Question

What is the spacing between crystal planes that diffract X-rays with a wavelength of 1.541 nm at an angle \(\theta \) of 15.55⁰ (first order reflection)?

Short answer

The spacing between crystal planes that diffract X-rays with a wave length of \(1.541\;{\rm{nm}}\) at an angle of \({15.55^\circ }\) is \(2.875\;{\rm{nm}}\).

Step-by-step solution

Define the 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 ionic radius of NaH

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 \(n\lambda = 2d\sin \theta \) as follows:

\(\begin{aligned}{}\lambda & = 2d\sin \theta d\\ &= \frac{{n\lambda }}{{2\sin \theta }}\\ &= \frac{{1 \times 1.541\;{\rm{nm}}}}{{2\sin {{15.55}^\circ }}}\\ &= 2.875\;{\rm{nm}}\end{aligned}\)