The second order Bragg diffrction of $X - r a y s$ with $λ = 1.00 A^{\circ}$ 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

2.00 A^{\circ}

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1.00 A^{\circ}

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0.575 A^{\circ}

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1.15 A^{\circ}

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Solution

The correct option is D $1.15 A^{\circ}$
According to Bragg's equation.
$n λ = 2 d sin θ$
As. $n = 2, λ = 1.00 A^{\circ}, θ = 60^{\circ}, d = ?$
$\Rightarrow d = \frac{1}{sin 60^{\circ}} = \frac{2}{\sqrt{3}} = 1.15 A^{\circ}$

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The second order Bragg diffrction of X−rays with λ=1.00A∘ from a set of parallel planes in a metal occurs at an angle 60∘. The distance between the scattering planes in the crystal is

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The second order Bragg diffrction of $X - r a y s$ with $λ = 1.00 A^{\circ}$ 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

The correct option is D $1.15 A^{\circ}$
According to Bragg's equation.
$n λ = 2 d sin θ$
As. $n = 2, λ = 1.00 A^{\circ}, θ = 60^{\circ}, d = ?$
$\Rightarrow d = \frac{1}{sin 60^{\circ}} = \frac{2}{\sqrt{3}} = 1.15 A^{\circ}$