Question: The current of a beam of electrons, each with a speed of $\mathbf{1}\mathbf{.}\mathbf{200}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, is $\mathbf{9}\mathbf{.}\mathbf{000}\mathbf{}\mathbf{mA}$. At one point along its path, the beam

encounters a potential barrier of height $\mathbf{4}\mathbf{.}\mathbf{719}\mathbf{}\mathbf{\mu V}$and thickness $\mathbf{200}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{nm}$. What is the transmitted current?

The electron beam speed in region 1,$v=1200\mathrm{m}/\mathrm{s}$

The height of the potential step$V_b=-4.719\mathrm{\mu V}$

The electric current $l_0=9.000\mathrm{mA}$

The thickness of the energy barrier $L=200.0nm$

A potential energy barrier is a region where a traveling particle

will have increased potential energy ${\mathit{U}}_{\mathbf{b}}$.

The particle can pass through the barrier if its total energy$\mathit{E}\mathbf{>}{\mathit{U}}_{\mathbf{b}}$.

The electron energy in region 1 can be given as

$E=\frac{1}{2}m{v}^2$

Where $m$is mass and $v$is speed

$\begin{array}{rcl}E& =& \frac{1}{2}\times 9.1\times {10}^{-31}\mathrm{kg}\times {\left(1200\mathrm{m}/\mathrm{s}\right)}^2\\ & =& 6.56\times {10}^{-25}\mathrm{J}\\ & =& 4.0995\mathrm{\mu eV}\end{array}$

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

Substituting the values for electron particles,

$\begin{array}{rcl}b& =& \sqrt{\frac{8{\mathrm{\pi }}^2\times \left(9.1\times {10}^{-31}\mathrm{kg}\right)\times \left(4.719\mathrm{\mu eV}-4.0995\mathrm{\mu eV}\right)\times \left(1.6\times {10}^{-25}\mathrm{J}/\mathrm{\mu eV}\right)}{\left(6.626\times {10}^{-34}\mathrm{Js}\right)}}\\ & =& 4.0298\times {10}^6{\mathrm{m}}^{-1}\end{array}$

Therefore, the transmission coefficient is

$\begin{array}{rcl}T& =& exp\left(-2\times 4.0928\times {10}^6{\mathrm{m}}^{-1}\times 200\times {10}^{-9}\mathrm{m}\right)\\ & =& \exp \left(-1.612\right)\\ & =& 0.1995\end{array}$

Therefore, the transmission coefficient is $T=0.1995$

So, the transmitted current is

$\begin{array}{l}l=T{I}_0\\ =(0.1995)(9.000\mathrm{A})\\ =1.795\mathrm{mA}\end{array}$

Hence the transmitted current is $0.1795\mathrm{mA}$.