Exercise 39 gives the condition for resonant tunneling through two barriers separated by a space of width 2 s, expressed in terms of a factor $\mathit{\beta }$given in Exercise 30. (a) Suppose that in some system of units, k and $\mathit{\alpha }$ are both$\mathbf{2}\mathbf{\pi }$. Find two values of 2s that give resonant tunneling. What are these distances in terms of wavelengths of$\mathit{\psi }$? Is the term resonant tunneling appropriate?(b) Show that the condition has no solution if s = 0 and explain why this must be so. (c) If a classical particle wants to surmount a barrier without gaining energy, is adding a second barrier a good solution?

The value of the wave number and some factor $\alpha$is given as $2\mathrm{\pi }$

The width of the separation between the two barriers is given as 2 s.

The expression for factor $\beta$of the wave is given by,

$\beta ={\tan }^{-1}\left(\frac{2\alpha k}{\left(k^2-{\alpha }^2\right)}coth\left(\alpha L\right)\right).....\left(1\right)$

Where, $\alpha$= any factors of the wave particle

k = the wave number and

L = the length of the barrier.

The expression for the wavelength of the particle,

$\lambda =\frac{2\mathrm{\pi }}{k}.....\left(2\right)$

Here, $\lambda$is the wavelength of the particle.

The expression for the width of the separation between the two barriers, $2s=\frac{\beta }{k}....\left(3\right)$

Here, 2s is the width of the separation between the two barriers.

Substitute $2\mathrm{\pi }$for $\alpha$and k in equation (1).

$\begin{array}{rcl}\beta & =& {\tan }^{-1}\left(\frac{2\left(2\mathrm{\pi }\right)\left(2\mathrm{\pi }\right)}{{\left(2\mathrm{\pi }\right)}^2-{\left(2\mathrm{\pi }\right)}^2}coth\left(2\mathrm{\pi L}\right)\right)\\ & =& {\tan }^{-1}\left(\infty \right)\end{array}$

Simplify the above expression for $\beta$

$\beta =\pi /2and3\pi /2$

Substitute $2\pi$for k in equation (2)

$\begin{array}{rcl}\lambda & =& \frac{2\mathrm{\pi }}{2\mathrm{\pi }}\\ & =& 1\end{array}$

Substitute $2\mathrm{\pi }$for k and $\mathrm{\pi }/2$for $\mathrm{\beta }$in equation (3)

$\begin{array}{rcl}2s& =& \frac{\left(\mathrm{\pi }/2\right)}{2\mathrm{\pi }}\\ & =& \frac{1}{4}\end{array}$

Substitute $\begin{array}{rcl}& & 2\mathrm{\pi }\end{array}$for k and $\begin{array}{rcl}& & 3\pi /2\end{array}$for $\begin{array}{rcl}& & \mathrm{\beta }\end{array}$in equation (3).

$\begin{array}{rcl}2s& =& \frac{\left(3\mathrm{\pi }/2\right)}{2\mathrm{\pi }}\\ & =& \frac{3}{4}\end{array}$

Therefore, the widths of the separation between the barriers are one-fourth and three-fourth of the wavelength. So, the tunneling at separation one-half wavelength will be maximum as resonant conditions come into action.

From equation (3), you know that

$2s=\frac{\beta }{k}$

Now, if s = 0, either $\beta =0ork=\infty$, they both imply that argument of arctan in the equation given below is zero.

$\beta ={\tan }^{-1}\left(\frac{2\alpha k}{\left(k^2-{\alpha }^2\right)}coth\left(\alpha L\right)\right)$

We can also ignore arctan and rewrite the formula as

$2s=\frac{2\alpha }{(k^2-{\alpha }^2)}coth\left(\alpha L\right)$

The above-mentioned equation is zero only if k is infinite, which will imply that E > U₀.

Hence, tunneling is not expected at s = 0, there will be just a single barrier in that case.

As you know that tunneling is not shown by classical particles. Moreover, quantum mechanics is not used for particles having macroscopic size. Hence, adding a second barrier will not be a good solution.