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Expert-verified Found in: Page 134 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # 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?

The spacing of atomic planes is $\text{d}=\text{3.33}×{\text{10}}^{\text{-10}}\text{\hspace{0.17em}}m$.

See the step by step solution

## Step 1: 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$\lambda =\text{0.26}×{\text{10}}^{\text{-9}}\text{\hspace{0.17em}}m$ .

## Step 2: Bragg's equation.

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

${\text{2d sinθ =mλ…………(1)}}$

## Step 3: Constructive interference

Constructive interference occurs from two perspectives. Using the equation ,

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

From equation

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

## Step 4: 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,

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

Therefore the atomic plane spacing is $\text{d}=\text{3.33}×{\text{10}}^{\text{-10}}\text{\hspace{0.17em}}m$. ### Want to see more solutions like these? 