X-rays of wavelength 1-54-A strike a crystal and are observed to be deflected at a angle of 22-5- Assuming that n - 1- calculate the spacing between the planes of atoms that are responsible for this reflection-

Applying Bragg's equation, nλ = 2dsinθ Given, n = 1, λ = 1.54∘A, θ = 22.5^∘ Using relation nλ = 2d sinθ , d = 1.542sin 22.5^∘ = 1.542× 0.383 = ...

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Standard XII

Physics

Single Slit Diffraction

X-rays of wav...

Question

X-rays of wavelength $1.54 \circ A$ strike a crystal and are observed to be deflected at a angle of ${22.5}^{\circ}$ . Assuming that $n = 1$ , calculate the spacing between the planes of atoms that are responsible for this reflection.

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Solution

Applying Bragg's equation,
$n λ = 2 d sin θ$
Given, $n = 1, λ = 1.54 \circ A, θ = {22.5}^{\circ}$
Using relation $n λ = 2 d sin θ$ ,
$d = \frac{1.54}{2 sin {22.5}^{\circ}} = \frac{1.54}{2 \times 0.383} = 2.01 \circ A$ .

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X-rays of wavelength 1.54∘A strike a crystal and are observed to be deflected at a angle of 22.5∘. Assuming that n=1, calculate the spacing between the planes of atoms that are responsible for this reflection.

Applying Bragg's equation,nλ=2dsinθGiven, n=1,λ=1.54∘A,θ=22.5∘Using relation nλ=2dsinθ,d=1.542sin22.5∘=1.542×0.383=2.01∘A.

X-rays of wavelength $1.54 \circ A$ strike a crystal and are observed to be deflected at a angle of ${22.5}^{\circ}$ . Assuming that $n = 1$ , calculate the spacing between the planes of atoms that are responsible for this reflection.

Applying Bragg's equation,
$n λ = 2 d sin θ$
Given, $n = 1, λ = 1.54 \circ A, θ = {22.5}^{\circ}$
Using relation $n λ = 2 d sin θ$ ,
$d = \frac{1.54}{2 sin {22.5}^{\circ}} = \frac{1.54}{2 \times 0.383} = 2.01 \circ A$ .