Question

Supposea barrier qualifies as wide, and width are such that $\frac{\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}\mathbf{m}{\mathbf{U}}_{\mathbf{0}}}}{\mathbf{h}}\mathbf{=}\mathbf{5}$ ,

(a) Calculate the transmission probabilities when$\raisebox{1ex}{${\displaystyle \mathbf{E}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\mathbf{U}}_{\mathbf{0}}}$}\right.$is $0.4$and when it is$\mathbf{0}\mathbf{.}\mathbf{6}$

(b) Repeat part (a), but for the case where $\frac{\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}\mathbf{m}{\mathbf{U}}_{\mathbf{0}}}}{\mathbf{h}}$is50 instead of 5.

(C) Repeat part (a) but for $\frac{\mathbf{2}\mathbf{L}\sqrt{\mathbf{2}\mathbf{m}{\mathbf{U}}_{\mathbf{0}}}}{\mathbf{h}}\mathbf{=}\mathbf{500}$.

(d) How do your results support the claim that the tunnelling probability is a far more sensitive function ofwhen tunnelling probability is small?

Short answer

1. The transmission probability when $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.$is 0.4 and when it is 0.6 are $0.08,0.16$respectivelyfor$\frac{2L\sqrt{2m{U}_0}}{h}=5$
2. The transmission probability when is 0.4 and when it is0.6 are localid="1660053345006" $5.8\times {10}^{-17},7.1\times {10}^{-14}$ respectivelyfor localid="1660053349313" $\frac{2L\sqrt{2m{U}_0}}{h}=50$.
3. The transmission probability when localid="1660053354806" $\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{${\displaystyle U_0}$}\right.$is 0.4 and when it is 0.6 are localid="1660053358612" $2.4\times {10}^{-168},1.8\times {10}^{-137}$respectivelyfor localid="1660053367586" $\frac{2L\sqrt{2m{U}_0}}{h}=500$.
4. The lesser the probability of tunnelling is, the more sharply, it increases as E increases.

Step-by-step solution

Definition of torque

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.

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

Here

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

Given quantities

• 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$E=\frac{1}{1000}{U}_0$

a) Finding the relation for transmission probability for the first case.

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

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

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

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

$\begin{array}{rcl}T& \cong & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\gamma \sqrt{\left(1-\frac{E_1}{U_0}\right)}}\\ & =& 16\times 0.4\times \left(1-0.4\right)\times e^{-5\sqrt{1-0.4}}\\ & =& 16\times 0.4\times 0.6\times e^{-5\sqrt{0.6}}\\ & =& 0.08\\ & & \end{array}$

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

$\begin{array}{rcl}T& \cong & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\gamma \sqrt{\left(1-E_1/U_0\right)}}\\ & =& 16\times 0.6\times (1-0.6)\times e^{-5\sqrt{1-0.6}}\\ & =& 16\times 0.6\times 0.4\times e^{-5\sqrt{0.4}}\\ & =& 0.16\\ & & \end{array}$

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

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

b) Finding the relation for transmission probability for the second case.

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

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

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

$\begin{array}{rcl}T& \cong & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\gamma \sqrt{\left(1-E_1/U_0\right)}}\\ & =& 16\times 0.4\times (1-0.4)\times e^{-50\sqrt{1-0.4}}\\ & =& 16\times 0.4\times 0.6\times e^{-50\sqrt{0.6}}\\ & =& 5.8\times {10}^{-17}\\ & & \end{array}$

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

$\begin{array}{rcl}T& \cong & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\gamma \sqrt{\left(1-E_1/U_0\right)}}\\ & =& 16\times 0.6\times \left(1-0.6\right)\times e^{-50\sqrt{1-0.6}}\\ & =& 16\times 0.6\times 0.4\times e^{-50\sqrt{0.4}}\\ & =& 7.1\times {10}^{-14}\\ & & \end{array}$

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 $5.8\times {10}^{-17},7.1\times {10}^{-14}$respectively for$\frac{2L\sqrt{2m{U}_0}}{h}=50$

c) Finding the relation for transmission probability for the third case.

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

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

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

$\begin{array}{rcl}T& \cong & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\gamma \sqrt{\left(1-\frac{E_1}{U_0}\right)}}\\ & =& 16\times 0.4\times \left(1-0.4\right)\times e^{-500\sqrt{1-0.4}}\\ & =& 16\times 0.4\times 0.6\times e^{-500\sqrt{0.6}}\\ & =& 2.41\times {10}^{-168}\\ & & \end{array}$

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

$\begin{array}{rcl}T& \cong & 16\frac{E_1}{U_0}\left(1-\frac{E_1}{U_0}\right)e^{-\gamma \sqrt{\left(1-\frac{E_1}{U_0}\right)}}\\ & =& 16\times 0.6\times \left(1-0.6\right)\times e^{-500\sqrt{1-0.6}}\\ & =& 16\times 0.6\times 0.4\times e^{-500\sqrt{0.4}}\\ & =& 1.77\times {10}^{-137}\\ & & \end{array}$

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 $2.41\times {10}^{-168},1.77\times {10}^{-137}$respectively for $\frac{2L\sqrt{2m{U}_0}}{h}=500$.

d) Drawing observation from the obtained information

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.