91影视

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.

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

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 $伪L=\frac{\sqrt{2m(U_0-E)}}{\mathrm{鈩�}}L鈮�1$and $T=\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{{\sinh }^2\left(伪L\right)+4\left(E/U_0\right)\left(1-E/U_0\right)}$.

We know that,

.$伪=\frac{\sqrt{2m\left(U_0-E\right)}}{\mathrm{鈩�}}$

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

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

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

$伪L=\frac{\sqrt{2m(U_0-E)}}{\mathrm{鈩�}}L鈮�1$and $T=\frac{4\left(E/U_0\right)\left(1-E/U_0\right)}{{\sinh }^2\left(伪L\right)+4\left(E/U_0\right)\left(1-E/U_0\right)}$.