where (a= 2.125 $A^0$). Consider the case of second order maximum and λ=0.592 $A^0$ -

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 7M

Year: Dec 2012

Bragg's Law:-

Consider a regular arrangement of atoms with inter-planar distance "d". An X-ray beam is incident upon the surface at glancing angle "θ"

Let us assume that the path difference between the scattered rays BC and EF as ∆=nλ thus producing constructive interference.

Now,

∆ = PE + EQ

∴∆ = BE.sin θ + BE.sin θ

∴∆ = 2BE.sin θ

∴∆ = 2d.sin θ

But, ∆ = nλ

∴nλ = 2d.sinθ

This, is the required expression for Bragg's Law

Numerical

Given:-

Miller Indices: (1 0 0) = (h k l)

a = 2.125 $A^0$

λ = 0.592 $A^0$

n = 2 (Second order maxima)

To Find:- Glancing Angle (θ)

Solution:-

We know that,

$d = \frac{a}{\sqrt{h^2 + k^2 + I^2}}$

$d = \frac{2.125}{\sqrt{1^2 + 0^2 + 0}}$

$d = 2.125$

Using Bragg's Law,

∴ nλ = 2d.sin⁡θ

∴2 x 0.592 = 2 x 2.125 x sin⁡θ

$∴θ = 16.176^0$