Question

In Fig. 36-48, let a beam of x-rays of wavelength 0.125 nm be incident on an NaCl crystal at angle θ = 45.0° to the top face of the crystal and a family of reflecting planes. Let the reflecting planes have separation d = 0.252 nm. The crystal is turned through angle ϕ around an axis perpendicular to the plane of the page until these reflecting planes give diffraction maxima. What are the (a) smaller and (b) larger value of ϕ if the crystal is turned clockwise and the (c) smaller and (d) larger value of ϕ if it is turned counter-clockwise

Short answer

(a)$\varphi =15.26°$ in clockwise for $m=2$.

(b)$\varphi =30.64°$ in clockwise for $m=1$.

(c)$\varphi =3.1°$ in counter-clockwise for $m=3$.

(d)$\varphi =37.8°$ in counter-clockwise for $m=4$.

Step-by-step solution

Bragg’s law

If the incident angle of the wave $\left(\mathbf{\theta }\right)$, measured from the surfaces of the reflecting plane and the radiation's wavelength $\left(\mathbf{\lambda }\right)$, fulfil Bragg's equation, diffraction maxima (caused by constructive interference) is achieved. The Bragg’s equation is given as-

$\mathbf{2}\mathbf{dsin\theta }\mathbf{=}\mathbf{m\lambda }\left(\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}...\right)$

Where$\mathbf{d}$is the interplanar distance and is the order of the maximum intensity.

Given Data

The wavelength of light used: $\lambda =0.125\text{nm}$

Plane separation: $d=0.252\text{nm}$

Angle of incidence: $\theta =45°$

Calculate the value of ϕ for each subpart.

If the crystal is rotated by an angle $\varphi$, the new angle of incident is$\theta \pm \varphi$ depending on the clockwise or counter-clockwise rotation.

Using the values given in the question,

For $m=1$, in clockwise direction, the angle $\theta -\varphi$is

role="math" localid="1663051508257" $$\begin{gathered} \theta -\varphi ={\sin }^{-1}\frac{m\lambda }{2d} \\ 45°-\varphi ={\sin }^{-1}\frac{\left(0.125nm\right)\left(1\right)}{2\left(0.252nm\right)} \\ \varphi =45°-{\sin }^{-1}0.248 \\ \varphi =30.64° \end{gathered}$$

For $m=2$, in clockwise direction, the angle $\theta -\varphi$is

$$\begin{gathered} \theta -\varphi ={\sin }^{-1}\frac{m\lambda }{2d} \\ 45°-\varphi ={\sin }^{-1}\frac{\left(0.125nm\right)\left(2\right)}{2\left(0.252nm\right)} \\ \varphi =45°-{\sin }^{-1}0.496 \\ =15.26° \end{gathered}$$

For $m=3$, in counter-clockwise direction, the angle $\theta +\varphi$is

$$\begin{gathered} \theta +\varphi ={\sin }^{-1}\frac{m\lambda }{2d} \\ 45°+\varphi ={\sin }^{-1}\frac{\left(0.125nm\right)\left(3\right)}{2\left(0.252nm\right)} \\ \varphi ={\sin }^{-1}0.744-45° \\ =3.1° \end{gathered}$$

For $m=4$, in counter-clockwise direction, the angle $\theta +\varphi$is

$$\begin{gathered} \theta +\varphi ={\sin }^{-1}\frac{m\lambda }{2d} \\ 45°+\varphi ={\sin }^{-1}\frac{\left(0.125nm\right)\left(4\right)}{2\left(0.252nm\right)} \\ \varphi ={\sin }^{-1}0.992-45° \\ =37.8° \end{gathered}$$

The smaller and larger values of$\varphi$ for clockwise rotation are$\varphi =15.26°$ and$\varphi =30.64°$ respectively. The smaller and larger values of$\varphi$ for counter-clockwise rotation are$\varphi =3.1°$ and$\varphi =37.8°$ respectively.