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Expert-verified Found in: Page 136 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # An electron moves along the x-axis with a well-defined momentum of ${\mathbf{5}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{25}}{\mathbit{k}}{\mathbit{g}}{\mathbf{.}}\mathbf{m}}{\mathbf{s}}$. Write an expression describing the matter wave associated with this electron. Include numerical values where appropriate.

A statement that describes the matter wave that this electron is linked with

$\psi \left(x,t\right)=Aexp\left(\left(7.54×{10}^{8}{m}^{-1}\right)x-\left(2.07×{10}^{15}{a}^{-1}\right)t\right)$

See the step by step solution

## Step 1: Given.

The momentum of the electron is $p=5×{10}^{-25}kg·m/s$

## Step 2: Concept Introduction

The equation of an electromagnetic wave can be expressed as,

${\mathbit{\psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{A}}{{\mathbit{e}}}^{\mathbf{i}\mathbf{\left(}\mathbf{k}\mathbf{x}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{\right)}}$ …………(1)

## Step 3: Calculate the value of k

As a result, we must determine the values of k and w .

$\begin{array}{rcl}p& =& hk\\ k& =& \frac{p}{h}\\ k& =& \frac{5×{10}^{-25}kg·m/s}{6.63×{10}^{-34}kg·{m}^{2}/s}\\ k& =& 7.54×{10}^{8}{m}^{-1}\dots \dots \dots \dots \dots \dots ..\left(2\right)\end{array}$

## Step 3: Calculate the value of ω

$\omega =\text{2.07}×{\text{10}}^{\text{15}}{s}^{-1}..............\left(1\right)$

Substitute values from equations (2) and (3) into equation (1), and we get,

$\psi \left(x,t\right)=Aexp\left(\left(7.54×{10}^{8}{m}^{-1}\right)x-\left(2.07×{10}^{15}{s}^{-1}\right)t\right)$. ### Want to see more solutions like these? 