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Question: For a crystal, the angle of diffraction ( \({{2\theta }}\) ) is 90 and the second order line has a d...

For a crystal, the angle of diffraction ( 2θ{{2\theta }} ) is 90 and the second order line has a d value of 2.28A02.28\mathop {\text{A}}\limits^0 . The wavelength (in A0\mathop {\text{A}}\limits^0 ) of X-rays used for Bragg’s diffraction is:
A.1.6121.612
B.2.002.00
C.2.282.28
D.4.004.00

Explanation

Solution

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:
nλ=2dsinθ{{n\lambda = 2dsin\theta }}
Here, n is an integer which represents the order of the diffraction, d is the distance between the atomic layers, θ{{\theta }} is the angle of incidence and λ{{\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 2.28A02.28\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, 2θ=90{{2\theta = 90^\circ }} . From this relation, we can calculate the angle of incidence θ{{\theta }}. So, we have:
2θ=90 θ=902 θ=45  {{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=2.28A0{{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,
2×λ=2×2.28×sin45 λ=2.28×sin45 λ=2.28×12 λ=1.612A0  {\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.