27E For the E>U0 potential barrie... [FREE SOLUTION]

Chapter 6: 27E (page 225)

For the E>U₀ potential barrier, the reflection, and transmission probabilities are the ratios:
$R = \frac{B^{} B}{A^{} A} T = \frac{F^{} F}{A^{} A}$
Where A, B, and F are multiplicative coefficients of the incident, reflected, and transmitted waves. From the four smoothness conditions, solve for B and F in terms of A, insert them in R and T ratios, and thus derive equations (6-12).

Short Answer

Expert verified

The required equations are obtained as:

$R = \frac{\sin^{2} (k^{'} L)}{\sin^{2} (k^{'} L) + 4 \frac{k^{' 2} k^{2}}{{(k^{' 2} - k^{' 2})}^{2}}} T = \frac{4 \frac{k^{' 2} k^{2}}{{(k^{' 2} - k^{' 2})}^{2}}}{\sin^{2} (k^{'} L) + 4 \frac{k^{' 2} k^{2}}{{(k^{' 2} - k^{' 2})}^{2}}}$

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.

Determine the equation as:

Dividing the 4^th condition by the 3^rd and eliminating F, we get:

$k = \frac{k^{'} ({Ce}^{{ik}^{'} L} - {De}^{{ik}^{'} L})}{({Ce}^{{ik}^{'} L} + {De}^{{ik}^{'} L})}$

This further gives $D = - \frac{k^{'} - k}{k + k^{'}} e^{2 {ik}^{'} L} C$

Put this in the condition in 1^st and 2^nd conditions, we get:

$\begin{array}{rcl} A + B & = & (1 - \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L}) C \\ \frac{k}{k^{'}} (A - B) & = & (1 + \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L}) C \end{array}$

Dividing these two $\frac{k (A - B)}{k^{'} (A + B)} = \frac{(1 - \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L})}{(1 + \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L})}$

Now, $B = - \frac{(1 + \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L}) - \frac{k}{k^{'}} (1 - \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L})}{(1 + \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L}) + \frac{k}{k^{'}} (1 - \frac{k - k^{'}}{k + k^{'}} e^{2 {ik}^{'} L})} A$

Multiplying by $k^{'} (k + k^{'})$ to numerator and denominator and solving:

$B = - \frac{(k^{' 2} - k^{2}) (1 - e^{2 {ik}^{'} L})}{{(k + k^{'})}^{2} e^{{ik}^{'} L} - (k - k^{'}) e^{{ik}^{'} L}} A$

Multiply by $e^{{ik}^{'} L}$ to numerator and denominator:

$B = - \frac{(k^{' 2} - k^{2}) (- 2 isin (k^{'} L))}{{(k + k^{'})}^{2} - (k - k^{'}) e^{2 {ik}^{'} L}} A \frac{B^{*} B}{A^{*} A} = \frac{2 i (k^{' 2} - k) \sin (k^{'} L)}{{(k + k^{'})}^{2} e^{{ik}^{'} L} - {(k - k^{'})}^{2} e^{- {ik}^{'} L}} \frac{- 2 i (k^{' 2} - k) \sin (k^{'} L)}{{(k + k^{'})}^{2} e^{- {ik}^{'} L} - {(k - k^{'})}^{2} e^{ikL}} \frac{B^{*} B}{A^{*} A} = \frac{4 {(k^{' 2} - k^{2})}^{2} \sin^{2} (k^{'} L)}{{(k + k^{'})}^{4} + {(k - k^{'})}^{4} - {(k + k^{'})}^{2} {(k - k^{'})}^{2} (e^{2 {ik}^{'} L} - e^{- 2 {ik}^{'} L})}$

Put $(e^{2 {ik}^{'} L} - e^{- 2 {ik}^{'} L}) = \cos 2 胃 = 1 - 2 \sin^{2} k^{'} L$ and solve as:

$\frac{B^{*} B}{A^{*} A} = \frac{4 {(k^{' 2} - k^{2})}^{2} \sin^{2} (k^{'} L)}{{(k + k^{'})}^{4} + {(k - k^{'})}^{4} - {(k + k^{'})}^{2} {(k - k^{'})}^{2} (1 - 2 \sin^{2} k^{'} L)}$

Arranging and solving you get:

$R = \frac{\sin^{2} (k^{'} L)}{\sin^{2} (k^{'} L) + 4 \frac{k^{' 2} k^{2}}{{(k^{' 2} - k^{' 2})}^{2}}}$

Since, data-custom-editor="chemistry" $R + T = 1$ . So, $T = \frac{4 \frac{k^{' 2} k^{2}}{{(k^{' 2} - k^{' 2})}^{2}}}{\sin^{2} (k^{'} L) + 4 \frac{k^{' 2} k^{2}}{{(k^{' 2} - k^{' 2})}^{2}}}$ .