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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 # A beam of electrons strikes a crystal at an angle${\mathbf{\text{θ}}}$with the atomic planes, reflects of many atomic planes below the surface, and then passes into a detector also making angle${\mathbf{\text{θ}}}$with the atomic planes. (a) If the minimum${\mathbf{\text{θ}}}$giving constructive interference is.${\mathbf{\text{35°}}}$ What is the ratio${\mathbf{\text{λ/d}}}$, 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 $\frac{\text{λ}}{\text{d}}\text{=1.15}$.

(b) Here can notbe considered as$\text{2}$ . So this is an invalid value of the$\text{sin θ}$.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.

See the step by step solution

## Step 1: Given data.

(a)

The angle of constructive interference${\text{θ}}_{\text{min}}=\text{35}°$.

## Step 2: Bragg's equation.

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

${\mathbf{\text{2d sinθ = mλ………………(1)}}}$

## Step 3: Ratio between wavelength and spacing

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

$\begin{array}{l}\text{2dsinθ=mλ}\\ \text{2sinθ=}\frac{\text{mλ}}{\text{d}}\\ \frac{\text{λ}}{\text{d}}\text{=}\frac{\text{2sinθ}}{\text{m}}\end{array}$

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

$\begin{array}{c}\frac{\text{λ}}{\text{d}}\text{=}\frac{\text{2}×\text{sinθ}}{\text{m}}\\ \text{=}\frac{\text{2}×\text{sin}\left(\text{35}°\right)}{\text{1}}\\ \text{=1.15}\end{array}$

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

## Step 4: Condition for the constructive interference(b)

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

So,

$\begin{array}{l}\text{2dsin θ=mλ}\\ \text{sin θ=}\frac{\text{mλ}}{\text{2d}}\\ \text{1}\ge \frac{\text{mλ}}{\text{2d}}\\ \text{m}\le \frac{\text{2}}{\frac{\text{λ}}{\text{d}}}\end{array}$

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

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

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

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. ### Want to see more solutions like these? 