The second order Bragg diffraction of X-rays with λ=1.0Ao from a set of parallel planes in a metal occurs at an angle of 60°. The distance between the scattering planes in the crystal is

The second order Bragg diffraction of X-rays with $λ = 1 .0 A^{o}$ from a set of parallel planes in a metal occurs at an angle of 60°. The distance between the scattering planes in the crystal is

A
$0 .575 A^{o}$
B
$1 .00 A^{o}$
C
$2 .00 A^{o}$
D
$1 .15 A^{o}$

Solution:

$nλ = 2 dsin θ 2 \times 1 = 2 \times dsin 60 \Rightarrow d = 1 .15 A^{o} ∵ (\sin 60 = \frac{\sqrt{3}}{2})$

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