Question: Obtain equation (6.18) from(6.16) and (6.17).

The transmission or tunneling probability can be calculated using transmitted intensity and the incident intensity.

In the case of tunneling barriers being wide, it can be found as follows.

$\mathit{T}\mathbf{\cong }\mathbf{16}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}\left(1-\frac{E}{U_0}\right){\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}\mathbf{m}\left(U_0-E_1\right)\mathbf{\text{/}}\mathbf{\hslash }}}$

Here E is the jump energy, U₀is barrier energy, L is the length of the tunnel, and m is the mass of the particle.

The given values are $\alpha L=\frac{\sqrt{2m(U_0-E)}}{\hslash }L\gg 1$and $T=\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{{\sinh }^2\left(\alpha L\right)+4\left(E/U_0\right)\left(1-E/U_0\right)}$.

We know that,

.$\alpha =\frac{\sqrt{2m\left(U_0-E\right)}}{\hslash }$

Use the hyperbolic relation of$Sinh\left(\alpha L\right)=\frac{e^{\alpha L}}{2}$ for $\alpha L\gg 1$in the transmission formula as:

$$\begin{gathered} T=\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{{\sinh }^2\left(\alpha L\right)+4\left(E/U_0\right)\left(1-E/U_0\right)} \\ =\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{\frac{e^{2\alpha L}}{4}+4\left(E/U_0\right)\left(1-E/U_0\right)} \\ =\frac{16\left(E/U_0\right)\left(1-E/U_0\right)}{e^{2\alpha L}} \\ =16\left(E/U_0\right)\left(1-E/U_0\right)e^{-2\alpha L} \\ =16\frac{E}{U_0}\left(1-\frac{E}{U_0}\right)e^{-2L\sqrt{2m\left(U_0-E\right)\text{/}\hslash }} \end{gathered}$$

Therefore, the equation is obtained from$T=16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-2L\sqrt{2m\left(U_0-E_1\right)\text{/}\hslash }}$

$\alpha L=\frac{\sqrt{2m(U_0-E)}}{\hslash }L\gg 1$and $T=\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{{\sinh }^2\left(\alpha L\right)+4\left(E/U_0\right)\left(1-E/U_0\right)}$.