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Calculate the wavelength of X-rays, which produces a diffraction angle \small 2\theta equal to \small 16.8^{\circ} for a crystal. Assume first order diffraction with inter particle distance in crystal of 0.2nm.

Option: 1

58.4\times10^{-11}m


Option: 2

4.3\times10^{-11}m


Option: 3

3.7\times10^{-11}m


Option: 4

5.8\times10^{-11}m


Answers (1)

Bragg’s Equation

This equation gives a simple relationship between the wavelength Of X-rays and the distance between the planes in the crystal and the angle Of reflection. This equation can be written as: 

\\\mathrm{n} \lambda=2 \mathrm{d} \sin \theta\\\\ \mathrm{Here} \\\mathrm{n\:=\:Order\:of\:reflection;\:in\:general\:it\:is\:taken\:as\:1.}\\\\\mathrm{\lambda\: =\:Wavelength\:of\:X-rays}\\\\\mathrm{d\: =\:Distance\:between\:two\:layers\:of\:the\:crystals}\\\\\mathrm{\theta\: =\: Angle\:of\:incident\:light}

 

As for a given set of lattice planes the value of 'd' is fixed so the possibility of getting maximum reflection depends only on θ. If we increase θ gradually a number of positions will be observed at which there will be maximum reflection.

Now,

n\lambda=2d\sin\theta


We have given,
n = 1,  d=0.2\times10^{-9}m

Thus, \theta=16.8/2=8.4^{\circ}

\mathrm{Therefore \: \: \lambda=\frac{2\times0.2\times10^{-9}\times\sin 8.4}{1}}
Thus, \lambda=5.84\times10^{-11}m
Therefore, Option(4) is correct

Posted by

Ramraj Saini

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