Question

The second order bragg diffraction of X rays with $\lambda = 1.00{A^0}$ from a set of parallel planes in a metal occurs at an angle of ${60^ \circ }$ . The distance between the scattering planes in the crystal is:
A.O.577
B.1.00
C.2.00
D.1.15

Answer 1

Hint: Bragg’s law gives the angles for coherent and incoherent scattering from a crystal lattice. When X-rays are incident on an atom, they make the electronic cloud move, as does any electromagnetic wave.
Formula used:
 $2d\sin \theta = n\lambda $
Where,
  $\theta $ Is the angle of diffraction
n is the order of diffraction
 $\lambda $ Is the wavelength of X rays

Complete step by step answer:
Bragg’s law is a special case or Laue diffraction. It gives the angles for coherent and incoherent scattering from a crystal lattice. Moreover, the movement in charges re-radiates the waves with the same frequency. This phenomenon is further known as Reighleigh’s scattering.
Moreover, a similar process occurs upon scattering the neutron waves from the nuclei or by a coherent spin interaction with the unpaired electron. Further, these re-emitted wave fields interfere with each other either constructively or destructively and hence, producing a diffraction pattern on the film. So, the resulting wave interference pattern is the basis of diffraction analysis and this was further known as Bragg’s diffraction.
Now, using the above formula, we can find out the value of d i.e. distance
By applying the formula,
We get,
$\Rightarrow 2d\sin 60 = 2 \times 1{A^ \circ }$
$\Rightarrow \sin 60 \circ = \dfrac{{\sqrt 3 }}{2}$
Therefore, d= $1.15{A^\circ }$

Hence, option D is correct.

Note:
X ray diffraction is one of the earliest methods for studying the structure of solids. In the process of diffraction, electromagnetic waves of a given frequency but different phases interact to produce constructive interference (bright spots on the film exposed to the light) and destructive interference (dark spots).