512$ nm is the separation between planes. Solving for $\theta$ gives $\theta = \sin^{-1}\left(\frac{\lambda}{2d}\right) = \sin^{-1}\left(\frac{0.14}{2*0.512}\right) \approx 7.86^\circ$.

X-Rays of Wavelength 0 ∘\circ∘ B)7 ∘\circ∘ C)9 ∘\circ∘ D)1 ∘\circ∘

X-Rays of Wavelength 0 $\circ$
B)7 $\circ$
C)9 $\circ$
D)1 $\circ$