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Q36E

Expert-verifiedFound in: Page 136

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 ****${\mathbf{5}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{25}}{\mathit{k}}{\mathit{g}}{\mathbf{.}}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. Write an expression describing the matter wave associated with this electron. Include numerical values where appropriate.**

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

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

The momentum of the electron is $p=5\times {10}^{-25}kg\xb7m/s$

**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)**

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

$\begin{array}{rcl}p& =& hk\\ k& =& \frac{p}{h}\\ k& =& \frac{5\times {10}^{-25}kg\xb7m/s}{6.63\times {10}^{-34}kg\xb7{m}^{2}/s}\\ k& =& 7.54\times {10}^{8}{m}^{-1}\dots \dots \dots \dots \dots \dots ..\left(2\right)\end{array}$

$E=h\omega \phantom{\rule{0ex}{0ex}}\omega =\frac{\text{E}}{\text{h}}\phantom{\rule{0ex}{0ex}}\omega =\frac{\raisebox{1ex}{${p}^{2}$}\!\left/ \!\raisebox{-1ex}{$2m$}\right.}{h}\phantom{\rule{0ex}{0ex}}\omega =\frac{{\text{p}}^{\text{2}}}{\text{2mh}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

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

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