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Q13E

Expert-verifiedFound in: Page 134

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 $\text{d}=\text{3.33}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}}m$.

The angle of constructive interference${\text{\theta}}_{\text{1}}=\text{23.0}\xb0$.

The angle of constructive interference${\text{\theta}}_{\text{2}}=\text{51.4}\xb0$.

The wavelength of X-rays is$\lambda =\text{0.26}\times {\text{10}}^{\text{-9}}\text{\hspace{0.17em}}m$ .

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

**${\text{2d sin\theta =m\lambda \u2026\u2026\u2026\u2026(1)}}$**

Constructive interference occurs from two perspectives. Using the equation ,

$\begin{array}{c}{\text{2d sin\theta}}_{\text{1}}\text{=m\lambda \u2026\u2026\u2026\u2026(2)}\\ {\text{2d sin\theta}}_{\text{2}}\text{=(m+1)\lambda \u2026\u2026\u2026\u2026(3)}\end{array}$

From equation

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

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

$\begin{array}{l}{\text{2dsin\theta}}_{\text{2}}\text{=(m+1)}\lambda \\ {\text{2dsin\theta}}_{\text{2}}\text{=}\left(\frac{{\text{2dsin\theta}}_{\text{1}}}{\text{\lambda}}\text{+1}\right)\lambda \\ {\text{2dsin\theta}}_{\text{2}}{\text{=2dsin\theta}}_{\text{1}}\text{+}\lambda \\ \text{2d}\left({\text{sin\theta}}_{\text{2}}{\text{-sin\theta}}_{\text{1}}\right)\text{=}\lambda \end{array}$

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

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

$\begin{array}{c}\text{d}=\frac{\text{0.26}\times {\text{10}}^{\text{-9}}\text{\hspace{0.17em}}m}{\text{2}(\text{sin}(\text{51.4}\xb0)-\text{sin}(\text{23.0}\xb0))}\\ =\text{3.33}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}m}\end{array}$

Therefore the atomic plane spacing is $\text{d}=\text{3.33}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}}m$.

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