Question

# In Fig. $$36-48,$$ let a beam of x rays of wavelength $$0.125 \mathrm{nm}$$ be incident on an $$\mathrm{NaCl}$$ crystal at angle $$\theta=45.0^{\circ}$$ to the top face of the crystal and a family of reflecting planes. Let the reflecting planes have separation $$d=0.252 \mathrm{nm} .$$ The crystal is turned through angle $$\phi$$ 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 $$\phi$$ if the crystal is turned clockwise and the(c) smaller and(d) larger value of $$\phi$$ if it is turned counterclockwise?

Solution

## We want the reflections to obey the Bragg condition $$2 d \sin \theta=m \lambda$$, where $$\theta$$ is the angle between the incoming rays and the reflecting planes, $$\lambda$$ is the wavelength, and $$m$$ is an integer. We solve for $$\theta$$$$\theta=\sin ^{-1}\left(\dfrac{m \lambda}{2 d}\right)=\sin ^{-1}\left(\dfrac{\left(0.125 \times 10^{-9} \mathrm{m}\right) m}{2\left(0.252 \times 10^{-9} \mathrm{m}\right)}\right)=0.2480 \mathrm{m}$$(a) For $$m=2$$ the above equation gives $$\theta=29.7^{\circ} .$$ The crystal should be turned$$\phi=45^{\circ}-29.7^{\circ}=15.3^{\circ} \text { clockwise }$$(b) For $$m=1$$ the above equation gives $$\theta=14.4^{\circ} .$$ The crystal should be turned$$\phi=45^{\circ}-14.4^{\circ}=30.6^{\circ} \text { clockwise }$$(c) For $$m=3$$ the above equation gives $$\theta=48.1^{\circ} .$$ The crystal should be turned$$\phi=48.1^{\circ}-45^{\circ}=3.1^{\circ} \text { counterclockwise. }$$(d) For $$m=4$$ the above equation gives $$\theta=82.8^{\circ} .$$ The crystal should be turned$$\phi=82.8^{\circ}-45^{\circ}=37.8^{\circ} \text { counterclockwise }$$Note that there are no intensity maxima for $$m>4,$$ as one can verify by noting that $$m \lambda /2 d$$ is greater than 1 for $$m$$ greater than 4Physics

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