Question:Consider a potential energy barrier like that of Fig. 38-17but whose height ${\mathbf{U}}_{\mathbf{b}}$is and $\mathbf{6}\mathbf{}\mathbf{eV}$whose thickness Lis $\mathbf{0}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{nm}$. What is the energy of an incident electron whose transmissioncoefficient is $\mathbf{0}\mathbf{.}\mathbf{0010}$?

The thickness of the energy barrier $L=0.70\mathrm{nm}$

The height of the potential barrier $U_b=6\mathrm{eV}$

Transmission coefficient is $T=0.0010$

When a particle's potential energy changes at a boundary, it can reflect even though it would not normally reflect under classical theory.

The transmission coefficient of a particle can be given as.

$T=e^{-2bL}$

Where, $L$is the length of the potential barrier.

And $b$can be given as,

$b=\sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}$

Where the mass of the particle is $m$and $E$is incident energy and $U_b$is the height of the barrier

Therefore,

The transmission coefficient can be given as

$T=exp\left[-2L\sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}\right]$

Taking anti-log both sides of the equation

role="math" localid="1663154511508" $\begin{array}{rcl}lnT& =& \left[-2L\sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}\right]\\ \sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}& =& -\frac{lnT}{2L}\\ U_b-E& =& \frac{1}{2m}{\left(\frac{hlnT}{4\pi L}\right)}^2\\ E& =& U_b-\frac{1}{2m}{\left(\frac{hlnT}{4\pi L}\right)}^2\end{array}$

Substituting the values into the formula

$\begin{array}{rcl}E& =& \left(6\mathrm{eV}\times 1.6\times {10}^{-16}\mathrm{J}/\mathrm{eV}\right)-\frac{1}{2\times 9.1\times {10}^{-31}\mathrm{kg}}\left(\frac{6.626\times {10}^{-34}\mathrm{Js}\times \ln 0.0010}{4\times 3.14\times 0.70\times {10}^{-9}\mathrm{m}}\right)\\ & =& 8.16\times {10}^{-19}\mathrm{J}\\ & =& 5.1\mathrm{eV}\end{array}$

Hence the energy of the incident electron is $5.1\mathrm{eV}$.