The Solid State - Result Question 29
####29. The second order Bragg diffraction of X-rays with $\lambda=1.00 \AA$ 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
(a) $0.575 \AA$
(b) $1.00 \AA$
(c) $2.00 \AA$
(d) $1.15 \AA$
[1998]
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Solution:
- (d) Given: Order of Bragg diffraction $(n)=2$; Wavelength $(\lambda)=1 \AA$ and angle $(\theta)=60^{\circ}$. We know from the Bragg’s equation $n \lambda=2 d \sin \theta$
or $2 \times 1=2 d \sin 60^{\circ}$
$\Rightarrow 2 \times 1=2 . d . \frac{\sqrt{3}}{2} \Rightarrow d=\frac{2}{\sqrt{3}}=1.15 \AA$
(where $d=$ difference between the scattering planes)
