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Q14E

Expert-verifiedFound in: Page 134

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**A beam of electrons strikes a crystal at an angle${\mathbf{\text{\theta}}}$****with the atomic planes, reflects of many atomic planes below the surface, and then passes into a detector also making angle${\mathbf{\text{\theta}}}$****with the atomic planes. (a) If the minimum${\mathbf{\text{\theta}}}$****giving constructive interference is****.${\mathbf{\text{35\xb0}}}$ What is the ratio${\mathbf{\text{\lambda /d}}}$****, Where is the spacing between atomic planes? (b) At what other angles, if any, would constructive interference occur?**

(a) The ratio of given minimum angle is $\frac{\text{\lambda}}{\text{d}}\text{=1.15}$.

(b) Here can notbe considered as$\text{2}$ . So this is an invalid value of the$\text{sin \theta}$.As a result, the only integer that exists is$\text{m=1}$ .This is the smallest angle at which constructive interference was detected. As a result, constructive interference is not visible from any other angles.

**(a)**

The angle of constructive interference${\text{\theta}}_{\text{min}}=\text{35}\xb0$.

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

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

Using equation$\text{(1)}$ .

$\begin{array}{l}\text{2dsin\theta =m\lambda}\\ \text{2sin\theta =}\frac{\text{m\lambda}}{\text{d}}\\ \frac{\text{\lambda}}{\text{d}}\text{=}\frac{\text{2sin\theta}}{\text{m}}\end{array}$

${\text{\theta}}_{\text{min}}$ is the minimum angle for the constructive interference, so$\text{m=1}$ .

$\begin{array}{c}\frac{\text{\lambda}}{\text{d}}\text{=}\frac{\text{2}\times \text{sin\theta}}{\text{m}}\\ \text{=}\frac{\text{2}\times \text{sin}(\text{35}\xb0)}{\text{1}}\\ \text{=1.15}\end{array}$

Therefore the ratio of$\frac{\text{\lambda}}{\text{d}}$ is $\text{1.15}$in the spacing between the atomic planes.

From Equation$\text{(1)}$ the minimum value for $\text{sin\theta}$is$\text{1}$ .

So,

$\begin{array}{l}\text{2dsin \theta =m\lambda}\\ \text{sin \theta =}\frac{\text{m\lambda}}{\text{2d}}\\ \text{1}\ge \frac{\text{m\lambda}}{\text{2d}}\\ \text{m}\le \frac{\text{2}}{\frac{\text{\lambda}}{\text{d}}}\end{array}$

Insert$\frac{\text{\lambda}}{\text{d}}\text{=1.15}$

$\begin{array}{c}\text{m}\le \frac{\text{2}}{\frac{\text{\lambda}}{\text{d}}}\\ \le \frac{\text{2}}{\text{1.15}}\\ \text{=1.74}\end{array}$

Here m can notbe considered as $\text{2}$. So this is an invalid value of the.$\text{sin \theta}$

As a result, the only integer that exists is$\text{m=1}$ .

This is the smallest angle at which constructive interference was detected. As a result, constructive interference is not visible from any other angles.

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