Question

A free particle is represented by the plane wave function$\mathit{\psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}\mathit{e}\mathit{x}\mathit{p}\mathbf{\left[}\mathbf{i}\left(1.58\times {10}^{12}x-7.91\times {10}^{16}t\right)\mathbf{\right]}$where SIunits are understood. What are the particle’s momentum, Kinetic energy, and mass? (Note: In nonrelativistic quantum mechanics, we ignore mass/internal energy, so the frequency is related to kinetic energy alone.)

Short answer

Particle’s Momentum of Plane Wave Function 1.66x10^-22 kg^.m/s

Particle’s Momentum of Kinetic Energy 8.30x10^-18 J

Particle’s Momentum of Mass 1.66x10^-27s

Step-by-step solution

Given information

The given wavefunction is$\psi (x,t)=Aexp\left[\mathrm{i}(1.58\times {10}^{12}\mathrm{x}-7.91\times {10}^{16}\mathrm{t})\right]$

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)

Calculations for momentum

Compare the given wavefunction with equation (1), and we get

$$\begin{gathered} k=1.58\times {10}^{12}{m}^{-1} \\ \omega =7.91\times {10}^{16}{s}^{-1} \end{gathered}$$

We use the following formula to find the momentum, p :

$\begin{array}{rcl}p& =& \mathcal{h}k\\ & =& \left(1.05\times {10}^{-34}J·s\right)\times (1.58\times {10}^{12}{m}^{-1})\\ & =& 1.66\times {10}^{-22}kg·m/s\end{array}$

Calculate the Kinetic energy.

To calculate the kinetic energy, , we use:

$\begin{array}{rcl}KE& =& \mathcal{h}\omega \\ & =& \left(1.05\times {10}^{-34}J·s\right)\times \left(7.91\times {10}^{16}{s}^{-1}\right)\\ & =& 8.30\times {10}^{-18}J\end{array}$

Calculate the mass.

For calculation of the Mass , we use:

$m=\frac{p^2}{2\times KE}$

Substitute the obtained values in the above expression and we get,

$\begin{array}{rcl}m& =& \frac{{\left(1.66\times {10}^{-22}kg·m/s\right)}^2}{2\times \left(8.3\times {10}^{-18}J\right)}\\ & =& 1.66\times {10}^{-27}s\end{array}$