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Calculate the angle at which second-order reflection will occur in an X-ray spectrometer when X-rays of 0.154nm are diffracted by the atoms of a crystal having distance b/w two planes at 404 pm

$\begin{array}{1 1}22.41^{\large\circ}\\32.41^{\large\circ}\\42.41^{\large\circ}\\12.41^{\large\circ}\end{array} $

1 Answer

For $n^{th}$ order diffraction
$n\lambda=2a\sin \theta$
$\sin \theta=\large\frac{n\lambda}{2a}$
Given : n=2,$\lambda=0.154nm=0.154\times 10^{-9}m,a=404pm=404\times 10^{-12}m$
$\sin \theta=\large\frac{2\times 0.154 \times 10^{-9}}{2\times 404\times 10^{-12}}$
$\sin \theta=0.3812$
$\therefore \theta=22.41^{\large\circ}$
answered Jul 31, 2014 by sreemathi.v

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