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The second order Bragg diffraction of X-rays with ={1.0}Å from

Question

The second order Bragg diffraction of X-rays with $λ = 1.0 \overset{˚}{A}$ 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

0.575 \overset{˚}{A}

1.00 \overset{˚}{A}

2.00 \overset{˚}{A}

1.17 \overset{˚}{A}

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The second order Bragg diffraction of X-rays with

λ = 1.0 \overset{˚}{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

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Text solutionVerified

According to Bragg's equation,

n λ = 2 d sin θ

n = 2

λ = 1

deflected angle

θ = 60^{\circ}

d = ? ltBrgt Distance between two plane of crystal

2 \times 1 = 2 \times d \times sin 60^{\circ}

2 \times 1 = 2 \times d \times \frac{3}{2}

d = \frac{2}{3} = \frac{2}{1.7} = 1.17 \overset{˚}{A}

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Question Text	The second order Bragg diffraction of X-rays with $λ = 1.0 \overset{˚}{A}$ 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
Answer Type	Text solution:1
Upvotes	150