Question

For a crystal, the angle of diffraction ( ${{2\theta }}$ ) is 90 and the second order line has a d value of $2.28\mathop {\text{A}}\limits^0 $ . The wavelength (in $\mathop {\text{A}}\limits^0 $ ) of X-rays used for Bragg’s diffraction is:
A.$1.612$
B.$2.00$
C.$2.28$
D.$4.00$

Answer 1

Hint:Bragg’s diffraction takes place when a subatomic particle or electromagnetic radiation waves possess wavelengths that are comparable to atomic spacing in a crystal lattice. The relationship between the incident wavelength and the angle of diffraction is given by the Bragg’s equation:
${{n\lambda = 2dsin\theta }}$
Here, n is an integer which represents the order of the diffraction, d is the distance between the atomic layers, ${{\theta }}$ is the angle of incidence and ${{\lambda }}$ is the wavelength of the incident X-ray beam.

Complete step by step answer:
We are given that the angle of diffraction for a crystal lattice is 90 degrees. Also given that the diffraction is of second order and that the second order line has a value of $2.28\mathop {\text{A}}\limits^0 $ . We need to find out the wavelength of the X – rays used for this Bragg's diffraction in units of Angstrom.
Thus, according to the given question, ${{2\theta = 90^\circ }}$ . From this relation, we can calculate the angle of incidence ${{\theta }}$. So, we have:
$
  {{2\theta = 90^\circ }} \\
   \Rightarrow {{\theta = }}\dfrac{{{{90^\circ }}}}{2} \\
   \Rightarrow {{\theta = 45^\circ }} \\
 $
Also, for second order diffraction, the value of ‘n’ is equal to 2. The distance between the atomic layers is ${{d}} = 2.28\mathop {\text{A}}\limits^0 $ . We can find out the wavelength of the X – rays used for this Bragg's diffraction by substituting all these values in Bragg's equation. Thus,
$
  {\text{2}} \times {{\lambda = 2}} \times {\text{2}}{\text{.28}} \times {\text{sin45}}^\circ \\
   \Rightarrow {{\lambda = 2}}{\text{.28}} \times {\text{sin45}}^\circ \\
   \Rightarrow {{\lambda = 2}}{\text{.28}} \times \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow {{\lambda = 1}}{\text{.612}}\mathop {\text{A}}\limits^{\text{0}} \\
 $

So, the correct option is A.

Note:
Bragg’s law of diffraction has numerous applications in the field of science. One of them is that it is applicable in the case of X-ray fluorescence spectroscopy or wavelength dispersive spectrometry where crystals of known spacings are used as analyzing crystals in the spectrometer.Thus from Bragg’s law, the following concluding ideas are obtained: the diffraction has three parameters which are the wavelength, the crystal orientation and the spacing of the crystal planes.