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  • Which of the following relation is correct for first order Bragg’s diffraction?Option: 1 <img alt="\sin \Theta =\frac{2a}{\lambda }(h^2+k^2+l^2

Which of the following relation is correct for first order Bragg’s diffraction?

Option: 1

\sin \Theta =\frac{2a}{\lambda }(h^2+k^2+l^2)


Option: 2

\sin \Theta =\frac{2a}{\lambda }(h^2+k^2+l^2)^\frac{1}{2}


Option: 3

\sin \Theta =\frac{\lambda}{2a }(h^2+k^2+l^2)^\frac{1}{2}


Option: 4

\sin \Theta =\frac{\lambda}{2a }(h^2+k^2+l^2)^2


Answers (1)

best_answer

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

Thus, d=\frac{\lambda}{2\sin\Theta }.........................(i)

Again, we have

d=\frac{a}{\sqrt{h^2+k^2+l^2}}.........................(ii)

On combining both these equations, we have

\frac{\lambda}{2\sin\Theta }=\frac{a}{\sqrt{h^2+k^2+l^2}}

Thus, \sin \Theta =\frac{\lambda}{2a}(h^2+k^2+l^2)^\frac{1}{2}

Therefore, Option(3) is correct.

Posted by

Anam Khan

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