Question

An electron moves along the x-axiswith a well-defined momentum of$\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathit{k}\mathit{g}\mathbf{.}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. Write an expression describing the matter wave associated with this electron. Include numerical values where appropriate.

Short answer

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

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

Step-by-step solution

Given.

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

Concept Introduction

The equation of an electromagnetic wave can be expressed as,

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

Calculate the value of k

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

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

Calculate the value of ω

$$\begin{gathered} E=h\omega \\ \omega =\frac{\text{E}}{\text{h}} \\ \omega =\frac{{\displaystyle \raisebox{1ex}{$p^2$}\!\left/ \!\raisebox{-1ex}{$2m$}\right.}}{h} \\ \omega =\frac{{\text{p}}^{\text{2}}}{\text{2mh}} \end{gathered}$$

$\begin{array}{rcl}\omega & =& \frac{{\text{p}}^{\text{2}}}{\text{2mh}}\\ & & \\ & =& \frac{{\displaystyle \text{25}\times {\text{10}}^{\text{- 50}}k{g}^2.\raisebox{1ex}{$m^2$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}{{\displaystyle \text{2}\times \left(\text{9.1}\times {\text{10}}^{\text{- 31}}\text{kg}\right)\times \left(\text{6.63}\times {\text{10}}^{\text{- 34}}kg.\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}}\\ & & \end{array}$

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

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

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