Question

Question: In Eq. keep both terms, putting $\mathbf{A}\mathbf{=}\mathbf{B}\mathbf{=}\mathbf{\psi }$. The

equation then describes the superposition of two matter waves of

equal amplitude, traveling in opposite directions. (Recall that this

is the condition for a standing wave.) (a) Show that ${\left|♆\left(x,t\right)\right|}^{\mathbf{2}}$ is

then given by ${\mathbf{|}\mathbf{♆}\left(x,t\right)\mathbf{|}}^{\mathbf{2}}\mathbf{=}\mathbf{2}{\mathit{\psi }}_{\mathbf{0}}^{\mathbf{2}}\left[1+cos2kx\right]$

(b) Plot this function, and demonstrate that it describes the square

of the amplitude of a standing matter wave. (c) Show that thenodes of this standing wave are located at $\mathbf{x}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\lambda }\mathbf{\right)}\mathbf{,}$where $\mathbf{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{\dots }$

and $\mathit{\lambda }$ is the de Broglie wavelength of the particle. (d) Write a similar

expression for the most probable locations of the particle.

Short answer

a) It is shown that${|♆(\mathrm{x},\mathrm{t})|}^2=2{\mathrm{\psi }}_0^2[1+\cos 2\mathrm{kx}]$

(b) The plot of the function is shown below, and it is demonstrated that it describes the square of the amplitude of a standing matter-wave.

(c) It is shown that nodes of the standing wave are located at$\mathrm{x}=(2\mathrm{n}+1)\left(\frac{1}{4}\mathrm{\lambda }\right)$,$\mathrm{n}=0,1,2,3,\dots$

(d) The most probable location can be given as$x=\frac{n\lambda }{2},n=0,1,2,3.....$

Step-by-step solution

Identifying the data given in the question.

The Equation is,

$♆\left(x,t\right)=A{e}^{i\left(kx-\omega t\right)}+B{e}^{i\left(kx-\omega t\right)}$

And$A=B=\psi$

Concept used to solve the question.

A matter wave can be described by a wave function,$\mathbf{♆}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}$

which can be separated into a space-dependent part$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and a time-dependent part${\mathbf{e}}^{\mathbf{-}\mathbf{i\omega t}}$, where$\mathit{\omega }$is the angular frequency of the wave.

Step 3:(a) Showing |♆(x,t)|2=2ψ02[1+cos2kx]

Given wave function is,

$♆\left(x,t\right)=A{e}^{i\left(kx-\omega t\right)}+B{e}^{i\left(kx-\omega t\right)}$

Substitutinglocalid="1663128338938" $A=B={\psi }_0$

$\begin{array}{rcl}♆\left(x,t\right)& =& {\psi }_0{e}^{i\left(kx-\omega t\right)}+{\psi }_0{e}^{i\left(kx-\omega t\right)}\\ & =& {\psi }_0{e}^{-i\omega t}\left(e^{ikx}+e^{-ikx}\right)\end{array}$

Using Euler’s formula $e^{i\varphi }=\cos \varphi +i\sin \varphi$

$\begin{array}{rcl}♆\left(x,t\right)& =& {\psi }_0{e}^{-i\omega t}\left(\left(\cos kx+i\sin kx\right)+\left(cioskx-i\sin kx\right)\right)\\ & =& {\psi }_0{e}^{-i\omega t}\left(2\cos kx\right)\end{array}$

Therefore,

localid="1663130532064" role="math" $$\begin{gathered} {\begin{array}{rcl}& & \left|♆\left(x,t\right)\right|\end{array}}^2\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \left|{\psi }_0{e}^{-i\omega t}\left(2\cos kx\right)\right|\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & \end{array} \\ \begin{array}{rcl}& & \left|♆\left(x,t\right)\right|\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & \left|{\psi }_0{e}^{-i\omega t}\right|\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & \left|\cos kx\right|\end{array}\begin{array}{rcl}& & {}^2\end{array} \\ \begin{array}{rcl}& & \end{array}{\begin{array}{rcl}& & \left|♆\left(x,t\right)\right|\end{array}}^2\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & \psi \end{array}\begin{array}{rcl}& & {}_0\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & \left(\cos kx\right)\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & \end{array} \\ \begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \left|♆\left(x,t\right)\right|\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & {\psi }_0\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & \left(\frac{1+\cos 2kx}{2}\right)\end{array}\begin{array}{rcl}& & \end{array} \\ \begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \left|♆\left(x,t\right)\right|\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & {\psi }_0\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & \left(1+\cos 2kx\right)\end{array} \end{gathered}$$

Hence it is proven that${|♆(\mathrm{x},\mathrm{t})|}^2=2{\mathrm{\psi }}_0^2[1+\cos 2\mathrm{kx}]$

 Step 4: (b) Plotting function and demonstrating that it describes the squareof the amplitude of a standing matter wave.

We know

${|♆(\mathrm{x},\mathrm{t})|}^2=2{\mathrm{\psi }}_0^2[1+\cos 2\mathrm{kx}]$

The wave function is

$♆\left(x,t\right)=A{e}^{i\left(kx-\omega t\right)}+B{e}^{i\left(kx-\omega t\right)}$

This represents the addition of two plane matter waves with equal amplitude and traveling in opposite directions.

Since we know two waves with the same amplitude traveling in opposite directions make a standing wave, this implies wave function role="math" localid="1663129415576" $♆\left(x,t\right)$is a standing matter wave

As we already calculated the amplitude of$♆\left(x,t\right)$is

${|♆(\mathrm{x},\mathrm{t})|}^2=2{\mathrm{\psi }}_0^2[1+\cos 2\mathrm{kx}]$

The graph of amplitude is shown below

(c) Finding nodes of standing matter wave

As we already calculate,

${|♆(\mathrm{x},\mathrm{t})|}^2=2{\mathrm{\psi }}_0^2[1+\cos 2\mathrm{kx}]$

As we know at nodes the wave functions amplitude is zero

This implies

$$\begin{gathered} {\left|♆\left(x,t\right)\right|}^2=0 \\ 2{{\psi }_0}^2\left[1+\cos 2kx\right]=0 \\ \cos 2kx=-1 \end{gathered}$$

Therefore,

$2kx=2\left(\frac{2\mathrm{\pi }}{\lambda }\right)=\left(2n+1\right)\mathrm{\pi },\mathrm{n}=0,1,2,3.....$

Solving for x

$x=\left(2n+1\right)\left(\frac{1}{4}\lambda \right),n=0,1,2,3.....$

Hence it is shown that nodes of the standing wave are located at

$x=\left(2n+1\right)\left(\frac{1}{4}\lambda \right),n=0,1,2,3.....$

(d) Find the most probable locations of the particle

As we already know,

The most probable position for finding the particle is where the amplitude is maximum

Since ${|♆(\mathrm{x},\mathrm{t})|}^2=2{\mathrm{\psi }}_0^2[1+\cos 2\mathrm{kx}]$

So, the most probable location will be when, $\cos 2kx=1$

Therefore,

$2kx=2\left(\frac{2\mathrm{\pi }}{\lambda }\right)=\left(2n+1\right)\mathrm{\pi },\mathrm{n}=0,1,2,3.....$

Solving for

$x=\frac{n\lambda }{2},n=0,1,2,3.....$

Hence the most probable location can be given as

data-custom-editor="chemistry" $x=\frac{n\lambda }{2},n=0,1,2,3.....$