In the figure below, let a beam of x rays of wavelength 0.125 nm be incident on

Question

In the figure below, let a beam of x rays of wavelength 0.125 nm be incident on an NaCl crystal at angle q = 45.0o 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 f 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 f if the crystal is turned clockwise and the (c) smaller and (d) larger value of f if it is turned counterclockwise?

Explanation / Answer

bragg's law

m lamdba = 2 d sin theta

so max at

theta = arcsin( m*0.125/(2*0.252))

={0., 14.3601, 29.7378, 48.0774, 82.7766}

turning clockwise decreases angle so

a) 45-29.7378=15.26

b) 45-14.36=30.64

c) 48.0774-45=3.0774

d) 82.7766-45=37.78

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