The second order Bragg's diffraction of rays with $λ$ = 1A from a set of parallel planes in a metal occurs at an angle of $60^{\circ}$ The distance between the scattering planes in the crystal is

0.575

\circ A

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1.00

\circ A

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2.00

\circ A

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1.15

\circ A

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Solution

The correct option is D 1.15 $\circ A$
$2 d s i n θ = n λ o r 2 \times d \times s i n 60^{\circ} = 2 \times 1 \circ A$
or, $s i n θ = n λ o r 2 \times d \times 0.8660 = 2$
or, $d = 1.15 \circ A (s i n 60^{\circ} = \frac{\sqrt{3}}{2} o r 0.8660)$

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The second order Bragg's diffraction of rays with λ = 1A 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 correct option is D 1.15 ∘A2d sinθ=nλ or 2×d×sin 60∘=2×1∘A or, sinθ=nλ or 2×d×0.8660=2 or, d=1.15∘A(sin 60∘=√32 or 0.8660)

The second order Bragg's diffraction of rays with $λ$ = 1A from a set of parallel planes in a metal occurs at an angle of $60^{\circ}$ The distance between the scattering planes in the crystal is

The correct option is D 1.15 $\circ A$
$2 d s i n θ = n λ o r 2 \times d \times s i n 60^{\circ} = 2 \times 1 \circ A$
or, $s i n θ = n λ o r 2 \times d \times 0.8660 = 2$
or, $d = 1.15 \circ A (s i n 60^{\circ} = \frac{\sqrt{3}}{2} o r 0.8660)$