Question

The setup depicted in Figure$\mathbf{\text{4.6}}$is used in a diffraction experiment using X-rays of$\mathbf{\text{0.26\hspace{0.17em}nm}}$wavelength. Constructive interference is noticed at angles of${\mathbf{\text{23.0}}}^{\mathbf{\text{o}}}$and,${\mathbf{\text{51.4}}}^{\mathbf{\text{o}}}$but none between. What is the spacing$\mathbf{d}$of atomic planes?

Short answer

The spacing of atomic planes is $\text{d}=\text{3.33}\times {\text{10}}^{\text{-10}}m$.

Step-by-step solution

Given Data.

The angle of constructive interference${\text{θ}}_{\text{1}}=\text{23.0}°$.

The angle of constructive interference${\text{θ}}_{\text{2}}=\text{51.4}°$.

The wavelength of X-rays is$\lambda =\text{0.26}\times {\text{10}}^{\text{-9}}m$ .

Bragg's equation.

When there is constructive interference, Bragg's equation can be used to describe the relationship between the quantities.

$\text{2dsinθ=mλ…………(1)}$

Constructive interference

Constructive interference occurs from two perspectives. Using the equation ,

$\begin{array}{c}{\text{2dsinθ}}_{\text{1}}\text{=mλ…………(2)}\\ {\text{2dsinθ}}_{\text{2}}\text{=(m+1)λ…………(3)}\end{array}$

From equation

$\begin{array}{c}{\text{2dsinθ}}_{\text{1}}\text{=mλ}\\ \text{m=}\frac{{\text{2dsinθ}}_{\text{1}}}{\lambda }\end{array}$

The Atomic plane spacing.

Substitute$\text{m}$ value in the equation.

$\begin{array}{l}{\text{2dsinθ}}_{\text{2}}\text{=(m+1)}\lambda \\ {\text{2dsinθ}}_{\text{2}}\text{=}\left(\frac{{\text{2dsinθ}}_{\text{1}}}{\text{λ}}\text{+1}\right)\lambda \\ {\text{2dsinθ}}_{\text{2}}{\text{=2dsinθ}}_{\text{1}}\text{+}\lambda \\ \text{2d}\left({\text{sinθ}}_{\text{2}}{\text{-sinθ}}_{\text{1}}\right)\text{=}\lambda \end{array}$

$\text{d=}\frac{\lambda }{\text{2}\left({\text{sinθ}}_{\text{2}}{\text{-sinθ}}_{\text{1}}\right)}$………………(4)

Therefore, using equation (4), we get the interplanar spacing such that,

$\begin{array}{c}\text{d}=\frac{\text{0.26}\times {\text{10}}^{\text{-9}}m}{\text{2}(\text{sin}(\text{51.4}°)-\text{sin}(\text{23.0}°))}\\ =\text{3.33}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}m}\end{array}$

Therefore the atomic plane spacing is $\text{d}=\text{3.33}\times {\text{10}}^{\text{-10}}m$.