An electron approaches a $1.0\mathrm{nm}$ wide potential-energy barrier of height $5.0eV$. What energy electron has a tunneling probability of (a) 10%, (b) 1.0%, and (c) 0.10%?

We have given,

length of potential barrier =$1\mathrm{nm}$

The potential =$5\mathrm{eV}$

probability =$10\%$

We have to find the energy of wave.

We know the probability of the tunneling can be

$$\begin{gathered} {\mathrm{P}}_{\mathrm{tunnel}}={\mathrm{e}}^{-2\mathrm{w}/\mathrm{\eta }} \\ \mathrm{\eta }=\frac{-2\mathrm{w}}{{\mathrm{lnP}}_{\mathrm{tunnel}}}...............\left(1\right) \end{gathered}$$

and we know the penetration distance is,

$\mathrm{\eta }=\frac{\mathrm{h}}{2\mathrm{\pi }\sqrt{2\mathrm{m}({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}})}$

then this equation will be equal to equation (1)

$$\begin{gathered} \frac{-2\mathrm{w}}{{\mathrm{lnP}}_{\mathrm{tunnel}}}=\frac{\mathrm{h}}{2\mathrm{\pi }\sqrt{2\mathrm{m}({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}})} \\ \frac{-2\times {10}^{-9}\mathrm{m}}{\ln (0.1)}=\frac{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}{2\mathrm{\pi }\sqrt{2\mathrm{m}(5\mathrm{eV}-\mathrm{E}})} \\ \sqrt{2\mathrm{m}(5\mathrm{eV}-\mathrm{E}})=1.2\times {10}^{-25} \\ (5\times 1.6\times {10}^{-19})-\mathrm{E}=0.0791\times {{10}^{-}}^{19} \\ \mathrm{E}=7.92\times {10}^{-19}\mathrm{J} \\ \mathrm{E}=4.95\mathrm{eV} \end{gathered}$$

We have to find the energy for the probability of tunneling is$1\%$.

We know that relation between the probability and the energy is

$$\begin{gathered} \frac{-2\mathrm{w}}{{\mathrm{lnP}}_{\mathrm{tunnel}}}=\frac{\mathrm{h}}{2\mathrm{\pi }\sqrt{2\mathrm{m}({\mathrm{U}}_{\mathrm{o}}-\mathrm{E}})} \\ \frac{-2\times {10}^{-9}\mathrm{m}}{\ln (0.01)}=\frac{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}{2\mathrm{\pi }\sqrt{2\mathrm{m}(5\mathrm{eV}-\mathrm{E}})} \\ \sqrt{2\mathrm{m}(5\mathrm{eV}-\mathrm{E}})=2.4\times {10}^{-25} \\ (5\times 1.6\times {10}^{-19})-\mathrm{E}=0.31\times {{10}^{-}}^{19} \\ \mathrm{E}=4.81\mathrm{eV} \end{gathered}$$

We have to find the energy of the system for probability is$0.1\%.$

We know the relationship between the probability and the energy is given by,

$$\begin{gathered} \frac{-2\times {10}^{-9}\mathrm{m}}{\ln (0.001)}=\frac{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}{2\mathrm{\pi }\sqrt{2\mathrm{m}(5\mathrm{eV}-\mathrm{E}})} \\ \sqrt{2\mathrm{m}(5\mathrm{eV}-\mathrm{E}})=3.6\times {10}^{-25} \\ (5\times 1.6\times {10}^{-19})-\mathrm{E}=0.71\times {{10}^{-}}^{19} \\ \mathrm{E}=4.56\mathrm{eV} \end{gathered}$$