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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Solid State
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X – Rays of wavelength $1.54A^{\large\circ}$ strike a crystal and are observed to be deflected at an angel $22.5^{\large\circ}$. Assuming that n = 1, calculate the spacing between the planes of atom that are responsible for this reflection.

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Applying Bragg’s equation
$n\lambda=2d\sin\theta$
Giving $n=1,\lambda=1.54A^{\large\circ},\theta=22.5^{\large\circ}$
Using relation $n\lambda=2d\sin \theta$
$d=\large\frac{1.54}{2\sin 22.5^{\large\circ}}=\frac{1.54}{2\times 0.383}$$=2.01A^{\large\circ}$
Hence (B) is the correct answer.
answered Jun 2, 2014 by sreemathi.v
 

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