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Q17E

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Found in: Page 134

### Modern Physics

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

# Determine the Compton wavelength of the electron, defined to be the wavelength it would have if its momentum were${{\mathbf{\text{m}}}}_{{\mathbf{\text{e}}}}{\mathbf{\text{c}}}$.

Compton Wavelength of the electron

$\lambda =\text{2.43}×{\text{10}}^{\text{-12}}\text{m}$

See the step by step solution

## Step 1:Given and unknowns.

${\text{m}}_{\text{e}}{\text{=9.1×10}}^{\text{-31}}\text{kg --}$ the electron's mass

${\text{p=m}}_{\text{e}}\text{c --}$ the electron's momentum

## Step 2: Concept Introduction

The following equation can be used to describe the de Broglie wavelength.

${\mathbf{\text{p=}}}\frac{\mathbf{\text{h}}}{\mathbf{\text{λ}}}$…………………..(1)

## Step 3: Expression of wavelength.

Know that the electron's momentum is${\text{p=m}}_{\text{e}}\text{c}$ .

Get the wavelength expression as follows

$\text{p=}\frac{\text{h}}{\text{λ}}$

${\text{p=m}}_{\text{e}}\text{c}$

${\text{m}}_{\text{e}}\text{c=}\frac{\text{h}}{\text{λ}}$

$\text{λ=}\frac{\text{h}}{{\text{m}}_{\text{e}}\text{c}}$

## Step 3: Compton Wavelength of Electron.

Using the wavelength's derived expression, Obtain$\lambda$ as:

$\begin{array}{c}\lambda =\frac{\text{h}}{{\text{m}}_{\text{e}}\text{c}}\\ =\frac{\text{6.626}×{\text{10}}^{\text{-34}}\text{\hspace{0.17em}}J\cdot s}{\left(\text{9.1}×{\text{10}}^{\text{-31}}\text{\hspace{0.17em}}kg\right)×\left(\text{3.0}×{\text{10}}^{\text{8}}\text{\hspace{0.17em}}m/s\right)}\\ =\text{2.43}×{\text{10}}^{\text{-12}}\text{m}\end{array}$ $\begin{array}{c}\lambda =\frac{\text{h}}{{\text{m}}_{\text{e}}\text{c}}\\ =\frac{\text{6.626}×{\text{10}}^{\text{-34}}\text{\hspace{0.17em}}J\cdot s}{\left(\text{9.1}×{\text{10}}^{\text{-31}}\text{\hspace{0.17em}}kg\right)×\left(\text{3.0}×{\text{10}}^{\text{8}}\text{\hspace{0.17em}}m/s\right)}\\ =\text{2.43}×{\text{10}}^{\text{-12}}\text{m}\end{array}$

The Compton Wavelength of the electron is$\lambda =\text{2.43}×{\text{10}}^{\text{-12}}\text{m}$ .