Question

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

Short answer

Answer

The equation$T=16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-2L\sqrt{2m\left(U_0-E_1\right)\text{/}\hslash }}$ is derived from $\alpha L=\frac{\sqrt{2m(U_0-E)}}{\hslash }L\gg 1$and$T=\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{{\sinh }^2\left(\alpha L\right)+4\left(E/U_0\right)\left(1-E/U_0\right)}$

Step-by-step solution

Definition of transmission probability

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.

Given quantities 

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

Imposing the limiting value in the equation of transmission probability.

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