The diffraction of a crystal of Barium with X-rays of wavelength 2-29 angstroms gives a first order reflection at 27-8- - What is the distance between the diffracting planes In angstroms- Sin 27-8-0-4561

The correct option is B 2.512d Sin θ = nλ d=n λ/2Sin θ=1 × 2.29/2 × 0.4561=2.51 langstroms ...

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Standard XII

Chemistry

First Order Reaction

The diffracti...

Question

The diffraction of a crystal of Barium with X-rays of wavelength 2.29angstroms gives a first order reflection at ${27.8}^{\circ}$ . What is the distance between the diffracting planes (In angstroms)? (Sin 27.8 = 0.4561)

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2.51

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Solution

The correct option is C 2.51
$2 d S i n θ = n λ$
$d = \frac{n λ}{2 S i n θ} = \frac{1 \times 2.29}{2 \times 0.4561} = 2.51 l a n g s t r o m s$

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The diffraction of a crystal of Barium with X-rays of wavelength 2.29angstroms gives a first order reflection at 27.8∘. What is the distance between the diffracting planes (In angstroms)? (Sin 27.8 = 0.4561)

The correct option is C 2.512d Sinθ=nλ d=nλ2Sinθ=1×2.292×0.4561=2.51 langstroms

The diffraction of a crystal of Barium with X-rays of wavelength 2.29angstroms gives a first order reflection at ${27.8}^{\circ}$ . What is the distance between the diffracting planes (In angstroms)? (Sin 27.8 = 0.4561)

The correct option is C 2.51
$2 d S i n θ = n λ$
$d = \frac{n λ}{2 S i n θ} = \frac{1 \times 2.29}{2 \times 0.4561} = 2.51 l a n g s t r o m s$