91影视

Energy contained by the object by the virtue of its motion is called kinetic energy.

Potential energyis the energy contained by an object by the virtue of its position.

Classical particles would continue to the right, their kinetic energy abruptly increasing from E to 4E $\left(\left[\mathrm{E}-\left(-3\mathrm{E}\right)\right]\right)$.

Consider the function as:

${\mathrm{蠄}}_{\mathrm{inc}}={\mathrm{e}}^{\mathrm{ikx}}$

To the left of the drop,$\mathrm{x}<0$solve as:

$\begin{array}{rcl}\mathrm{蠄}& =& {\mathrm{蠄}}_{\mathrm{inc}}+{\mathrm{蠄}}_{\mathrm{refl}}\\ & =& {\mathrm{e}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}\end{array}$

To the right of the drop, (x>0):$\mathrm{蠄}={\mathrm{Be}}^{{\mathrm{ik}}^{\text{'}}\mathrm{x}}$

Here,

$\begin{array}{rcl}{\mathrm{k}}^{\text{'}}& =& \frac{\sqrt{2\mathrm{m}\left[\mathrm{E}-\left(-3\mathrm{E}\right)\right]}}{\mathrm{h}}\\ & =& 2\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\end{array}$

Here,data-custom-editor="chemistry" $\mathrm{蠄}$ must be continuous at data-custom-editor="chemistry" $\mathrm{x}=0$.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{o}}+{\mathrm{Be}}^{\mathrm{o}}={\mathrm{Ce}}^{\mathrm{o}} \\ 1+\mathrm{B}=\mathrm{C} \end{gathered}$$

Here,$\frac{d\mathrm{蠄}}{d\mathrm{x}}$ must be continuous at$\mathrm{x}=0$.

$$\begin{gathered} {\mathrm{ike}}^{\mathrm{o}}-{\mathrm{ikBe}}^{\mathrm{o}}=-{\mathrm{伪颁别}}^{\mathrm{o}} \\ \mathrm{k}\left(1-\mathrm{B}\right)={\mathrm{k}}^{\text{'}}\mathrm{C} \end{gathered}$$

From the first and second conditions solve as:

$$\begin{gathered} \mathrm{k}\left(1-\mathrm{B}\right)={\mathrm{k}}^{\text{'}}\left(1+\mathrm{B}\right) \\ \mathrm{B}=\frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}} \\ \mathrm{B}=\frac{{\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}-{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-\left(-3\mathrm{E}\right)\right)}}{\mathrm{h}}}}{{\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}+{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-\left(-3\mathrm{E}\right)\right)}}{\mathrm{h}}}} \\ \mathrm{B}=-\frac{1}{3} \end{gathered}$$

So, data-custom-editor="chemistry" $\mathrm{C}=\frac{2}{3}$.

Hence the required wave functions are given by,

$$\begin{gathered} {\mathrm{蠄}}_{\mathrm{refl}}=-\frac{1}{3}{\mathrm{e}}^{-\mathrm{ikx}} \\ {\mathrm{蠄}}_{\mathrm{x}>0}=\frac{2}{3}{\mathrm{e}}^{{\mathrm{ik}}^{,}\mathrm{x}} \end{gathered}$$

The required probability can be calculated by

$\frac{{\mathrm{B}}^{*}\mathrm{B}}{{\mathrm{A}}^{*}\mathrm{A}}=\frac{1}{9}$