Chapter 41, Problem 59CP

(a)

To determine

Prove that the reflection coefficient for the given condition is $R=\frac{{\left(k_1-k_2\right)}^2}{{\left(k_1+k_2\right)}^2}$.

Answer to Problem 59CP

Proof for the reflection coefficient for the given condition is $R=\frac{{\left(k_1-k_2\right)}^2}{{\left(k_1+k_2\right)}^2}$ shown below.

Explanation of Solution

Write the Schrodinger’s equation.

    $\frac{{\partial }^2\psi }{\partial x^2}=-\frac{2m}{{\hslash }^2}\left(E-U\right)\psi$                                                   (I)

Here, $\psi$ is the wavefunction, $m$ is the mass, $E$ is the energy, $U$ is the potential energy and $\hslash$ is $h/2\pi$, $h$ is Plank’s constant

The solutions of the equation I for the region I in the above figure is

    ${\psi }_1=A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}$                                                 (II)

The solutions of the equation I for the region II in the above figure is

    ${\psi }_2=C{e}^{i{k}_{2}x}$                                                 (III)

Here, $A$, $B$ and $C$ are the constants.

Check the solution for the region I satisfies the equation I.

    $\begin{array}{c}\frac{{\partial }^2{\psi }_1}{\partial x^2}=-\frac{2m}{{\hslash }^2}E{\psi }_1\\ \frac{{\partial }^2}{\partial x^2}\left(A{e}^{i{k}_{1}x}\right)+\frac{{\partial }^2}{\partial x^2}\left(B{e}^{-i{k}_{1}x}\right)=-\frac{2m}{{\hslash }^2}E\left(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}\right)\\ -k_{{}_1}^2\left(A{e}^{i{k}_{1}x}\right)-k_1^2\left(B{e}^{-i{k}_{1}x}\right)=-\frac{2m}{{\hslash }^2}E\left(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}\right)\\ -k_{{}_1}^2\left(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}\right)=-\frac{2m}{{\hslash }^2}E\left(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}\right)\end{array}$

The above equation will be true only if the $k_{{}_1}^2=\frac{2m}{{\hslash }^2}E$, then substitute the above value in the following energy equation.

    $\begin{array}{c}E=\frac{p^2}{2m}\\ =\frac{{\left(\hslash k_1\right)}^2}{2m}\end{array}$

Here, $p$ is the momentum.

Rewrite the above equation to find the value of $k_1$.

    $k_1=\frac{\sqrt{2mE}}{\hslash }$                                                 (IV)

Therefore the equation I is satisfied for the region I.

Check the equation I for the region II.

    $\begin{array}{l}\frac{{\partial }^2{\psi }_2}{\partial x^2}=-\frac{2m}{{\hslash }^2}\left(E-U\right){\psi }_2\\ \frac{{\partial }^2}{\partial x^2}\left(C{e}^{i{k}_{2}x}\right)=-\frac{2m}{{\hslash }^2}\left(E-U\right)\left(C{e}^{i{k}_{2}x}\right)\\ -k_{{}_2}^2\left(C{e}^{i{k}_{2}x}\right)=-\frac{2m}{{\hslash }^2}\left(E-U\right)\left(C{e}^{i{k}_{2}x}\right)\end{array}$

The above equation will be true only if the $k_{{}_2}^2=\frac{2m}{{\hslash }^2}\left(E-U\right)$, then substitute the above value in the following energy equation.

    $\begin{array}{c}E=\frac{p^2}{2m}+U\\ =\frac{{\left(\hslash k_2\right)}^2}{2m}\end{array}$

Rewrite the above equation to find the value of $k_2$.

    $k_2=\frac{\sqrt{2m\left(E-U\right)}}{\hslash }$                                                 (V)

Therefore the equation I is satisfied for the region II.

Then apply the boundary conditions such as matching the function and derivatives at x=0.

    ${\left({\psi }_1\right)}_0={\left({\psi }_2\right)}_0$  gives $A+B=C$

    ${\left(\frac{d{\psi }_1}{dx}\right)}_0={\left(\frac{d{\psi }_2}{dx}\right)}_0$ gives $k_1\left(A-B\right)=k_{2}C$

From the above equations,

    $B=\frac{1-k_2/k_1}{1+k_2/k_1}A$ and $C=\frac{2}{1+k_2/k_1}A$                                                (VI)

Write the equation for probability.

    $R=\frac{B^2}{A^2}$                                                 (VII)

Conclusion:

Substitute equation VI in VII.

    $\begin{array}{c}R=\frac{{\left(1-k_2/k_1\right)}^2}{{\left(1+k_2/k_1\right)}^2}\\ =\frac{{\left(k_1-k_2\right)}^2}{{\left(k_1+k_2\right)}^2}\end{array}$

Therefore, the proof for the reflection coefficient for the given condition is $R=\frac{{\left(k_1-k_2\right)}^2}{{\left(k_1+k_2\right)}^2}$ shown above.

(b)

To determine

Find the probability of particle being reflected.

Answer to Problem 59CP

The probability of particle being reflected is $0.0920$.

Explanation of Solution

Write the equation for ration of $k_2$ to $k_1$.

    $\frac{k_2}{k_1}=\sqrt{\frac{E-U}{E}}$                                                  (VIII)

Conclusion:

Substitute $7.00\text{ eV}$ for $E$ and $5.00\text{ eV}$ for $U$ in equation VIII.

    $\begin{array}{c}\frac{k_2}{k_1}=\sqrt{\frac{2.00\text{ eV}}{7.00\text{ eV}}}\\ =0.535\end{array}$

Substitute $0.535$ for $\frac{k_2}{k_1}$ in the following probability equation.

  $\begin{array}{c}R=\frac{{\left(1-k_2/k_1\right)}^2}{{\left(1+k_2/k_1\right)}^2}\\ =\frac{{\left(1-0.535\right)}^2}{{\left(1+0.535\right)}^2}\\ =0.0920\end{array}$

Therefore, the probability of particle being reflected is $0.0920$.

(c)

To determine

Find the probability of particle being transmitted.

Answer to Problem 59CP

The probability of particle being transmitted is $0.908$.

Explanation of Solution

Write the equation for probability of transmitted.

    $T=1-R$                                                  (IX)

Conclusion:

Substitute $0.0920$ for $R$ in equation IX.

    $\begin{array}{c}T=1-0.0920\\ =0.908\end{array}$

Therefore, the probability of particle being transmitted is $0.908$.