The glancing angle in a X-ray diffraction is $30^{o}$ and the wavelength of X-rays is used is $20 n m$ . The interplanar spacing of the crystal diffracting these X-rays will be:

40 n m

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20 n m

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15 n m

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10 n m

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Solution

The correct option is B $20 n m$
$2 d s i n Θ = n λ o r d = \frac{n λ}{2 s i n Θ} = \frac{1 \times 20}{2 \times s i n 30^{o}} = 20 n m$

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The glancing angle in a X-ray diffraction is 30o and the wavelength of X-rays is used is 20 nm. The interplanar spacing of the crystal diffracting these X-rays will be:

The correct option is B 20 nm2dsinΘ=nλord=nλ2sinΘ=1×202×sin30o=20nm

The glancing angle in a X-ray diffraction is $30^{o}$ and the wavelength of X-rays is used is $20 n m$ . The interplanar spacing of the crystal diffracting these X-rays will be:

The correct option is B $20 n m$
$2 d s i n Θ = n λ o r d = \frac{n λ}{2 s i n Θ} = \frac{1 \times 20}{2 \times s i n 30^{o}} = 20 n m$