Question

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?

Short answer

a) The values of the width of the separation between the two 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.

b) Tunneling is not expected at s = 0, because at, s = 0, E > U₀ . There will be just a single barrier in that case.

c) No, 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.

Step-by-step solution

 Given data

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.

 Concept/Formula involved

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.

Step 3(a): Determining the width of the separation between 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.

Step 4(b): Solution when s = 0 

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.

Step 5(c): Tunneling of classical particle

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.