91影视

Classically, the particle shouldn鈥檛 reflect. It should slow down between the barriers. But quantum mechanically, we apply the Schr枚dinger equation and get the solution to be of the form:${\mathbf{e}}^{\mathbf{卤}\mathbf{ikx}}$. So,

$\begin{array}{rcl}{\mathbf{蠄}}_{\mathbf{x}\mathbf{<}\mathbf{0}}\left(x\right)& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{ikx}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{ikx}}\\ {\mathbf{蠄}}_{\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}}\left(x\right)& \mathbf{=}& {\mathbf{Ce}}^{{\mathbf{ik}}^{\mathbf{\text{'}}}\mathbf{x}}\mathbf{+}{\mathbf{De}}^{\mathbf{-}{\mathbf{ik}}^{\mathbf{\text{'}}}\mathbf{x}}\\ {\mathbf{蠄}}_{\mathbf{x}\mathbf{>}\mathbf{0}}\left(x\right)& \mathbf{=}& {\mathbf{Fe}}^{\mathbf{ikx}}\end{array}$

Here, 伪 in second equation signifies the real nature.data-custom-editor="chemistry" $\mathbf{蠄}$= Wave Functions, A,B,C,D,F are the arbitrary multiplicative constants.

Applying boundary conditions for 0<x<L and x<0:

If, L= Width of potential barrier, $\mathrm{k}=\frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{V}}_{\mathrm{o}}\right)}}{{\mathrm{h}}^2}$and ${\mathrm{k}}^{\text{'}}=\frac{\sqrt{2\mathrm{mE}}}{{\mathrm{h}}^2}$

$$\begin{gathered} {\mathrm{蠄}}_{\mathrm{x}<0}\left(0\right)={\mathrm{蠄}}_{0<\mathrm{x}<\mathrm{L}}\left(0\right): \\ {\mathrm{Ae}}^{\mathrm{ik}0}+{\mathrm{Be}}^{-\mathrm{ik}0}={\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}0}+{\mathrm{De}}^{-{\mathrm{ik}}^{,}0} \\ \mathrm{A}+\mathrm{B}=\mathrm{C}+\mathrm{D} \end{gathered}$$

Solve as:

$$\begin{gathered} \frac{d{\mathrm{蠄}}_{\mathrm{x}<0}}{d\mathrm{x}}{|}_{\mathrm{x}=0}=\frac{d{\mathrm{蠄}}_{0<\mathrm{x}<\mathrm{L}}}{d\mathrm{x}}{|}_{\mathrm{x}=0}: \\ {\mathrm{ikAe}}^{\mathrm{ik}0}-{\mathrm{ikBe}}^{-\mathrm{ik}0}={\mathrm{ik}}^{\text{'}}{\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}0}-{\mathrm{ik}}^{\text{'}}{\mathrm{De}}^{{\mathrm{ik}}^{\text{'}}0} \\ \mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)={\mathrm{k}}^{\text{'}}\left(\mathrm{C}-\mathrm{D}\right) \end{gathered}$$

Now applying boundary conditions for 0<x<L and x>L:

$$\begin{gathered} {\mathrm{蠄}}_{0<\mathrm{x}<\mathrm{L}}\left(\mathrm{L}\right)={\mathrm{蠄}}_{\mathrm{x}>0}\left(\mathrm{L}\right): \\ {\mathrm{Ce}}^{-{\mathrm{ik}}^{\text{'}}\mathrm{L}}+{\mathrm{De}}^{-{\mathrm{ik}}^{\text{'}}\mathrm{L}}={\mathrm{Fe}}^{\mathrm{ikL}} \end{gathered}$$

Also.

$$\begin{gathered} \frac{d{\mathrm{蠄}}_{0<\mathrm{x}<\mathrm{L}}}{d\mathrm{x}}{|}_{\mathrm{x}=\mathrm{L}}=\frac{d{\mathrm{蠄}}_{\mathrm{x}>0}}{d\mathrm{x}}{|}_{\mathrm{x}=\mathrm{L}}: \\ {\mathrm{ik}}^{\text{'}}\left({\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\mathrm{De}}^{{\mathrm{ik}}^{,}\mathrm{L}}\right)={\mathrm{ikFe}}^{\mathrm{ikL}} \end{gathered}$$

From the second condition:$\mathrm{C}=\frac{\mathrm{k}}{{\mathrm{k}}^{\text{'}}}\left(\mathrm{A}-\mathrm{B}\right)+\mathrm{D}$put in first to obtain:

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{C}=\frac{{\mathrm{k}}^{\text{'}}\left(\mathrm{A}+\mathrm{B}\right)+\mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)}{2{\mathrm{k}}^{\text{'}}} \\ \mathrm{D}=\frac{{\mathrm{k}}^{\text{'}}\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)}{2{\mathrm{k}}^{\text{'}}} \end{gathered}$$

Substitute these values in the definition of reflection, the coefficient of reflection is given by:

$\mathrm{R}=\frac{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)+4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^2-{\mathrm{k}}^{\text{'}2}\right)}^2}}}$

Now, substituting the values in the above equation,$\mathrm{k}=\sqrt{\frac{2\mathrm{m}\left(\mathrm{E}-{\mathrm{V}}_{\mathrm{o}}\right)}{{\mathrm{h}}^2}}$and${\mathrm{k}}^{\text{'}}=\sqrt{\frac{2\mathrm{mE}}{{\mathrm{h}}^2}}$

Hence,$\mathrm{R}=\frac{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}}\right]}{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}}\right]+4\left({\displaystyle \frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}}\right)\left[\left({\displaystyle \frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}}\right)+1\right]}$

Now, R is non-zero in general because the numerator contains sine term.

In order to consider no reflection, the numerator becomes zero and is equated to 苍蟺, the condition is:

$$\begin{gathered} \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{U}}_{\mathrm{o}}\right)}}{\mathrm{h}}\mathrm{L}=\mathrm{苍蟺} \\ \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{U}}_{\mathrm{o}}\right)}}{\mathrm{h}}\mathrm{L}=\mathrm{蟺} \end{gathered}$$OR

(n=1 for first transmission resonance)

$\mathrm{E}={\mathrm{U}}_{\mathrm{o}}+\frac{{\mathrm{蟺}}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$

Here, E= Kinetic energy, U_o= Potential energy, h= Modified Plank鈥檚 constant

$\mathrm{E}={\mathrm{U}}_{\mathrm{o}}+\frac{{\mathrm{蟺}}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$

This expression is not a quantization condition but it defines the transmission resonance condition such that it occurs only at certain energies no matter what incident energy is.

Hence, there should be no reflection when the incident wave is at one of the transmission resonances