Chapter 47: Problem 33

Question: An x-ray beam, containing radiation of two distinct wavelengths is scattered from a crystal, yielding the intensity spectrum shown in figure 26, chapter 47. The interplane spacing is $d~=~0.94~{\rm nm}$. Determine the wavlengths of the x-rays present in the beam.

Solution: For Bragg scattering we have that $2d\sin\theta~=~m\lambda$, where $\theta$ is measured from the plane surface and NOT the normal to the plane. Giebn that intensity maximium in the scattering are found at $\theta_1~\sim~0.8^o$, $\theta_2\sim1.2^o$, $\theta_3\sim 1.6^o$, and $\theta_4\sim 2.4^o$, we have that for m = 1

$$\lambda_1~=~2 d \sin\theta_1~=~0.0262~{\rm nm}$$

from which gives the peak at $\theta_3$ at m = 2. The we have at m = 1

$$\lambda_2~=~2 d \sin\theta_2~=~0.039~{\rm nm}$$

which gives the peak at $\theta_4$ at m = 2.