Given the situation of exercise 25, show that

(a) as ${\mathbf{U}}_{\mathbf{o}}\mathbf{\to }\mathbf{\infty }$, reflection probability approaches 1 and

(b) as $\mathbf{L}\mathbf{\to }\mathbf{0}$, the reflection probability approaches 0.

(c) Consider the limit in which the well becomes infinitely deep and infinitesimally narrow--- that is ${\mathbf{U}}_{\mathbf{o}}\mathbf{\to }\mathbf{\infty }$and data-custom-editor="chemistry" $\mathbf{L}\mathbf{\to }\mathbf{0}$but the product U₀L is constant. (This delta well model approximates the effect of a narrow but strong attractive potential, such as that experienced by a free electron encountering a positive ion.) Show that reflection probability becomes:

$\mathbf{R}\mathbf{=}{\left[1+\frac{2h^{2}E}{m{\left(U_{o}L\right)}^2}\right]}^{\mathbf{-}\mathbf{1}}$

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.

Write the given data:

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

Here,

R = reflection probability

E = kinetic energy

U = potential energy

m = mass of the particle

L= width of potential barrier

As U₀ approaches 0, the term$4\left(\frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}\right)\left[\left(\frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}\right)+1\right]$approaches 1, leaving the form,$\mathrm{R}=\frac{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}}\right]}{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}}\right]}$which answers 1.

Hence, for U₀ tends to be infinite, R tends to be 1.

As L approaches zero, the term ${\sin }^2\left[\frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}\right]$becomes 0 in numerator and denominator but the term $4\left(\frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}\right)\left[\left(\frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}\right)+1\right]$is finite and so overall, R becomes 0.

Hence, For L tends to 0, R tends to 0.

If U_oL is a finite constant, it will be very small, so will be$\sqrt{{\mathrm{U}}_{\mathrm{o}}\mathrm{L}}$ and also$\sqrt{{\mathrm{U}}_{\mathrm{o}}\mathrm{L}}\sqrt{\mathrm{L}}$

sine of x tends to x for small values of x. So, we can say that:

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

Now, putting E+U_o as U_o and$\left(\frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}\right)+1$as 1, we get:

$\mathrm{R}=\frac{{\displaystyle \frac{2{\mathrm{mU}}_{\mathrm{o}}\mathrm{L}}{\mathrm{h}}}}{{\displaystyle \frac{2{\mathrm{mU}}_{\mathrm{o}}\mathrm{L}}{\mathrm{h}}}+4{\displaystyle \frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}}}$giving$\mathrm{R}={\left[1+\frac{2{\mathrm{h}}^2\mathrm{E}}{\mathrm{m}{\left({\mathrm{U}}_{\mathrm{o}}\mathrm{L}\right)}^2}\right]}^{-1}$

Hence, For U₀L being a small constant,$\mathrm{R}={\left[1+\frac{2{\mathrm{h}}^2\mathrm{E}}{\mathrm{m}{\left({\mathrm{U}}_{\mathrm{o}}\mathrm{L}\right)}^2}\right]}^{-1}$