Question

Reflection and Transmission probabilities can be obtained from equations (6-12). The first step is substituting $\mathbf{-}\mathbf{i\alpha }$for${\mathbf{k}}^{\mathbf{\text{'}}}$. (a) Why? (b) Make the substitutions and then use definitions of k and α to obtain equation (6-16).

Short answer

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

Step-by-step solution

Concept involved

Tunneling is a phenomenon in which a wavefunction tunnels or propagates through a potential barrier. Reflection happens if the wavefunction is not enough for tunneling.

Given/known parameters

$\begin{array}{rcl}\mathrm{R}& =& \frac{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)+4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^2-{\mathrm{k}}^{\text{'}2}\right)}^2}}}\\ \mathrm{T}& =& \frac{4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^2-{\mathrm{k}}^{\text{'}2}\right)}^2}}}{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)+4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^2-{\mathrm{k}}^{\text{'}2}\right)}^2}}}\end{array}$

(a) Reason for the substitution

On substitution of$-\mathrm{i\alpha }$ for k^', the initial wave function changes, which overall changes the equation from one form to another; in this case, from (6-12) to (6-16).

(b) Substituting -iα for k' to obtain equation (6-16)

Since,

$$\begin{gathered} \mathrm{\alpha }=\frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}} \\ \mathrm{k}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}} \\ {\mathrm{\alpha }}^2+{\mathrm{k}}^2=\frac{\sqrt{2{\mathrm{mU}}_{\mathrm{o}}}}{\mathrm{h}} \end{gathered}$$

Putting these values in R and T values:

role="math" localid="1660048948839" $$\begin{gathered} \mathrm{R}=\frac{{\sin }^2\left(-\mathrm{i}{\displaystyle \frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}}}\mathrm{L}\right)}{{\sin }^2\left({\displaystyle \frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}}}\mathrm{L}\right)+4{\displaystyle \frac{-{\left({\displaystyle \frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}}}\right)}^2{\left({\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}\right)}^2}{{\left({\displaystyle \frac{\sqrt{2{\mathrm{mU}}_{\mathrm{o}}}}{\mathrm{h}}}\right)}^2}}} \\ \mathrm{T}=\frac{4{\displaystyle \frac{-{\left({\displaystyle \frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}}}\right)}^2{\left({\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}\right)}^2}{{\left({\displaystyle \frac{\sqrt{2{\mathrm{mU}}_{\mathrm{o}}}}{\mathrm{h}}}\right)}^2}}}{{\sin }^2\left({\displaystyle \frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}}}\mathrm{L}\right)+4{\displaystyle \frac{-{\left({\displaystyle \frac{\sqrt{2\mathrm{m}\left({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}\right)}}{\mathrm{h}}}\right)}^2{\left({\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}\right)}^2}{{\left({\displaystyle \frac{\sqrt{2{\mathrm{mU}}_{\mathrm{o}}}}{\mathrm{h}}}\right)}^2}}} \end{gathered}$$

Hence, on solving and rearranging, we can get the required equations as:

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