Calculate the wavelength of X-rays, which produces a diffraction angle $\dpi{100} \small 2\theta$ equal to $\dpi{100} \small 16.8^{\circ}$ for a crystal. Assume first order diffraction with inter particle distance in crystal of 0.2nm.Option: 1 Option: 2 Option: 3 Option: 4

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