Solid State Question 277
Question: The second order Bragg diffraction of X-rays with $ =1.00\overset{o}{\mathop{A}} $ from a set of parallel planes in a metal occurs at an angle $ 60{}^\circ $ . The distance between the scattering planes in the crystal is
Options:
A) $ 0.575\overset{o}{\mathop{A}} $
B) $ 1.00\overset{o}{\mathop{A}} $
C) $ 2.00\overset{o}{\mathop{A}} $
D) $ 1.15\overset{o}{\mathop{A}} $
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Answer:
Correct Answer: D
Solution:
[d] Order of Bragg diffraction (n) = 2: Wavelength $ ( \lambda ~ )=1\overset{o}{\mathop{A}} $ and angle $ ( \theta )=60{}^\circ $ . We know from the Bragg’s equation $ n\lambda =2d\sin \theta $ Or $ 2\times 1=2dsin{{60}^{{}^\circ }} $
$ \Rightarrow 2\times 1=2.d\frac{\sqrt{3}}{2}\Rightarrow d=\frac{2}{\sqrt{3}}=1.15\overset{o}{\mathop{A}} $ (where d = Distance between the scattering planes)
BETA
