Q102E A diffracrometer using X-rays wi... [FREE SOLUTION]

Chapter 10: Q102E (page 594)

A diffracrometer using X-rays with a wavelength of \(0.2287\;{\rm{nm}}\) produced first-order diffraction peak for a crystal angle \(\theta = {16.21^\circ }\). Determine the spacing between the diffracting planes in this crystal.

Short Answer

Expert verified

the spacing between crystal planes that diffract X-rays with a wave length of \(0.2287\;{\rm{nm}}\) at an angle of \({16.21^\circ }\) is \(0.4096\;{\rm{nm}}\)

Step by step solution

Define the spacing between crystal planes that diffract X-rays with a wave length ionic radius

Calculate the spacing between crystal planes that diffract X-rays with a wave length of \(1.541\;{\rm{nm}}\) at an angle of \({15.55^\circ }\) by using the following formula:

\(n\lambda = 2d\sin \theta \)

Here \(n\) is an integer, \(\lambda \) is the wave length of \(X\)-ray,\(\theta \) is the angle of diffracted beam and \(\theta \) is the angle of diffracted beam.

Find the spacing between crystal planes that diffract X-rays with a wave length ionic radius

Calculate the spacing between crystal planes that diffract X-rays with a wave length of \(0.2287\;{\rm{nm}}\) at an angle of \({16.21^\circ }\) by using \(n\lambda = 2d\sin \theta \) as follows:

\(\begin{aligned}{}n\lambda &= 2d\sin \theta d\\ &= \frac{{n\lambda }}{{2\sin \theta }}\\& = \frac{{1 \times 0.2287\;{\rm{nm}}}}{{2 \times \sin {{16.21}^\circ }}}\\ &= 0.4096\;{\rm{nm}}\end{aligned}\)