Question

The second order Bragg diffraction of X-rays with $\lambda = 1.0 \mathring{A}$ from a set of parallel planes in a metal occurs at an angle $60^\circ$. The distance between the scattering planes in the crystals is

• A. $0.575 \mathring{A}$
• B. $1.00 \mathring{A}$
• C. $2.00 \mathring{A}$
• D. $1.17 \mathring{A}$

Answer 1

Answer: (D)

According to Bragg- equation,
$\, \, \, \, \, \, \, \, \, \, \, \, n\lambda$=2d sin $\theta$
$\, \, \, \, \, \, \, \, \, \, \, \, $n=2
$\, \, \, \, \, \, \, \, \, \, \, \, \, lambda=1$
deflected angle$\theta=60^\circ$
$\, \, \, \, \, \, \, \, \, \, \, \, \, d=?$
Distance between two plane of crystal
$\, \, \, \, \, \, \, \, \, \, \, \, \, 2 \times 1 = 2 \times d \times sin 60^\circ$
$\, \, \, \, \, \, \, \, \, \, \, \, \, 2 \times 1 = 2 \times d \times \frac{\sqrt{3}}{2}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, d=\frac{2}{\sqrt{3}}=\frac{2}{1.7}=1.17\mathring{A}$