91Ó°ÊÓ

The ratio of the transmitted intensity to the incident intensity is known as the transmission probability or tunnelling probability.

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

$\begin{array}{rcl}\mathbf{T}& â‰\dots & \mathbf{16}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}\left(\mathbf{1}\mathbf{-}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}\right){\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}\mathbf{m}\frac{\left({\mathbf{U}}_{\mathbf{0}}\mathbf{-}\mathbf{E}\right)}{\mathbf{h}}}}\\ & \mathbf{=}& \mathbf{16}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}\left(\mathbf{1}\mathbf{-}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}\right){\mathbf{e}}^{\mathbf{-}\mathbf{γ}\sqrt{\mathbf{1}\mathbf{-}\frac{\mathbf{E}}{{\mathbf{U}}_{\mathbf{0}}}}}\\ & & \end{array}$

Here

.$\mathbf{γ}\mathbf{=}\frac{\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}{\mathbf{mU}}_{\mathbf{0}}}}{\mathbf{h}}$

• The given values are$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.$is 0.4 and 0.6
• For the first case$\frac{2L\sqrt{2m{U}_0}}{h}=5$
• For the second case,$\frac{2L\sqrt{2m{U}_0}}{h}=50$
• For the third case,$\frac{2L\sqrt{2m{U}_0}}{h}=500$
• Also, for the second case$E=\frac{1}{1000}{U}_0$

Estimate the value of transmission probability for the first case, using the given value,.

$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.4,0.6$

$\begin{array}{rcl}\frac{2L\sqrt{2m{U}_0}}{h}& =& 5\\ \mathrm{γ}& =& 5\\ & & \end{array}$

For role="math" localid="1660052007690" $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.4$

$\begin{array}{rcl}T& â‰\dots & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\mathrm{γ}\sqrt{\left(1-\frac{E_1}{U_0}\right)}}\\ & =& 16×0.4×\left(1-0.4\right)×e^{-5\sqrt{1-0.4}}\\ & =& 16×0.4×0.6×e^{-5\sqrt{0.6}}\\ & =& 0.08\\ & & \end{array}$

For $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.6$

$\begin{array}{rcl}T& â‰\dots & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\mathrm{γ}\sqrt{\left(1-E_1/U_0\right)}}\\ & =& 16×0.6×(1-0.6)×e^{-5\sqrt{1-0.6}}\\ & =& 16×0.6×0.4×e^{-5\sqrt{0.4}}\\ & =& 0.16\\ & & \end{array}$

Therefore, the transmission probability when $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.$is 0.4 and when it is 0.6 are $0.08,0.16$respectively for

$\frac{2L\sqrt{2m{U}_0}}{h}=5$

Estimate the value of transmission probability for the second case, using the given value,$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.4,0.6$.

$\begin{array}{rcl}\frac{2L\sqrt{2m{U}_0}}{h}& =& 50\\ \mathrm{γ}& =& 50\\ & & \end{array}$

For$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.4$

$\begin{array}{rcl}T& â‰\dots & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\mathrm{γ}\sqrt{\left(1-E_1/U_0\right)}}\\ & =& 16×0.4×(1-0.4)×e^{-50\sqrt{1-0.4}}\\ & =& 16×0.4×0.6×e^{-50\sqrt{0.6}}\\ & =& 5.8×{10}^{-17}\\ & & \end{array}$

For$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.6$

$\begin{array}{rcl}T& â‰\dots & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\mathrm{γ}\sqrt{\left(1-E_1/U_0\right)}}\\ & =& 16×0.6×\left(1-0.6\right)×e^{-50\sqrt{1-0.6}}\\ & =& 16×0.6×0.4×e^{-50\sqrt{0.4}}\\ & =& 7.1×{10}^{-14}\\ & & \end{array}$

Therefore, the transmission probability when$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.$ is 0.4 and when it is 0.6 are $5.8×{10}^{-17},7.1×{10}^{-14}$respectively for$\frac{2L\sqrt{2m{U}_0}}{h}=50$

Estimate the value of transmission probability for the second case, using the given value,$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.4,0.6$.

$\begin{array}{rcl}\frac{2L\sqrt{2m{U}_0}}{h}& =& 500\\ \mathrm{γ}& =& 500\\ & & \end{array}$

For$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.4$

$\begin{array}{rcl}T& â‰\dots & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\mathrm{γ}\sqrt{\left(1-\frac{E_1}{U_0}\right)}}\\ & =& 16×0.4×\left(1-0.4\right)×e^{-500\sqrt{1-0.4}}\\ & =& 16×0.4×0.6×e^{-500\sqrt{0.6}}\\ & =& 2.41×{10}^{-168}\\ & & \end{array}$

For$\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.=0.6$

$\begin{array}{rcl}T& â‰\dots & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\mathrm{γ}\sqrt{\left(1-\frac{E_1}{U_0}\right)}}\\ & =& 16×0.6×\left(1-0.6\right)×e^{-500\sqrt{1-0.6}}\\ & =& 16×0.6×0.4×e^{-500\sqrt{0.4}}\\ & =& 1.77×{10}^{-137}\\ & & \end{array}$

Therefore, the transmission probability when $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.$is 0.4 and when it is 0.6 are $2.41×{10}^{-168},1.77×{10}^{-137}$respectively for $\frac{2L\sqrt{2m{U}_0}}{h}=500$.

On comparing the values of obtained transmission probabilities, it varies quite strongly with the energy, when it is small.

So, transmission probability is a far more sensitive function of Energy when the probability is small.

Thus, the lesser the probability of tunnelling is, the more sharply, it increases as E increases.