91影视

A particle is defined by the wave function:${\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$for$\mathit{x}\mathbf{<}\mathbf{0}$and$\mathit{C}{\mathit{e}}^{\mathbf{4}\mathbf{x}}$for$\mathit{x}\mathbf{>}\mathbf{0}$. For the given wave function to becontinuous, at$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{B}\mathbf{=}\mathit{C}$

To the left of the step $x<0$solve as:

$$\begin{gathered} 蠄={蠄}_{inc}+{蠄}_{ref} \\ 蠄=e^{ikx}+B{e}^{-ikx} \end{gathered}$$

To the right of the step $x>0$solve as:

$蠄=C{e}^{ik\text{'}x}$

Considermust be continuous at $x=0$solve as:

$$\begin{gathered} e^0+B{e}^0=C{e}^0 \\ 1+B=C \end{gathered}$$

Consider $\frac{d蠄}{dx}$must be continuous at $x=0$.

$$\begin{gathered} ik{e}^0-ikB{e}^0=-伪C{e}^0 \\ k\left(1-B\right)=k\text{'}C \end{gathered}$$

From the first and second conditions solve as:

$$\begin{gathered} k\left(1-B\right)=k\text{'}\left(1+B\right) \\ \frac{k-k\text{'}}{k+k\text{'}}=\frac{\frac{\sqrt{2mE}}{h}-\frac{\sqrt{2m\left(E-\frac{3}{4}E\right)}}{h}}{\frac{\sqrt{2mE}}{h}+\frac{\sqrt{2m\left(E-\frac{3}{4}E\right)}}{h}} \end{gathered}$$

Divide by $\frac{\sqrt{2mE}}{h}$everywhere:

$$\begin{gathered} B=\frac{1-\sqrt{\left(1-\frac{3}{4}\right)}}{1+\sqrt{\left(1-\frac{3}{4}\right)}}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}} \\ B=\frac{1}{3} \end{gathered}$$

Putting this in C, we get $C=\frac{4}{3}$

(a)

Write the components of the function as:

$$\begin{gathered} {蠄}_{refl}=B{e}^{-ikx} \\ =\frac{1}{3}{e}^{-ikx} \end{gathered}$$

Also:

$$\begin{gathered} {蠄}_{x>0}=C{e}^{ik\text{'}x} \\ =\frac{4}{3}{e}^{ik\text{'}x} \end{gathered}$$

$Here,k=\frac{\sqrt{2mE}}{h}andk\text{'}=\frac{\sqrt{2m\left(\frac{1}{4}E\right)}}{h}.$

(b)

If it is given that, ${蠄}_{inc}=e^{ikx}$and ${蠄}_{ref}=B{e}^{-ikx}$solve as:

$$\begin{gathered} \frac{{\left|{蠄}_{refl}\right|}^2}{{\left|{蠄}_{inc}\right|}^2}=\frac{{\left(\frac{1}{3}\right)}^2}{1^2} \\ =\frac{1}{9} \end{gathered}$$

From equation (6-7), $R=\frac{1}{9}$