Question

An x-ray beam of wavelength A undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is

2323 ^ { \circ }

, and an x-ray beam of wavelength 61 pm undergoes third-order reflection when its angle of incidence to that face is

6060 ^ { \circ }

. Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength A.

Solution

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Answered 2 years ago
Answered 2 years ago
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For Bragg diffraction we know that

2dsinθ=mλ2d\sin\theta=m\lambda

So the interplaner spacing dd is given by

d=mλ2sinθ=3×(61×1012)2×sin(60°)=1.05×1010 md=\frac{m\lambda}{2\sin\theta}=\frac{3\times(61\times 10^{-12})}{2\times \sin(60\text{\textdegree})}=1.05\times10^{-10}\ \mathrm{m}

So the interplanner spacing is 1.05 A˚\text{\AA}.

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